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# Math Help - Calculus heavy physics problem

1. ## Calculus heavy physics problem

A 5.00 crate is suspended from the end of a short vertical rope of negligible mass. An upward force is applied to the end of the rope, and the height of the crate above its initial position is given by +(0.61 )

What is the magnitude of the force when 3.60 ?

Alright, I have done about 20 physics problems that are due next week, and this is the last one. It is very heavy on calculus, which i am not good at. In fact, I don't even really know where to start on this one. Thanks for the help.

2. ## Re: Calculus heavy physics problem

Originally Posted by CrusaderKing1
A 5.00 crate is suspended from the end of a short vertical rope of negligible mass. An upward force is applied to the end of the rope, and the height of the crate above its initial position is given by +(0.61 )

What is the magnitude of the force when 3.60 ?

Alright, I have done about 20 physics problems that are due next week, and this is the last one. It is very heavy on calculus, which i am not good at. In fact, I don't even really know where to start on this one. Thanks for the help.
$y(t) = 2.80t + 0.61t^3$

surely you know how to take derivatives of polynomial expressions ...

$a(t) = y''(t)$

$F(t) = m \cdot a(t)$

3. ## Re: Calculus heavy physics problem

Originally Posted by skeeter
$y(t) = 2.80t + 0.61t^3$

surely you know how to take derivatives of polynomial expressions ...

$a(t) = y''(t)$

$F(t) = m \cdot a(t)$
Yes, If the question said use the derivative, and then find F(t), I would have understood it. I am very bad at starting problems!

Ok, I did derive the function and received 20.6848 for acceleration, with then multiplied by 5.00 kg = 103.424

Would the units be in N? Probably N right? Yes, I'm not very good at this yet.

Thanks for your help! Much appreciated.

4. ## Re: Calculus heavy physics problem

Originally Posted by CrusaderKing1
Yes, If the question said use the derivative, and then find F(t), I would have understood it. I am very bad at starting problems!

Ok, I did derive the function and received 20.6848 for acceleration, with then multiplied by 5.00 kg = 103.424

Would the units be in N? Probably N right? Yes, I'm not very good at this yet.

Thanks for your help! Much appreciated.
incorrect

$a(6) = 21.96 \, m/s^2$ ... recheck your calculations

at this level of calculus-based physics, you are expected to know the relationship between position, velocity, and acceleration ...

$v(t) = y'(t)$

$a(t) = v'(t) = y''(t)$

5. ## Re: Calculus heavy physics problem

Originally Posted by skeeter
incorrect

$a(6) = 21.96 \, m/s^2$ ... recheck your calculations

at this level of calculus-based physics, you are expected to know the relationship between position, velocity, and acceleration ...

$v(t) = y'(t)$

$a(t) = v'(t) = y''(t)$
65.88 N

I didn't realize that I need to derive the function twice.

I have calculus and physics this semester, and had only high school trig. before 5 weeks ago.

The value is definitely right, but the program(masteringphysics) won't accept "N" as the correct units.

would kg*m/s2 work?

Thanks.

6. ## Re: Calculus heavy physics problem

Originally Posted by CrusaderKing1
65.88 N

I didn't realize that I need to derive the function twice.

I have calculus and physics this semester, and had only high school trig. before 5 weeks ago.

The value is definitely right, but the program(masteringphysics) won't accept "N" as the correct units.

would kg*m/s2 work?
Thanks.
Not being familiar with what the "machine" wants, I can't really say ... it's worth a shot.

7. ## Re: Calculus heavy physics problem

Originally Posted by skeeter
Not being familiar with what the "machine" wants, I can't really say ... it's worth a shot.
yes, I guess so.


text: 2020-10-01 13:56:23 +0200 commented answer Defining 4jm Wigner symbols Perfect, thanks for your help! 2020-10-01 13:55:54 +0200 answered a question Defining 4jm Wigner symbols Perfect, thanks! 2020-10-01 13:55:31 +0200 received badge ● Supporter (source) 2020-10-01 13:55:28 +0200 received badge ● Scholar (source) 2020-09-30 23:23:55 +0200 received badge ● Nice Question (source) 2020-09-30 23:01:38 +0200 received badge ● Student (source) 2020-09-30 22:15:41 +0200 asked a question Defining 4jm Wigner symbols Hi everyone, I am a neophyte and I can't understand my error in spite of the many tutorials I have watched. I want to define the 4jm Wigner symbols, which should basically be defined as: var('m') wigner_4j(j1,j2,j3,j4,j,m1,m2,m3,m4) = sum(wigner_3j(j1,j2,j,m1,m2,m)*wigner_3j(j,j3,j4,-m,m3,m4),m,-j,j)  I get plenty of errors that I can't fix. It is weird to me because the overall logic seemed good, as it works for instance when I define in a similar manner: var('n') b(p,q) = sum(binomial(p,n)*binomial(q,n),n,0,min(p,q))  Thanks for your help!

format: plain


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## University Physics with Modern Physics (14th Edition)

$v = (5000 ~rpm)(2\pi ~r)(\frac{1~min}{60~s})$ $v = \frac{500~\pi~r}{3}~m/s$ We can use the radial acceleration to find the required radius $r$. $a_{rad} = \frac{v^2}{r} = 3000~g$ $\frac{(\frac{500~\pi~r~}{3}~)~^2}{r} = 3000~g$ $r = \frac{(9)(3000)(9.80~m/s^2)}{(500~\pi ~s^{-1})^2}$ $r = 0.107~m$ Since the required diameter is 0.214 meters, while the bench space is claimed to be 0.127 meters, the claim seems unrealistic.


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# Operational Amplifier Problem

I'm trying to find $$\frac{V2}{V1}$$ as a function of x.

I found this circuit but noone gave good explanation. I think I understand the way this potentiometer works, but I just can't work out the equations to get result. Also I tried to simulate this circuit and here is result (for simulation I assumed x=1) :

• Why is R3 connected to the op-amp output? Study the original circuit carefully. – Andy aka Jun 5 '19 at 12:13
• Because Potentiometer R could be viewed as 2 resistors xR and (1-x)R . Since 1 of them is short circuited, there is just resistor xR left. Since I assumed x=1 then xR=R and in simulation that resistance is named R3 – Aleksandar Simonović Jun 5 '19 at 12:17
• Nice observation sir. I missed that. – Aleksandar Simonović Jun 5 '19 at 12:20
• No, The pot can't be viewed as xR and (1 - x)R since the wiper is connected to one end. It can only be considered as xR and 0R. The dot at the bottom of R3 should not be there. – Transistor Jun 5 '19 at 12:30
• I wrote that (1-x)R is short circuited so that is same sa 0R – Aleksandar Simonović Jun 5 '19 at 12:47

This is easier than it looks. I won't solve it all the way, but lets call the voltage at the top of the pot $$\V_a\$$ and the voltage at the bottom of the pot, $$\V_b\$$. Lets call the positive input $$\V_{1+}\$$ and the negative, $$\V_{1+}\$$. The input terminals of the op amp are $$\V_-\$$ and $$\V_+\$$.

Using the rule that no current enters the op amp, it is straightforward to show that $$\V_a=2V_- - V_{1+}\$$. Using the rule the $$\V_- = V_+\$$, it is straightforward to show that $$\V_b = 2V_- - V_{1-}\$$. Now, we know the voltage at either side of the pot, and the value of the pot, so $$\i_{pot} = \frac{(V_a-V_b)}{(1-X)R} = \frac{V_{1-} - V_{1+}}{(1-X)R}\$$

Application of Kirchoff's current law to the node at the top of the pot should give you $$\V_2\$$

• Kirchhoff's current law to node: $0 = \frac{(V_a-V_2)}{R} + \frac{V_{1-} - V_{1+}}{(1-X)R} + \frac{(V_{1+} - V_-)}{R}$ .. $0 = \frac{(V_a-V_2)}{R} - \frac{V_1}{(1-X)R} + \frac{(V_{1+} - V_-)}{R}$ – Aleksandar Simonović Jun 5 '19 at 18:31

Well, I think it is a tricky circuit - and it is not a simple task to derive the gain formula. Therefore, I recommend to split the calculation in separate steps:

(1) Set the lower input (we call it vi2) to zero vi2=0 and find the inverting gain as a function of the upper voltage (vi1): Vout1=f(vi1). This is relatively simple by applying the star-to-triangle conversion method for the negative feedback loop.

(2) Set the upper input to zero (vi1=0) and find the non-inv. gain as a function of the lower voltage vi2: Vout2=f(vi2). However, in this case, we have to face a negative as well as a positive feedback loop. Hence, the total feedback factor Hf again must be calculated in two sepate steps: Hf=Hf1+Hf2 (sum of both feedback factors, one of which must have a negative sign).

(3) Vout=Vout1+Vout2 with V1=Vi1-Vi2


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# Affine Varieties - Single Points and maximal ideals

#### Peter

##### Well-known member
MHB Site Helper
In Dummit and Foote Chapter 15, Section 15.3: Radicals and Affine Varieties on page 679 we find the following definition of affine variety: (see attachment)

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Definition. A nonempty affine algebraic set $$\displaystyle V$$ is called irreducible if it cannot be written as [TEX] V = V_1 \cup V_2 [/TEX] where [TEX] V_1 [/TEX] and [TEX] V_2 [/TEX] are proper algebraic sets in $$\displaystyle V$$.

An irreducible affine algebraic set is called an affine variety.

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Dummit and Foote then prove the following results:

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Proposition 17. The affine algebraic set $$\displaystyle V$$ is irreducible if and only if [TEX] \mathcal{I}(V) [/TEX] is a prime ideal.

Corollary 18. The affine algebraic set $$\displaystyle V$$ is a variety if and only if its coordinate ring $$\displaystyle k[V]$$ is an integral domain.

---------------------------------------------------------------------------------------------------------------------------------------

Then in Example 1 on page 681 (see attachment) D&F write:

"Single points in [TEX] \mathbb{A}^n [/TEX] are affine varieties since their corresponding ideals in [TEX] k[A^n] [/TEX] are maximal ideals."

I do not follow this reasoning.

Can someone please explain why the fact that ideals in [TEX] k[A^n] [/TEX] that correspond to single points are maximal

imply that single points in [TEX] A^n [/TEX] are affine varieties.

Presumably Proposition 17 and Corollary 18 are involved but I cannot see the link.

I would appreciate some help.

Peter

[This has also been posted on MHF]

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Problems

# Partition of the set

Consider a set consisting of the first n natural numbers: Nn = {1, 2, ..., n}. Partition - a representation of this set as the union of one or more non-empty sets. Examples of partitions for n = 5 are:

{1, 2, 3, 4, 5} = {1, 2, 3} U {4, 5}

{1, 2, 3, 4, 5} = {1, 3, 5} U {2, 4}

{1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}

{1, 2, 3, 4, 5} = {1} U {2} U {3} U {4} U {5}

In total there are 52 partitions of N5. Note that the partitions that differ only by order of the merged sets did not differ.

Partition of the set Nn can be ordered lexicographically.

In order to determine the order, first define the lexicographic order on subsets of Nn. We say that a setANnlexicographically less than the setBNn and write A < B, if one of the following statements:

• There existsi, such thatiA, iB, for all j < i: jA iff jB, and there is ak > i, such thatkB;
• AB and i < j for all iA and jB \ A.

It is obvious that this relation is a total order on subsets of Nn. Now we define the canonical representation of the partition as a representation in that combines multiple lexicographically.

Partitions are ordered lexicographically as follows. Partition ofNn = A1 U A2 U ... U Aklexicographically less than the partition ofNn = B1 U B2 U ... U Bl, if there is an i, asA1 = B1, A2 = B2, ..., Ai-1 = Bi-1 and Ai < Bi.

According to the partition of Nn find the following in the lexicographic order decomposition.

Input

The input file contains several descriptions of the tests. Each description is the canonical representation of the partition. The first line contains the description of n and k - number of elements in the partitions the set and the number of parts in a partition (1n200). Subsequent k rows contain elements of the partition. Elements of each set in ascending order.

Describe the tests are separated by blank lines. The last line of input contains two zero. This test should not be handled.

Sum of n over all the descriptions do not exceed 2000.

Output

For each test, the following output in lexicographic order decomposition. If the partition in the input file is the latest in leksigoraficheskom order, display the first in lexicographic order. Use the same format as the input file. Separate partitions from each other by blank lines.

Time limit 1 second
Memory limit 64 MiB
Input example #1
5 2
1 2 3
4 5

5 2
1 3 5
2 4

5 1
1 2 3 4 5

5 5
1
2
3
4
5

0 0

Output example #1
5 2
1 2 3 4
5

5 4
1 4
2
3
5

5 2
1 2 3 5
4

5 4
1
2
3
4 5


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Use this URL to cite or link to this record in EThOS: https://ethos.bl.uk/OrderDetails.do?uin=uk.bl.ethos.544524
Title: Radiation dosimetry of conventional and laser-driven particle beams
Author: Kirby, Daniel James
ISNI:       0000 0004 2713 8982
Awarding Body: University of Birmingham
Current Institution: University of Birmingham
Date of Award: 2011
Availability of Full Text:
 Access from EThOS: Access from Institution:
Abstract:
The measurement of radiation dose in radiotherapy is vital in ensuring the accuracy of treatments. As more advanced techniques using protons and ions emerge, they pose challenges to ensure the same level of accuracy of dosimetry is achieved as for conventional X-ray radiotherapy. A relatively new method of particle acceleration using ultra-high intensity lasers and thin metallic targets has sparked a large effort to investigate the possible application of this technology in radiotherapy, which in turn requires accurate methods of dosimetry to be carried out and is the main motivation for this work. Accurate dosimetry was initially performed here using an air ionisation chamber, various models of GafChromic film and a PMMA phantom in 15 and 29 MeV protons and 38 MeV $$\alpha$$-particles from the Birmingham cyclotron. In developing an accurate protocol for absorbed dose-to-water at these relatively low proton energies, new data was generated on the proton energy response of GafChromic films. This enabled accurate dosimetry of a prototype laser-particle source, and provided improvements to a method of spectroscopic measurement in the resultant mixed field of multi-energy protons, electrons and X-rays. Monte Carlo simulations using MCNPX but mainly FLUKA were performed throughout to support and verify experimental measurements.
Supervisor: Not available Sponsor: Not available
Qualification Name: Thesis (Ph.D.) Qualification Level: Doctoral
EThOS ID: uk.bl.ethos.544524  DOI: Not available
Keywords: QC Physics ; R Medicine (General)
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# A Double Convex Lens Has Focal Length 25 Cm. the Radius of Curvature of One of the Surfaces is Double of - Science

Sum

A double convex lens has focal length 25 cm. The radius of curvature of one of the surfaces is double of the other. Find the radii, if the refractive index of the material of the lens is 1.5.

#### Solution

Given,
Double convex lens of focal length, f = 25 cm
Refractive index of the material, μ = 1.5
As per the question, the radius of curvature of one surface is twice that of the other.
i.e. R1= R and R2= −2R
(according to sign conventions)

Using the lens maker formula
$\frac{1}{f} = (\mu - 1)\left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$' we have:
$\frac{1}{25} = \left( 1 . 5 - 1 \right)\left[ \frac{1}{R} - \left( \frac{1}{- 2R} \right) \right]$

$\Rightarrow \frac{1}{25} = 0 . 5\left[ \frac{3R}{2} \right]$

$\Rightarrow \frac{1}{25} = \left[ \frac{3R}{4} \right]$

$\Rightarrow R = 18 . 75 \text{ cm }$

R1 = 18.75 cm and R2 = 37.5 cm

Hence, the required radii of the double convex lens are 18.75 cm and 37.5 cm.

Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 18 Geometrical Optics
Q 48 | Page 415


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# The probability of a 10-year flood to occur at least once in the next 5 years is

1. 35%
2. 41%
3. 60%
4. 65%

Option 2 : 41%

## Detailed Solution

Concept:

The probability for the annual maximum flood magnitude to occur = $$\frac{1}{{Return\;period\;}}$$

Calculation:

$$\begin{array}{l} p = \frac{1}{T} = 0.1\\ q = 1 - p = 0.9 \end{array}$$

Probability of flood at least once in next 5 years

= 1 – no flood in next 5 years

= 1 – q5

= 1 – (0.9)5 = 0.409

# To estimate the magnitude of a flood peak (Qp) by the formula $${Q_P} = \frac{{124A}}{{\sqrt {A + 10.4} }}$$, where A = catchment area in km2 . This formula is termed as:-

1. Dicken's formula
2. Inglis formula
3. Ryves formula
4. Rational formula

Option 2 : Inglis formula

## Detailed Solution

Explanation

Inglis Method

This formula is used only in Maharashtra. Here three different cases are taken into consideration.

$${Q_P} = \frac{{124A}}{{\sqrt {A + 10.4} }} \cong 123\sqrt A$$

where

QP = Peak discharge in m3/s.

A = Area in km2.

Important Points

Ryve’s Method

This formula is used only in Southern India.

$${Q_P} = {C_R}{A^{\frac{2}{3}}}$$

Rational Method

This method is based on the principle of the relationship between rainfall and runoff. Hence, it can be considered to be similar to empirical method.

$${Q_P} = \frac{1}{{36}}k{P_c}{A^1}$$

Dicken’s Method

It was earlier adopted only in Northern India but now it can be used in most of the States in India after proper modification of the constant.

$${Q_P} = {C_D}{A^{\frac{3}{4}}}$$

# The Rainfall Intensity of Light Rain is:

1. Upto 2.5 mm/Hr
2. Upto 3.0 mm/Hr
3. Upto 5.00 mm/Hr
4. Upto 7.5 mm/Hr

Option 1 : Upto 2.5 mm/Hr

## Detailed Solution

Concept:

Precipitation- It is the fall of water in various forms on the earth from the clouds.

The usual forms of precipitation are as follows:

Forms of precipitation

Characteristics

1. Rain

Water droplets of size 0.5mm to 6mm

Based on its intensity it is classified as:

 Intensity (mm/hr) Type 0 – 2.5 Light 2.5 – 7.5 Medium >7.5 Heavy
1. Snow

Ice crystals of density 0.1 g/cc

1. Drizzle

Water droplets of size < 0.5 mm and intensity < 1 mm/hr

1. Glaze

Water droplets in contact with cold water surface (0℃)

1. Sleet

Frozen raindrops

1. Hail

Lumps of Ice > 8 mm

Note:

Rainfall intensity is measured by Rain Gauge, Pluviometer, Ombrometer or Hyetometer.

# The instrument used for measuring evaporation is:

1. hygrometer
2. atmometer
3. lysimeter
4. luxmeter

Option 2 : atmometer

## Detailed Solution

Explanation:.

• Atmometer is a device that is used to measure the rate of water evaporation.  It is also known as the evaporimeter.
• Rotameter is used to measure discharge but the current meter is used to measure velocity in open channels.

Important Points

Instruments used in the measurement

 Instrument Use Psychrometer Relative humidity Hygrometer Humidity Tensiometer Capillary potential Anemometer Wind velocity Ombrometer/pluviometer Rainfall depth Lysimeter Evapotranspiration Atmometer Evaporation Phytometer Transpiration Permeameter Hydraulic conductivity Rainfall simulator Infiltration capacity

Confusion point:

The pitot tube is used to measure the velocity of fluid whereas an anemometer is used to measure gas and air velocity

# During a storm, the rates of rainfall observed at a frequency of 15 min for one hour are 12.5, 17.5, 22.5 abd 7.5 cm/h. If the phi-index is 7.5 cm/h, then the total run-off will be

1. 22.50
2. 15.00
3. 30.00
4. 7.50

Option 4 : 7.50

## Detailed Solution

Concept:

The ϕ – index is a rate of infiltration in which, the rate of infiltration exceeds the value at which volume of runoff become equals to the volume of rainfall.

$${ϕ_{\left( {index} \right)}} = \frac{{{Total\; Infiltration}}}{{Total\;time\;of\;the\;storm}}$$

$${ϕ_{\left( {index} \right)}} = \frac{{{P_{total}} - Q}}{{Total\;time}}$$

where

PTotal = Total precipitation

Q = Runoff

Calculation:

ϕ - index = 75 cm/hr

ϕ - index is the average rate of rainfall such that the volume of rainfall in excess of that rate is equal to the volume of surface runoff

So runoff (R) in cm is,

$$=(12.5-7.5) \dfrac{15}{60}+ (17.5 - 7.5) \dfrac{15}{60}+ (22.5 - 7.5) \dfrac{15}{60}$$

$$=\dfrac{5\times 15}{60}+\dfrac{10 \times 15}{60}+ \dfrac{15 \times 15}{60}$$

$$R=\dfrac{15}{60}(5+10+15)$$

$$=\dfrac{15}{60}\times 30$$

∴ R = 7.5 cm

# A canal is 80 km long and has an average surface width of 15 m. If the evaporation measured in a class A pan is 0.5 cm/day, the volume of water evaporated in a month of 30 days is (in m3)

1. 12600
2. 18000
3. 180000
4. 126000

Option 4 : 126000

## Detailed Solution

Concept:

The volume of water evaporated (V) = E × L × B

Where, E = Evaporation measured per day, L = Stretch of evaporation  & B = Average surface width

Calculation:

Given, L = 80 km = 8 × 104

B = 15 m

Evaporation = 0.5 cm / day

Pan co-efficient = 0.7

Total volume of evaporation (V) = 8 × 104 × 15 × 0.5 x 10-2 × 0.7 × 30 m3

∴ V = 126000 m3

# The rainfall figures for successive 30 – minute intervals are 35, 40, 120, 85, 45, 45 and 30 mm/hr. If the ϕ index is 35 mm/hour, determine W-index.

1. 24.3 mm/hour
2. 44.3 mm/hour
3. 34.3 mm/hour
4. 54.3 mm/hour

Option 3 : 34.3 mm/hour

## Detailed Solution

Concept:

ϕindex = $$\frac{P_e-R}{t_e}$$ and Windex = $$\frac{P-R}{t_r}$$

Where,

Pe = effective rainfall causing runoff, R = runoff,

te = duration of effective rainfall, P = total rainfall and tr = total duration of rainfall

Calculation:

Given,

ϕindex = 35 mm/hr

∵  We know that,  runoff occurs when the intensity of rainfall (i) >  ϕindex

∴ From rainfall intensity data, the intensity of 30 mm/hr < 35 mm/hr, hence ignored in the calculation of ϕindex

⇒ Pe = (35 + 40 + 120 + 85 + 45 + 45) × 0.5

⇒ Pe = 185 mm and te = 3 hr

∵ We know that, ϕindex = $$\frac{P_e-R}{t_e}$$

⇒ 35 = $$\frac{185-R}{3}$$

R = 80 mm

Now, P = (35 + 40 + 120 + 85 + 45 + 45 + 30) × (30/60)

⇒ P = 200 mm and tr = 3.5 hr

∵ We know that, Windex = $$\frac{P-R}{t_r}$$

⇒ Windex = $$\frac{200-80}{3.5}$$

⇒ Windex = 34.28 mm/hr

# Hydrograph is the graphical representation of

1. runoff and time
2. surface runoff and time
3. ground water flow and time
4. rainfall and time

Option 1 : runoff and time

## Detailed Solution

Hydrograph: It is a continuous plot of instantaneous discharge (runoff) v/s time.

It results from a combination of physiographic and meteorological conditions in a watershed and represents the integrated effects of climate, hydrologic losses, surface runoff, interflow, and groundwater flow.

Detailed analysis of hydrographs is usually important in flood damage mitigation, flood forecasting, or establishing design flows for structures that convey floodwaters.

Factors that influence the hydrograph shape and volume

i. Meteorological factors

ii. Physiographic or watershed factors and

iii. Human factors

# The main factor which affects infiltration capacity is

1. Thickness of saturated layer
2. Depth of surface retention
3. Soil moisture
4. All of the above

Option 4 : All of the above

## Detailed Solution

Infiltration:

Infiltration is the flow of water through the soil surface into a porous medium under gravity action and pressure effects.

The infiltration capacity is the maximum rate at which water can be absorbed by a given soil per unit area under given conditions.

The main factors that influence the infiltration are:

1. The soil type i.e. texture, structure, hydrodynamic characteristics. The soil characteristics influence capillary forces and adsorption.

2. The soil coverage and vegetation: They increase the time of water penetration in soil.

3. The topography and morphology of slopes: Steep slope has low infiltration capacity as water flow rapidly due to gravity.

4. Rain intensity: High rainfall implies more rate of infiltration.

5. Soil humidity or moisture:  Dry soil has more infiltration as compared to wet soil.

6. Soil compaction: High compaction reduces infiltration capacity.

7. Soil thickness: Higher the thickness soil can hold more water.

8. Surface retention or runoff: More surface run-off means less water is available to infiltrate into soil.

# A direct runoff hydrograph due to a storm idealized into a triangular shape has a peak flow rate of 60 m3/s occurring at 25 hours from its start. If the base width of this hydrograph is 72 hours, and the catchment area is 777.6 km2, the runoff from the storm is

1. 1 cm
2. 2 cm
3. 5 cm
4. 10 cm

Option 1 : 1 cm

## Detailed Solution

Concept:

Hydrograph is used to measure runoff discharge.The area under the discharge and time curve gives the volume of runoff.

Calculation:

Catchment area = 777.6 km2 = 777.6 × 106 m2

Total volume of water = Area of Hydrograph

Total volume $$= \frac{1}{2} \times 72 \times 60 \times 3600$$

= 7776000 m3

Runoff storm $$= \frac{{Volume}}{{Area}} = \frac{{7776000\;{m^2}}}{{777.6 \times {{10}^6}{m^2}}}$$

Runoff storm = 10-2 m

∴ Runoff storm = 1 cm

# Which of the following factors doesn't affect the water infiltration of a formation?

1. Thickness of a saturated layer
2. Wind velocity
3. Temperature
4. Vegetative cover

Option 2 : Wind velocity

## Detailed Solution

Concept:

Infiltration: It is the flow of water into the ground through the soil surfaces.

Infiltration Capacity: The maximum rate at which a given soil at a given time can absorb water is defined as infiltration capacity. It is denoted as fp and its unit is cm/hr.

Hence, if the intensity of rainfall is more than the infiltration capacity then, infiltration rate will be equal to infiltration capacity (fp) and rest of the amount of rainfall will be in the form of runoff.

Important points related to fp:

The intensity of rainfall is “i”. Then, three conditions arise from here:

i) If the actual rate of infiltration is f such that f = fp and i > fp then certain amount of runoff will be observed.

ii) If f < fp then infiltration rate will be will be f and no runoff will be observed.

Factor affecting infiltration rate

(1) Soil moisture

(2) Type of soil medium

(3) Permeability

(4) Vegetal cover

(5) Surface fines or washing of fines

(6) Compaction of soil

(7) Depth of surface detention

(8) Temperature

Wind velocity is not affected in infiltration but in evaporation wind velocity lead a main role

initially evaporation rate is increases and some time later evaporation rate is constant.

# Rainfall of intensity 20 mm/h occurred over a watershed of area 1 km2 for duration of six hours. It measures a direct runoff for a volume of 30,000 m3 in the stream. Find the precipitation NOT available for runoff in this case?

1. 5 mm
2. 9 cm
3. 17.5 mm
4. 3 cm

Option 2 : 9 cm

## Detailed Solution

Concept:

Direct Runoff: It is the part of runoff that enters the stream immediately after the precipitation. It includes surface runoff, prompt interflow, and precipitation on the channel surface. It is sometimes termed as direct storm runoff or storm runoff.

The volume of runoff = Area of catchment × Depth of rainfall

Precipitation: It is any liquid or frozen water that forms in the atmosphere and falls back to the earth. Precipitation is any product of the condensation of atmospheric water vapour that falls under gravity from clouds. The main forms of precipitation include drizzle, rain, sleet, snow, ice pellets, graupel and hail.

Infiltration Capacity: It is the rate at which water infiltrates into ground is called infiltration capacity. For consistency, in hydrological calculations, a constant value of infiltration rate for the entire storm duration is adopted.

Total infiltrated volume = Total rainfall – Runoff

Calculation:

Given that:

Rainfall intensity = 20 mm/hr = 0.02 m/hr

Watershed area = 1 km2

Total duration = 6 hour

Direct runoff = 30000 m3

Total rainfall = Rainfall intensity × Total duration area

Total rainfall = (0.02 × 6) × 1 × 106 = 1,20,000 m3

Precipitation not available for runoff means precipitation that is infiltrated.

Total infiltrated volume = Total rainfall – Runoff

Total infiltrated volume = 1,20,000 – 30,000 = 90,000 m3

For 1 km2 watershed area,

$$The\;infiltrated\;depth = \frac{{90,000}}{{1 \times {{10}^6}}} = 0.09\;m$$

∴ Infiltrated depth is 9 cm.

# Which of the following statements is/are correct?1. Isochrones are curves of equal pore water pressure2. Isochrones depict the variation of the pore water pressure along with the depth of the soil sample3. Isochrones vary with time

1. 1 only
2. 2 and 3 only
3. 1 and 2 only
4. 1, 2 and 3

Option 2 : 2 and 3 only

## Detailed Solution

Explanation:

Isochrones:

Isochrones is a line on the map which connects points having an equal time of travel of the surface runoff to the catchment outlet. These are some property of Isochrones:

• Isochrones vary with time, It's most commonly used to depict travel times, such as drawing a 30 minute travel time perimeter around a start location. The isochrone below joins up all points within a 45-minute drive from the origin.
• Isochrones depict the variation of the pore water pressure along with the depth of the soil sample.
• Isochrones are mainly used for transport planning, property search, sales territory planning, etc.

Important Points

 Isobar Equal Pressure line Isohyets Equal Rainfall line Isopleths Equal Evapotranspiration line Isotherm Equal Temperature line Isonif Equal Snowfall line

# In the given hydrograph the peak or crest segment is indicated by:-

1. BP
2. BC
3. AB
4. PC

Option 2 : BC

## Detailed Solution

Explanation:

Elements of flood hydrograph:

Flood hydrograph has three characteristic regions-

1). Rising limb AB: Curve Joining point A, the starting point of the rising curve and point B, the point of inflection. It is concave in shape and the shape depends both upon the rainfall and the catchment parameters.

2). Peak or Crest segment BC: Between the two points of inflection B and C with a peak P. For large catchments it generally occurs sometimes after the rainfall has ended.

3). Falling limb/ Recession Limb: Curve CD starting from the second point of inflection C and the shape depends only on catchment parameters as the rainfall has ceased by then.

# In the cloud names such as Nimbostratus and Cumulonimbus, what is the meaning of the prefix ‘nimbus’?

1. Light
2. Heat
3. Rain Bearing
4. None of the above

Option 3 : Rain Bearing

## Detailed Solution

The correct answer is Rain Bearing.

• Cloud types receive names based on their position in the atmosphere, their overall shape, and the weather in which they appear.
• Nimbus, for example, means "rain-bearing" in Latin and is added to cloud names as a prefix or suffix when they produce precipitation of any kind.
• Nimbostratus clouds, for example, are typically thick, low clouds that form a dense black and yield steady snow or rain.

Nimbostratus Clouds:

• Nimbostratus clouds are low-level, rain clouds that form a uniform layer.
• They are dark in color and produce steady, prolonged precipitation.
• They represent the strengthening and thickening of a stratus cloud layer.
• In contrast with the intense, short-lived rains associated with unstable cumulonimbus clouds, nimbostratus clouds typically produce light or moderate rainfall of a longer duration.

Cumulonimbus Clouds:

• They are formed by the upward movement of warm, moist air.
• In their mature stage, they also produce strong downdrafts of cold air.
• Cumulonimbus clouds are considered low-level clouds, even though their vertical development may extend high into the atmosphere.
• These clouds produce intermittent, intense precipitation and are associated with severe weather.

# Which of the following is a non-recording rain gauge?

1. Weighing type rain gauge
2. Tipping Bucket rain gauge
3. Simon’s rain gauge
4. Float type rain gauges

Option 3 : Simon’s rain gauge

## Detailed Solution

Non – recording type rain gauges:

1) Symon’s gauge: This is the non-recording type rain gauge commonly used in India by meteorological department.

Recording type rain gauges:

1) Natural syphon or float type: This type of rain gauge is adopted as a standard recording type rain gauge in India.

2) Weighing bucket type

3) Tipping bucket type

# A storm with 150 mm precipitation produces a direct runoff of 8.7 cm, with incremental hourly rainfall values being 0.6, 1.35, 2.25, 3.45, 2.7, 2.4 1.5 and 0.75 cm/hr. Estimate the ϕ index of the storm.

1. 0.925 cm/hr

2. 0.825 cm/hr
3. 0.625 cm/hr
4. 0.725 cm/hr

Option 2 : 0.825 cm/hr

## Detailed Solution

Concept:

ϕ index of a catchment is defined as the constant infiltration capacity that would yield the actual total runoff for a given rainfall amount.

Mathematically, ϕindex =  $$\frac{P_e -R}{t_e}$$

Where,

Pe = effective rainfall causing runoff, R = runoff, and te = effective rainfall period

As we know that, for runoff to occur, Rainfall intensity (i) > ϕindex

Calculation:

Given,

P = 15 cm, R = 8.7 cm, duration of rainfall = 8 hrs

Trial 1: Considering all rainfall values in a period of 8 hrs causes runoff.

∵ We know that, ϕindex = $$\frac{P_e -R}{t_e}$$

⇒ ϕindex = $$\frac{15 - 8.7}{8}$$

∴ ϕindex = 0.7875 cm/hr

$$\because$$ We know that, for runoff to occur, i > ϕindex,

$$\Rightarrow$$ rainfall intensities 0.6 and 0.75 cm/hr are less than ϕindex = 0.7875 cm/hr

$$\therefore$$ for trial 2, rainfall intensities 0.6 and 0.75 cm/hr are ignored.

Trial 2:

Total rainfall, P = 1.35 + 2.25 + 3.45 + 2.7 + 2.4 + 1.5 = 13.65 cm, R = 8.7 cm

Duration of rainfall, t = 6 hrs

$$\Rightarrow$$ ϕindex = $$\frac{13.65 - 8.7}{6}$$

$$\therefore$$ ϕindex = 0.825 cm/hr

# Evaporation is measured using

1. Rain gauge
2. An open pan
3. Iysimeter
4. Infiltrometer

Option 2 : An open pan

## Detailed Solution

Concept:

Evaporation is a cooling process in which latent heat of vaporization must be provided by the water body. Vaporization rate increases due to faster winds but if the wind velocity is sufficient to remove all the vapor above the water surface, then any further increase in wind velocity will not affect the evaporation rate. Cetyl Alcohol is used to reduce the evaporation rate.

Evaporation is measured using an open pan. The most commonly used evaporation pan is Class A evaporation pan.

Class A evaporation pan is cylindrical in shape having 120.7 cm diameter and has a depth of 25 cm.

This pan rest on a carefully leveled, wooden base and is mostly enclosed by a metal fence to prevent animals drinking from it.

Important Points

Instrument for measurement of evapotranspiration → Iysimeter

Instrument for measurement of capillary potential → Tensiometer

Instrument for measurement of humidity → Hygrometer

Instrument for measurement of radiation from sun → Pyranometer

Instrument for measurement of wind velocity → Anemometer

# A 6-hour rainfall of 6 cm at a place A was found to have a return period of 40 years. The probability that a 6 hour rainfall of this or larger magnitude will occur at least once in 20 successive years is:

1. 0.308
2. 0.015
3. 0.605
4. 0.397

Option 4 : 0.397

## Detailed Solution

Concept:

• Return Period (T): It is the average time interval between the occurrence of rainfall of magnitude equal to or in excess of a specific magnitude.
• Probability (P): The probability of rainfall whose magnitude is equal to or in excess of specific magnitude having a return period of T is given as, $$\text{P} = \frac{1}{T}$$
• q = probability of rainfall not occurring in a given year = 1 - P
• Probability of rainfall not occurring at all in 'n' successive years = qn
• Probability of rainfall occurring at least once in 'n' successive years = 1 - qn

Calculation:

Given,

n = 20 years, T = 40 years

∵ We know that, $$\text{P} = \frac{1}{T}$$

⇒ $$\text{P} = \frac{1}{40}$$ = 0.025

∴ q = 1 - P = 0.975

∵ We know that the Probability of rainfall occurring at least once in 'n' successive years = 1 - qn

⇒ Probability of rainfall occurring at least once in '20' successive years = 1 - 0.97520 = 0.397

1. 0.38
2. 0.42
3. 0.48
4. 0.52


text:
# How do I calculate d20 success probability using the Halfling 'lucky' trait with advantage?

Here is a comprehensive DPR calculator, and here is the mathematics behind it. I'm trying to follow along with the equations.

At the bottom of the second page are formulas for success probability $$\L\$$ of a Halfling (who has luck) in normal circumstances and with advantage and disadvantage. With advantage it is $$L_{adv} = P_{adv} + \left(\frac{2}{20}(1-P) - \frac{1}{400}\right)P,$$ where:

• $$\P\$$ is the probability of succeeding on any single roll and
• $$\P_{adv} = 1 - (1-P)^2\$$ is the probability of succeeding with advantage (ie not failing both rolls).

In my attempt for deriving this (below), I have a sign error. Please can someone explain where I've gone wrong/show a correct derivation?

To succeed you require:

• succeeding outright while advantaged, OR
• rolling a $$\1\$$ with die $$\a\$$ AND failing with die $$\b\$$, AND THEN succeeding the reroll, OR
• rolling a $$\1\$$ with die $$\b\$$ AND failing with die $$\a\$$, AND THEN succeeding the reroll, OR
• rolling two $$\1\$$s AND THEN succeeding a reroll: $$L_{adv} = P_{adv} + \frac{1}{20}*(1-P)*P + \frac{1}{20}*(1-P)*P + \frac{1}{400}*P\\= P_{adv} + \left(\frac{2}{20}(1-P) + \frac{1}{400}\right)P$$
• @Anagkai Please consider putting that into a full answer below – Someone_Evil Jun 1 '20 at 12:25


text:
# Can I not use the align & operator within a \frac?

This code is causing LaTeX to throw an error: Missing } inserted. It appears to be because I put a & inside a \frac command (on line 4). I feel as though it should be possible to accomplish my goal.

\begin{align*}
A & + B \\
\frac{A}{C} & + \frac{B}{C} \\
\frac{A & + B}{C}
\end{align*}

Has anyone been frustrated by this before? How did you get around it?

-
You can't put alignment points within the \frac. –  Peter Grill Jan 16 '12 at 19:22
align work like an array environment. So the separation with & is used to separate cells. In this case it isn't possible to because the cell isn't a closed object. –  Marco Daniel Jan 16 '12 at 19:26

The & in the align environment are used to specify alignment points. These can not be within a \frac. So, depending on if you want to align on the right hand side of the expression, or on the + sign, you would use the code below to produce:

The second situation is a bit trickier. The way I solved it was to use \makebox to reserve a space equal to the largest text that goes in there and use the [c] option to center the text. However, if you want all three + to be aligned, then note that the spacing on either side of the = is not identical.

In this particular situation, you could choose to not align the + sign of the first expression, and then the spacing for all three = would be correct (as in the third image above). So it is a trade off between aligning the three +, or getting the correct spacing on either side of the equal.

\documentclass{article}
\usepackage{amsmath}
\usepackage{calc}

\begin{document}
If you want the right side aligned:
\begin{align*}
A &+ B \\
\frac{A}{C} &+ \frac{B}{C} \\
\frac{A + B}{C} &+
\end{align*}
%
\newcommand*{\LargeFrac}{\frac{A}{C} + \frac{B}{C}}%
\newcommand*{\Center}[1]{\makebox[\widthof{$\LargeFrac$}][c]{$\displaystyle #1$}}%
%
\noindent
If you want the plus sign aligned:
\begin{align*}
\Center{A + B} & = x\\
\Center{\frac{A}{C} + \frac{B}{C}} & = x \\
\Center{\frac{A + B}{C}} & = x
\end{align*}
%
If you want correct spacing around the equal:
\begin{align*}
A + B & = x\\
\Center{\frac{A}{C} + \frac{B}{C}} & = x \\
\Center{\frac{A + B}{C}} & = x
\end{align*}
\end{document}

-

Maybe you can use a derivation of array:

\documentclass{article}
\usepackage{amsmath}
\usepackage{array}
\begin{document}
\begin{align*}
\begin{array}{@{}r@{\mathrel{+}}l@{}}
A & B \\
\frac{A}{C} & \frac{B}{C} \\
\scriptstyle A  & \scriptstyle B \\[-2.5pt]\cline{1-2}
\multicolumn{2}{c}{\scriptstyle C}
\end{array}
\end{align*}
\end{document}
`

-


text:
## 2.3 Bound functions

int min(int x, int y)
Return the minimum of x and y. $\text{min}(x, y) = \begin{cases} x & \text{if } x < y\\ y & \text{otherwise} \end{cases}$

int max(int x, int y)
Return the maximum of x and y. $\text{max}(x, y) = \begin{cases} x & \text{if } x > y\\ y & \text{otherwise} \end{cases}$


text:
# Electronic Thesis/Dissertation

## Insight into Nucleation Mechanisms of Tetrahedral Materials from Advanced Molecular Dynamics Simulations Open Access

This dissertation studies the nucleation mechanisms of ice, clathrate hydrate and silicon clathrate which all belong to tetrahedral materials and carry significant importance to modern society. Because of the stochastic nature and the ultra-fine scale of nucleation, the mechanisms through which these important tetrahedral materials form from liquid remain poorly understood. Our goal is to address the current knowledge gap between experiment and theory on the nucleation mechanisms by conducting molecular dynamics (MD) studies. To overcome the rare event nature of nucleation, an advanced sampling method Forward Flux Sampling (FFS) is integrated with classical MD simulations. This integration allows obtaining not only nucleation rate explicitly but also an ensemble of nucleation trajectories with their correct statistical weights. By analyzing the ensemble of trajectories obtained from FFS, we reveal the important details of nucleation at the molecular level, particular at the early stage of nucleation. By combining Backward Flux Sampling (BFS) with FFS, we can also compute the free energy profile of nucleation explicitly, which allows a comparison against the classical nucleation theory (CNT). We began our investigation by studying heterogeneous ice nucleation, which is the most relevant form of ice formation. In this part of study, we aim to understand the key microscopic factors that control ice formation, including surface hydrophilicity, surface crystallinity, and surface geometry. Our simulations reveal that heterogeneous ice nucleation on graphitic surfaces is controlled by the coupling of surface crystallinity and surface hydrophilicity. In particular, our analysis shows that the crystalline graphitic lattice with an appropriate hydrophilicity may indeed template ice basal plane by forming a strained ice layer, thus significantly enhance its ice nucleation efficiency. The templating effect is further found to transit from within the first contact layer of water to the second as the surface hydrophilicity increases, which yields an oscillating distinction between the crystalline and amorphous graphitic surfaces in their ice nucleation efficiency. With this understanding, we then shifted our focus on the role of surface geometry, where we find that an atomically sharp, concave wedge can further promote ice nucleation. Remarkably, our molecular analysis shows a significant enhancement of ice nucleation can emerge both when the geometry of a wedge matches the ice lattice and when such lattice match does not exist. In particular, a 45\si{\degree} wedge is found to greatly enhance ice nucleation by facilitating the formation of special topological defects that consequently catalyse the growth of regular ice. We then investigated the nucleation of gas hydrate, a binary compound composed of water and natural gas, e.g., methane. To facilitate the application of FFS in studying gas hydrate nucleation, we proposed a new order parameter: Half-Cage Order Parameter (H-COP), based on the topological signature of hydrate structure, and conducted a $p_{B}$ histogram analysis to verify the effectiveness of the H-COP order parameter. Our analyses show that the nucleation of M-hydrate (M can be methane or carbon dioxide) starts in the vicinity of water-M interface, and gradually transit from amorphous to crystalline structures. This is the direct support to the proposed two-step nucleation mechanism of methane hydrate. However, within the ensemble of nucleation trajectories, we also identified some direct crystallization pathways without going through a amorphous phase in the nucleation of both M-hydrate and L-hydrate. Remarkably the free energy profile of L-hydrate nucleation, which is obtained independent of nucleation theory, is found to fit well against the CNT, albeit with noticeable differences when the nucleus is small. Therefore, the CNT-like free energy profile and the existence of multiple nucleation pathways indicate the near degeneracy of these pathways in their free energy profiles and highlight the complexity of hydrate nucleation. On the basis of these results, we further proposed a new perspective of gas hydrate nucleation, which can be considered to be an entropically driven, kinetic process that proceeds via multiple pathways that have similar free energy profiles. Finally, inspired by our insight gained in studying gas hydrate nucleation, we proposed a novel synthesis route to obtain inert gas silicon clathrate, which has an attractive opto-electronic property for energy application, but remains as an experimental challenge for synthesis. We thoroughly examined this proposal through high-throughput computational studies that show the novel phases of silicon could form spontaneously from liquid silicon in the presence of noble gases under high pressure and high temperature. In particular, our results show that a medium size of noble gas, e.g., Ar, can trigger the nucleation and growth of inert-gas silicon clathrate, whereas a small noble gas such as He is able to induce the formation of an unconventional, inclusion-type compound $\text{Si}_{2}\text{He}$. Our findings, along with the gained molecular insights, thus strongly suggest it is viable to experimentally synthesize novel silicon phases with noble gas through high pressure and high temperature.


text:
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 Trudy MIAN: Year: Volume: Issue: Page: Find

 Tr. Mat. Inst. Steklova, Forthcoming paper (Mi tm4072)

Schatten class operators in a representation space of Canonical Commutation Relations

A. S. Holevo

Steklov Mathematical Institute of Russian Academy of Sciences, Moscow

Abstract: Schatten classes $\mathfrak{S}_{p}$ of operators in a Hilbert space – the noncommutative analog of the classical $l_{p}$ spaces – play a significant role in recent studies on quantum information theory. In the present paper we give sufficient conditions for an operator to belong to $\mathfrak{S}_{p}$ in the case where the Hilbert space is the space of an irreducible representation of the Canonical Commutation Relations.

UDC: 519.248.3
Revised: October 14, 2019
Accepted: January 27, 2020


text:
Drake
Contact surface constraints

Consider two points pᵢ and pⱼ on rigid bodies i and j, respectively, and assume that at a certain configuration of the two bodies, ᶜq, the two points are coincident at a single location in space, p(ᶜq).

To constrain the motion of pᵢ and pⱼ to the contact surface as the bodies move, one can introduce the constraint g(q) = n(q)ᵀ(pᵢ(q) - pⱼ(q)), where n(q) is the common surface normal expressed in the world frame. gₚ(q) is a unilateral constraint, meaning that complementarity constraints are necessary (see Constraint types):

0 ≤ gₚ  ⊥  λ ≥ 0

#### Differentiating gₚ(.) with respect to time

As usual, gₚ(.) must be differentiated (with respect to time) once to use the contact constraint in velocity-level constraint formulation or twice to use it in a force/acceleration-level formulation. Differentiating gₚ(q) once with respect to time yields:

ġₚ(q,v) = nᵀ(ṗᵢ - ṗⱼ) + ṅᵀ(pᵢ - pⱼ);

one more differentiation with respect to time yields:

g̈ₚ(q,v,v̇) = nᵀ(p̈ᵢ - p̈ⱼ) + 2ṅᵀ(ṗᵢ - ṗⱼ) + n̈ᵀ(pᵢ - pⱼ).

The non-negativity condition on the constraint force magnitudes (λ ≥ 0) keeps the contact force along the contact normal compressive, as is consistent with a non-adhesive contact model.

A more substantial discussion on the kinematics of contact can be found in [Pfeiffer 1996], Ch. 4.

Figure 1: Illustration of the interpretation of non-interpenetration constraints when two boxes are interpenetrating (right). The boxes prior to contact are shown at left, and are shown in the middle figure at the initial time of contact; the surface normal n̂ is shown in this figure as well. The bodies interpenetrate over time as the constraint becomes violated (e.g., by constraint drift). Nevertheless, n̂ is tracked over time from its initial direction (and definition relative to the blue body). σ represents the signed distance the bodies must be translated along n̂ so that they are osculating (kissing).

#### Constraint softening

As discussed in discretization, the non-interpenetration constraint can be softened by adding a term to the time derivative of the equation:

0 ≤ ġₚ  ⊥  λ ≥ 0

resulting in:

0 ≤ ġₚ + γλ  ⊥  λ ≥ 0

for γ ≥ 0.

#### Constraint stabilization

As discussed in Constraint stabilization, the drift induced by solving an Index-0 DAE or DCP (that has been reduced from an Index-3 DAE or DCP) can be mitigated through one of several strategies, one of which is Baumgarte Stabilization. The typical application of Baumgarte Stabilization will use the second time derivative of the complementarity condition:

0 ≤ g̈ₚ  ⊥  λ ≥ 0

Baumgarte Stabilization would be layered on top of this equation, resulting in:

0 ≤ g̈ₚ + 2αġₚ + β²gₚ  ⊥  λ ≥ 0

#### Effects of constraint softening and stabilization on constraint problem data

• Constraint regularization/softening is effected through gammaN
• Constraint stabilization is effected through kN
• Note that neither Baumgarte Stabilization nor constraint regularization/ softening affects the definition of g(.)'s Jacobian operators, N_mult and either N_minus_muQ_transpose_mult (for ConstraintAccelProblemData) or N_transpose_mult (for ConstraintVelProblemData)


text:
# A few questions about the calculations behind the metabin function, part of the R 'meta' package for meta analysis of binomial data

I am new to meta analysis, and am reading about how to perform meta analysis with binomial data.

I'm looking for some clarification on some of the results given by the metabin function, which is part of the meta package in R.

Here's an example:

> test.dat <- data.frame(exp.events = c(3, 4, 5), exp.n = c(10, 11, 12),
control.events = c(14, 7, 2), control.n = c(20, 11, 7))

> result <- metabin(event.e = exp.events, n.e = exp.n, event.c = control.events,
n.c = control.n, data = test.dat, method.tau = "PM")

This code yields:

> result
RR           95%-CI %W(fixed) %W(random)
1 0.4286 [0.1594; 1.1525]      49.5       36.4
2 0.5714 [0.2322; 1.4060]      37.1       43.3
3 1.4583 [0.3784; 5.6205]      13.4       20.3

Number of studies combined: k = 3

RR           95%-CI     z p-value
Fixed effect model   0.6195 [0.3441; 1.1154] -1.60  0.1105
Random effects model 0.6224 [0.3343; 1.1588] -1.50  0.1348

Quantifying heterogeneity:
tau^2 = 0.0213; H = 1.03 [1.00; 3.19]; I^2 = 5.2% [0.0%; 90.1%]

Test of heterogeneity:
Q d.f. p-value
2.11    2  0.3481

Details on meta-analytical method:
- Mantel-Haenszel method
- Paule-Mandel estimator for tau^2

Question 1:

How exactly is the $$Q$$ statistic of 2.11 calculated from the numbers in the results?

From the literature I've seen that the $$Q$$ statistic defined as the weighted sum of squared difference between the individual study effects and pooled effect.

However I don't obtain $$Q = 2.11$$ when I perform this calculation with both the fixed and random effects models:

> weights.random <- c(.364, .433, .203)
> weights.fixed <- c(.495, .371, .134)
> RR <- c(.4286, .5714, 1.4583)
> log.RR <- log(RR)

> sum(weights.random*(RR - .6224)**2)
[1] 0.1566395

> sum(weights.fixed*(RR - .6195)**2)
[1] 0.113178

Nor do I obtain $$Q = 2.11$$ when I perform this calculation with the relative risks on the log scale:

> sum(weights.random*(log.RR - log(.6224))**2)
[1] 0.2009899
> sum(weights.fixed*(log.RR - log(.6195))**2)
[1] 0.1678128

Question 2

In the metabin() function example above I set method.tau = "PM". This is the Paule-Mandel method for estimating between-study variance $$\tau^{2}$$.

The documentation for this package cites this paper by Paule and Mandel when they discuss this method for estimating $$\tau^{2}$$.

However, in that paper Paule and Mandel define their method only for continuous data. The paper does not discuss how their method for estimating between study variance $$\tau^{2}$$ would be performed in the case of binomial data.

How is the package able to use method.tau = "PM"?

• The weights have been normalised to percentages. – mdewey Jan 15 '19 at 18:10
• Thanks for the response. Could you explain further? How does the fact that the weights are normalized affect the calculation of Q? – XRK Jan 15 '19 at 20:32

The problem here is that the weights have been normalised to add up to 100 so that they are percentage of total weight. The actual fixed effects weights are:

4.666667 3.500000 1.263158

which can be obtained by hand calculation or extracted from the object returned by metabin(). They are in w.fixed and w.random. Dividing them by their sum gives:

0.4948837 0.3711628 0.1339535

agreeing with the output in the question.

Questions about R code for meta-analysis might be better posed on the R mailing list for meta-analysis https://stat.ethz.ch/mailman/listinfo/r-sig-meta-analysis// as they might be considered off-topic here.

• Thanks for your response! I'll ask the mailing list. – XRK Jan 16 '19 at 20:47


text:
# Prime is already confusing, now what?

Find the sum of primes $p$ such that $p^2+11$ has exactly six different divisors (including 1 and the number itself).

×


text:
# Why does the constant term have a P value in statistical programs?

I notice when running regressions in programs such as Stata that the constant term has a p value. Why and what does this p value represent?

• Why should it not? – Tim Jan 5 '20 at 19:40

The standard error, p-value, and confidence intervals about regression constants represent the uncertainty in the estimated value of the constant.

Let's take bivariate OLS linear regression as an example. One of the primary questions we answer with linear regression is what straight line* best predicts the trend in mean values of $$y$$ over values of $$x$$?

When we recall that a straight line is defined by two parameters, a $$y$$ intercept, and a slope describing change in $$y$$ given a 1 unit increase in $$x$$, we can recognize the regression constant term (let's call it $$\beta_{0}$$) as the former, and the coefficient of your predictor variable (let's call it $$\beta_{x}$$) as the latter, we can then see one place where statistical inference enters a linear regression: what are the estimated values of $$\beta_{0}$$ and $$\beta_{x}$$, and what are the uncertainties about these estimates?

The question mentioned p-values, and that specifically means hypothesis tests are involved, so setting aside confidence intervals, what hypotheses are being tested? These are:

1. $$H_{0}: \beta_{0} = 0\text{, with } H_{\text{A}}: \beta_{0} \ne 0$$; and
2. $$H_{0}: \beta_{x} = 0\text{, with } H_{\text{A}}: \beta_{x} \ne 0$$

You can see that hypothesis (1) gives some interpretability to a rejection decision attached to the p-value for a regression constant: is there evidence that the intercept has a different value than 0? If a p-value is low enough (with respect to $$\alpha$$), then the answer is yes, otherwise the answer is there is not evidence that the intercept has a value different than 0. Aside: this is, of course, different than asking if there is evidence that the intercept is equivalent to 0 (say, by using TOST in Stata with the tostregress command which is part of the tost package).

As an example, let's look at some Stata regression output, where we regress automobile model weight onto automobile length:

. sysuse auto
(1978 Automobile Data)

. regress weight length

Source |       SS           df       MS      Number of obs   =        74
-------------+----------------------------------   F(1, 72)        =    613.27
Model |  39461306.8         1  39461306.8   Prob > F        =    0.0000
Residual |  4632871.55        72  64345.4382   R-squared       =    0.8949
Total |  44094178.4        73  604029.841   Root MSE        =    253.66

------------------------------------------------------------------------------
weight |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
length |   33.01988   1.333364    24.76   0.000     30.36187    35.67789
_cons |  -3186.047   252.3113   -12.63   0.000     -3689.02   -2683.073
------------------------------------------------------------------------------
`

We can see that the estimated intercept term ($$\hat{\beta}_{0}$$) is –3186.047, and that the reported p-value is less than 0.001. Therefore, with an $$\alpha = 0.05$$, we would reject $$H_{0}: \beta_{0} = 0$$ in favor of $$H_{\text{A}}: \beta_{0} \ne 0$$, and conclude that we found evidence that the intercept was significantly different (less) than zero, at the $$\alpha=0.05$$ level. Aside: remember that what we want to make an inference about are $$\beta_{0}$$ and $$\beta_{x}$$, but what we actually have to work with are $$\hat{\beta}_{0}$$ and $$\hat{\beta}_{x}$$, so we need statistical inference to get to the former from the latter.

One final point: I began this answer by assuming "OLS least squares." It may be obvious that this answer also applies to multiple OLS linear regression where there is more than one predictor variable. But it is worth saying that this also applies to other forms of regression, including regressions with nonlinear link functions like logistic regression, to generalized linear models, to nonlinear least squares regression, etc: whenever you are estimating parameters (whether or not those parameters are model constants), you have the opportunity to perform statistical inference about the values of those parameters. Aside: while the model constant term in these examples defines the $$y$$ intercept of a line, and, for example, defines the $$y$$ intercept of the sigmoid logistic curve in logistic regression, in some models (notably autoregressive time series models) there may not be any $$y$$ intercept per se, even if there is a model constant term estimated—in other words the meaning of $$\beta_{0}$$ may not always be an intercept term. Second aside: while your question is about frequentist statistical inference, Bayesian statistical inference also cares about constant terms.

* Or at least a straight line segment over observed range of $$x$$ values.

• When the p value of the variable is non significant, but the p value of the constant is highly significant...What does this mean? There is no significant relationship between the two variables, but if there was, we would be able to predict it with significance? – Paze Jan 5 '20 at 19:23
• @Paze Hi! I am saving the question in parts as I write it. I think the question in your comment is answered now... I have a few more additions to make, so even more questions may get answered. :) – Alexis Jan 5 '20 at 19:33
• @Paze Alright, I am all done answering the question know. :) By the way, welcome to CV! – Alexis Jan 5 '20 at 20:06
• @Paze And in just in case here's an even more explicit answer to the question in your comment: a non-significant slope ($\hat{\beta}_{x}$) means there is not evidence that the slope is different than 0. Another way of saying that is there is not evidence that the regression line is different than a horizontal line (i.e. $y$ does not change over values of $x$). However, even a horizontal line has a $y$ intercept. – Alexis Jan 5 '20 at 20:18

The intercept $$\beta_0$$ is interpreted as the predicted value of $$y$$ when $$x$$ is zero. The p-value corresponds to the test of $$H_0: \beta_0 = 0$$. Sometimes this is an interesting test, and other times it isn't. In general, you might ask the question, "For a given value of $$x$$, does the predicted value for $$y$$ differ from zero?". The test of $$\beta_0$$ answers this question for $$x=0$$. If you subtract a value $$k$$ from all of your $$x$$ values and then rerun the regression, the test of the intercept now answers this question for $$x=k$$.


text:
# You're reading: Posts Tagged: Jim Propp

### My adventures in 3D printing: Spherical pseudo-cuboctahedron

At work we’ve got a 3D printer. In this series of posts I’ll share some of the designs I’ve made.

This shape is a “spherical pseudo-cuboctahedron”, prompted by a request from Jim Propp on the math-fun mailing list.

It has 24 vertices, 12 edges and 14 faces. That doesn’t satisfy Euler’s formula $V – E + F = 2$, so it can’t be a proper polyhedron – hence “pseudo-cuboctahedron”.

However, if you push all the vertices onto the surface of a sphere, all the edges are spherical arcs, it sort of works.

While designing this object, I got fed up with OpenSCAD‘s awkward control syntax, and switched to Python. I wrote Python code to produce the coordinates of points along the edges, which the SolidPython library turned into something that OpenSCAD can cut out of a sphere.


text:
UT Invitational Programming Contest Fall 2019

#### Start

2019-11-23 19:00 UTC

## UT Invitational Programming Contest Fall 2019

#### End

2019-11-24 00:00 UTC
The end is near!
Contest is over.
Not yet started.
Contest is starting in -190 days 2:10:16

5:00:00

0:00:00

# Problem IPigeon X

$M$ bad pigeons in a mansion
Wrist on Milly Rock them diamonds on me dancin’
When you workin’ hard then your money start expandin’
I got model pigeon wanna peck me like some candy
And them seeds come in handy
Last name Savage pigeon, no I’m not Randy

$21$ Savage has $M$ bad pigeons in a mansion with $N$ rooms. Each pigeon is in a different room. He also has $M$ seed stashes stored in rooms across the mansion. The mansion can be represented as a weighted undirected graph where each room is a node and each corridor is an edge between two nodes. $21$ Savage desires to pair pigeons with a seed stash so that the total distance each pigeon has to travel to reach its seed is minimized.

Help $21$ Savage match pigeons with seed!

## Input

The first line of input has two space separated integers $N$ and $M$, the number of rooms and the number of pigeons and seed stashes. It is guaranteed that $1 \le N \le 500$ and $1 \le 2M \le N$.

$N$ lines follow. Each line has $N$ space separated integers. The $j$th entry on the $i$th line represents the length of the corridor from room $i$ to room $j$. It is guaranteed that the length of the corridor from room $i$ to room $j$ is the same as the length of the corridor from room $j$ to room $i$. It is also guaranteed that the length of the corridor from room $i$ to room $i$ for all $i$ is $0$. Finally the lengths $l_{i,j}$ of the corridor between room $i$ and room $j$ satisfies $0 \le l_{i,j} \le 10\, 000$ for all $i$ and $j$.

The next line has $M$ space separated unique integers representing the room locations of the $M$ pigeons. It is guaranteed that all integers $F_ i$ satisfy $0 \le P_ i < N$.

The final line has $M$ space separated unique integers representing the room locations of the $M$ seed stashes. It is guaranteed that all integers $F_ i$ satisfy $0 \le F_ i < N$.

## Output

Output a single integer, the minimum total distance that all pigeons have to travel to eat a unique seed stash.

Sample Input 1 Sample Output 1
4 2
0 2 2 2
2 0 2 2
2 2 0 2
2 2 2 0
0 1
2 3

4


text:
# Examples of using tidynamics¶

The examples below show how one can use tidynamics to compute correlations and mean-squared displacements (MSD).

The example “Command-line interface to tidynamics” runs in the command-line. It performs the two basic operations of tidynamics (autocorrelation function and MSD) using files as input data. The file random_steps_sample_0.txt.gz contains example data for computing the autocorrelation function and the file random_walk_sample_0.txt.gz contains example data for computing the MSD.

The following commands run the command-line tool:

python tidynamics_tool.py msd random_walk_sample_0.txt.gz random_walk_msd_0.txt
python tidynamics_tool.py acf random_steps_sample_0.txt.gz random_steps_acf_0.txt

The output is stored in the files random_walk_msd_0.txt and random_steps_acf_0.txt.

The other examples require matplotlib.

Gallery generated by Sphinx-Gallery


text:
# Dominant Random Strategies: Why Judges Are Sometimes Better Off Flipping Coins

In cases in which policy arguments for a choice can go both ways, judges, agencies, and other decision makers are sometimes better off flipping a coin than making an arbitrary, or even qualified choice.

There is a big debate over what methods judges should use to interpret written laws. The choice of interpretation often has grave consequences. In one case, a man’s prison sentence depended on the interpretation of “using” a gun. If “use” referred only to shooting a gun, he was innocent, but if it also included trading a gun for drugs, he would be guilty (see Smith v. US). The problem here is not the result of the case – the “justness” or “correctness” of the result is largely ignored in the decision; rather, the problem is that the Supreme Court, in this case among many others, fiercely debates the correct method of interpretation.

Without stating what they are, we can assume that the choice of method has policy implications – there are simply too many smart people debating the issue for it to be a moot point. It’s not very clear that courts apply any sort of consistent approach, though the debate seems to suggest that there should be one. Since there is a divide, and (apparently) consistency does not matter much, perhaps choosing which method to use when interpreting laws is a decision best made by flipping a coin.

Math that inspires the result:

Here’s one of my favorite math puzzles. Solve it now if you’re up to the challenge, or read on for a big hint.

What makes this problem interesting is that the when Alice picks the “two different real numbers by an unknown process”, she is picking them arbitrarily, not at random. It turns out that this makes a difference to our strategy, and this particular puzzle is a complex demonstration of the idea that arbitrary is not the same as random.

Let’s simplify things:

Suppose Alice picks a secret number from one to six inclusive, and you win $1 if you guess it correctly. It makes a difference if Alice chose her number arbitrarily or if she rolled a hidden die. Do you see why? If Alice rolls a dice, we can simply guess “1″ and be guaranteed a one in six chance of guessing correctly. But if Alice instead just chose 6 every time she played this game, we would be guaranteed a loss if we guess “1″. We can eliminate this uncertainty by using a random strategy. If we roll a dice instead of just calling our an arbitrary number, we can guarantee ourselves a one in six chance of guessing correctly regardless of how Alice chooses her number. All this talk of random strategies should give you a huge hint as to how to solve the math puzzle. Whereas in the simple example above, a random strategy simply guaranteed a one in six chance of winning, a random strategy in the math puzzle actually produces a greater than 50% chance of winning. Here is the solution (the first comment). An intuitive geometric solution exists, but it’s hard to explain over the Internet, so if the algebra doesn’t convince you… take my word for it – the solution works. How it applies to law: For decisions in which consistency does not matter, it follows from the above discussion that when offered a choice between two mutually exclusive judicial strategies, with equal policy arguments on both sides, we are better off flipping a coin. Notice that the “correct” strategy – that is, the one that best advances policy objectives – is an unknown known. Although we don’t know which strategy is better, one is necessarily better. This is analogous to Alice picking a number of her choice instead of rolling a dice. We don’t know what that number is, but it’s necessarily one of the six choices. Therefore, if we pick one strategy and stick with it, we risk being wrong 100% of the time (of course, we could also be right 100% of the time). If we instead flip a coin and choose our strategy at random, we are guaranteed the correct strategy 50% of the time. In some cases, if the consistency of methodology does not matter, this is the dominant strategy. Interestingly, a 50/50 coin flipping strategy may be superior even in cases in which we think that policy arguments lean towards one side. This occurs because of society’s natural risk aversion (more on this later – for now assume that society uses the natural logarithm as its utility function). Here is a quantitative demonstration (please excuse my outlandish numbers, and the several implicit assumptions I’m making here – an extreme example is sufficient and within my mathematical capability; also note the most important assumption is that the costs and benefits don’t compound): Suppose policy considerations give us 60% confidence that Strategy A is the correct strategy, and that using the correct strategy keeps the status quo, but using the incorrect one costs a relevant administrative agency$1 each decision. Suppose there are two such decisions a year, and that the relevant agency has an annual budget of \$3.

The Expected Utility of using Strategy A for both decisions is

$0.4\cdot \ln (\frac{1}{3}) + 0.6\cdot \ln (1) = -1.1,$ whereas the utility of flipping a coin is

$0.25\cdot \ln (\frac{1}{3}) + 0.5\cdot \ln (\frac{2}{3}) + 0.25\cdot \ln (1) = -.58.$

In this example, sticking to a single strategy is nearly twice as bad as flipping a coin, even though we were 60% sure that the single strategy we chose was correct. This suggests that flipping a weighted coin might be better than both options.

Intuitively, the result occurs because we prefer moderate outcomes to extreme outcomes. You can see how this result might extend to more reasonable scenarios (smaller relative costs each decision, but more decisions / year).

So, going back to the legal starting point, we see that if a judge (or agency?) is faced with a decision between two methodologies that arrive at different results, and there is no consistency reason for him to pick one or the other, he might be better off flipping a coin.

Categories: Random.

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The sum of the first 'n' natural numbers is a 3 digit number in which all the digits are the same. How many numbers have been summed?

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When I type a sequence of letters my calculator gives the product of all the numbers in the corresponding memories. What numbers should I store so that when I type 'ONE' it returns 1, and when I type. . . .

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Find the numbers in this sum

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What is the largest number you can make using the three digits 2, 3 and 4 in any way you like, using any operations you like? You can only use each digit once.

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A combination mechanism for a safe comprises thirty-two tumblers numbered from one to thirty-two in such a way that the numbers in each wheel total 132... Could you open the safe?

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A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.

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Can you arrange the digits 1,2,3,4,5,6,7,8,9 into three 3-digit numbers such that their total is close to 1500?

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If you wrote all the possible four digit numbers made by using each of the digits 2, 4, 5, 7 once, what would they add up to?

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Ben’s class were making cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?

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This challenge is to make up YOUR OWN alphanumeric. Each letter represents a digit and where the same letter appears more than once it must represent the same digit each time.

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There are over sixty different ways of making 24 by adding, subtracting, multiplying and dividing all four numbers 4, 6, 6 and 8 (using each number only once). How many can you find?

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Whenever two chameleons of different colours meet they change colour to the third colour. Describe the shortest sequence of meetings in which all the chameleons change to green if you start with 12. . . .

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Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.

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In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?

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The Scot, John Napier, invented these strips about 400 years ago to help calculate multiplication and division. Can you work out how to use Napier's bones to find the answer to these multiplications?

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You have four jugs of 9, 7, 4 and 2 litres capacity. The 9 litre jug is full of wine, the others are empty. Can you divide the wine into three equal quantities?

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The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?

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Use 4 four times with simple operations so that you get the answer 12. Can you make 15, 16 and 17 too?

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Find out what a Deca Tree is and then work out how many leaves there will be after the woodcutter has cut off a trunk, a branch, a twig and a leaf.

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Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.

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Fill in the missing numbers so that adding each pair of corner numbers gives you the number between them (in the box).

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Using 3 rods of integer lengths, none longer than 10 units and not using any rod more than once, you can measure all the lengths in whole units from 1 to 10 units. How many ways can you do this?

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On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?

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Fill in the numbers to make the sum of each row, column and diagonal equal to 34. For an extra challenge try the huge American Flag magic square.

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Using some or all of the operations of addition, subtraction, multiplication and division and using the digits 3, 3, 8 and 8 each once and only once make an expression equal to 24.

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For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target?

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Got It game for an adult and child. How can you play so that you know you will always win?

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This article suggests some ways of making sense of calculations involving positive and negative numbers.

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Use your logical reasoning to work out how many cows and how many sheep there are in each field.

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The value of the circle changes in each of the following problems. Can you discover its value in each problem?

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How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?

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Find at least one way to put in some operation signs (+ - x ÷) to make these digits come to 100.

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Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done?

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Can you substitute numbers for the letters in these sums?

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What is happening at each box in these machines?

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You have 5 darts and your target score is 44. How many different ways could you score 44?

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This group activity will encourage you to share calculation strategies and to think about which strategy might be the most efficient.

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Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column?

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Can you score 100 by throwing rings on this board? Is there more than way to do it?

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In this game for two players, the idea is to take it in turns to choose 1, 3, 5 or 7. The winner is the first to make the total 37.

### Sealed Solution

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Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes?

### Wild Jack

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A game for 2 or more players with a pack of cards. Practise your skills of addition, subtraction, multiplication and division to hit the target score.


text:
# Ex.5.3 Q3 Arithmetic Progressions Solution - NCERT Maths Class 10

Go back to  'Ex.5.3'

## Question

In an AP

(i) Given $$a = 5, d = 3, {a_n} = 50$$, find $$n$$ and $${S_n}$$.

(ii) Given $$a = 7, {a_{13}} = 35$$, find $$d$$ and $${S_{13}}$$.

(iii) Given $${a_{12}} = 37, d = 3$$, find $$a$$ and $${S_{12}}$$.

(iv) Given $${a_3} = 15, {S_{10}} = 125$$, find $$d$$ and $${a_{10}}$$.

(v) Given, find $$a$$ and $${a_9}$$.

(vi) Given $$a = 2, d = 8, {S_n} = 90$$, find $$n$$ and $${a_n}$$. $$d = 5, {S_9} = 75$$

(vii) Given $$a = 8, {a_n} = 62, {S_n} = 210$$, find $$n$$ and $$d$$.

(viii) Given $${a_n} = 4, d = 2, {S_n} = - 14$$, find $$n$$ and $$a$$.

(ix) Given $$a = 3, n = 8, S = 192$$, find $$d$$.

(x) Given $$l = 28, S = 144$$ and there are total $$9$$ terms. Find $$a$$.

Video Solution
Arithmetic Progressions
Ex 5.3 | Question 3

## Text Solution

(i) Given $$a = 5, d = 3, {a_n} = 50$$, find $$n$$ and $${S_n}$$.

What is Known?

$$a = 5, d = 3, {a_n} = 50$$

What is Unknown?

$$n$$ and $${S_n}$$.

Reasoning:

Sum of the first $$n$$ terms of an AP is Given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$ or $${S_n} = \frac{n}{2}\left[ {a + l} \right]$$, and $$n^\rm{th}$$ term of an AP is $$\,{a_n} = a + \left( {n - 1} \right)d$$

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms and $$l$$ is the last term.

Steps:

Given,

• First term, $$a = - 5$$
• Common difference, $$d = 3$$
• nth term, $$l = {a_n} = 50$$

As $$\,{a_n} = a + \left( {n - 1} \right)d$$

\begin{align}50&=5+\left( n-1 \right)3 \\45&=\left( n-1 \right)3 \\15&=n-1 \\n&=16 \\\end{align}

\begin{align}{S_n} &= \frac{n}{2}\left[ {a + l} \right]\\{S_{16}} &= \frac{{16}}{2}\left[ {5 + 50} \right]\\& = 8 \times 55\\ &= 440\end{align}

(ii) Given $$a = 7, {a_{13}} = 35$$, find $$d$$ and $${S_{13}}$$.

What is Known?

$$a = 7, {a_{13}} = 35$$,

What is Unknown?

$$d$$ and $${S_{13}}$$.

Reasoning:

Sum of the first $$n$$ terms of an AP is Given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$ or $${S_n} = \frac{n}{2}\left[ {a + l} \right]$$, and $$nth$$ term of an AP is $$\,{a_n} = a + \left( {n - 1} \right)d$$

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms and $$l$$ is the last term.

Steps:

Given,

• First term, $$a = 7$$
• $$13^\rm{th}$$ term, $$l = {a_{13}} = 35$$

As $$\,{a_n} = a + \left( {n - 1} \right)d$$

\begin{align}{{a}_{13}}&=a+\left( 13-1 \right)d \\35&=7+12d \\35-7&=12d \\d&=\frac{28}{12} \\ d& =\frac{7}{3} \\\end{align}

\begin{align}{S_n} &= \frac{n}{2}\left[ {a + l} \right]\\{S_{13}} &= \frac{{13}}{2}\left[ {7 + 35} \right]\\ &= \frac{{13}}{2} \times 42\\ &= 13 \times 21\\ &= 273\end{align}

(iii) Given $${a_{12}} = 37, d = 3$$, find $$a$$ and $${S_{12}}$$.

What is Known?

$${a_{12}} = 37, d = 3$$

What is Unknown?

$$a$$ and$${S_{12}}.$$

Reasoning:

Sum of the first $$n$$ terms of an AP is Given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$ or $${S_n} = \frac{n}{2}\left[ {a + l} \right]$$, and term of an AP is $$\,{a_n} = a + \left( {n - 1} \right)d$$

Where $$a$$ is the first term, is the common difference and $$n$$is the number of terms and $$l$$ is the last term.

Steps:

Given,

• $$12$$th term, $${a_{12}} = 37$$
• Common Difference, $$d = 3$$

As $$\,{a_n} = a + \left( {n - 1} \right)d$$

\begin{align}\,{a_{12}} &= a + \left( {12 - 1} \right)3\\37 &= a + 33\\a &= 4\end{align}

\begin{align}{S_n}& = \frac{n}{2}\left[ {a + l} \right]\\{S_{12}} &= \frac{{12}}{2}\left[ {4 + 37} \right]\\{S_{12}}& = 6 \times 41\\{S_{12}} &= 246\end{align}

(iv) Given $${a_3} = 15, {S_{10}} = 125$$, find $$d$$ and $${a_{10}}.$$

What is Known?

$${a_3} = 15, {S_{10}} = 125$$

What is Unknown?

$$d$$ and $${a_{10}}.$$

Reasoning:

Sum of the first $$n$$ terms of an AP is Given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$, and $$n^\rm{th}$$ term of an AP is $$\,{a_n} = a + \left( {n - 1} \right)d$$

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

Given,

• 3rd term, $${a_3} = 15$$
• Sum up to ten terms, $${S_{10}} = 125$$

As $$\,{a_n} = a + \left( {n - 1} \right)d$$

\begin{align}{a_3} &= a + \left( {3 - 1} \right)d\\15 &= a + 2d \qquad \qquad \dots\text{Equation (i )} \end{align}

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\ {S_{10}} &= \frac{{10}}{2}\left[ {2a + \left( {10 - 1} \right)d} \right]\\125 &= 5\left[ {2a + 9d} \right]\\ 25 &= 2a + 9d \qquad \qquad \dots\text{Equation (ii)}\end{align}

On multiplying equation (i) by $$2,$$ we obtain

$30 = 2a + 4d \qquad \qquad \dots \text{Equation (iii)}$

On subtracting equation (iii) from Equation (ii), we obtain

\begin{align} - 5 &= 5d\\d &= - 1\end{align}

From equation (i),

\begin{align}15 &= a + 2\left( { - 1} \right)\\15&= a-2\\a& = 17\end{align}

\begin{align}{a_{10}}& = a + \left( {10 - 1} \right)d\\{a_{10}}& = 17 + 9( - 1)\\{a_{10}} &= 17 - 9\\{a_{10}} &= 8\end{align}

(v) Given $$d = 5, {S_9} = 75$$, find $$a$$ and $${a_9}.$$

What is Known?

$$d = 5, {S_9} = 75$$

What is Unknown?

$$a$$ and $${a_9}.$$

Reasoning:

Sum of the first $$n$$ terms of an AP is Given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$, and $$n\rm{th}$$ term of an AP is $$\,{a_n} = a + \left( {n - 1} \right)d$$

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

Given,

• Common difference, $$d = 5$$
• Sum up to nine terms, $${S_9} = 75$$

As $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$

\begin{align}{S_9} &= \frac{9}{2}\left[ {2a + \left( {9 - 1} \right)5} \right]\\75 &= \frac{9}{2}\left( {2a + 40} \right)\\25 &= 3\left( {a + 20} \right)\\25 &= 3a + 60\\a &= \frac{{ - 35}}{3}\end{align}

We know that $$n\rm{th}$$ term of the AP series, $${a_n} = a + \left( {n - 1} \right)d$$

\begin{align}{a_9} &= a + \left( {9 - 1} \right) \times 5\\ &= \frac{{ - 35}}{3} + 8 \times 5\\ &= \frac{{ - 35}}{3} + 40\\ &= \frac{{ - 35 + 120}}{3}\\ &= \frac{{85}}{3}\end{align}

(vi) Given $$a = 2, d = 8, {S_n} = 90$$, find $$n$$ and $${a_n}.$$

What is Known?

$$a = 2, d = 8, {S_n} = 90$$

What is Unknown?

$$n$$ and $${a_n}.$$

Reasoning:

Sum of the first $$n$$ terms of an AP is Given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$ or $${S_n} = \frac{n}{2}\left[ {a + l} \right]$$, and $$n\rm{th}$$ term of an AP is$$\,{a_n} = a + \left( {n - 1} \right)d$$

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms and $$l$$ is the last term.

Steps:

Given,

• First term, $$a = 2$$
• Common difference, $$d = 5$$
• Sum up to nth terms, $${S_n} = 90$$

As $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$

\begin{align}90 = \frac{n}{2}\left[ {4 + \left( {n - 1} \right)8} \right]\\90{\rm{ }} = n{\rm{ }}[2 + (n - 1)4]\\ 90 = n\left[ {2 + 4n - 4} \right]\\90 = n\left[ {4n - 2} \right]\\90 = 4{n^2} - 2n\\4{n^2} - 2n - 90 = 0\\4{n^2} - 20n + 18n - 90 = 0\\4n\left( {n - 5} \right) + 18\left( {n - 5} \right) = 0\\\left( {n - 5} \right)\left( {4n + 18} \right) = 0\end{align}

Either $$(n - 5) = 0$$ or $$(4n + 18) = 0$$

$$n = 5\,$$ or $$n = \frac{{ - 9}}{2}$$

However, $$n$$ can neither be negative nor fractional.

Therefore, $$n = 5\,$$

\begin{align}{a_n} &= a + \left( {n - 1} \right)d\\{a_5} &= 2 + \left( {5 - 1} \right)8\\{a_5} &= 2 + 4 \times 8\\{a_5} &= 2 + 32\\{a_5} &= {\rm{ }}34\end{align}

(vii) Given $$a = 8, {a_n} = 62, {S_n} = 210$$, find and $$d$$.

What is Known?

$$a = 8, {a_n} = 62, {S_n} = 210$$

What is Unknown?

$$n$$ and $$d.$$

Reasoning:

Sum of the first $$n$$ terms of an AP is Given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$ or $${S_n} = \frac{n}{2}\left[ {a + l} \right]$$, and $$nth$$ term of an AP is $$\,{a_n} = a + \left( {n - 1} \right)d$$

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$is the number of terms and $$l$$ is the last term.

Steps:

Given,

• First term, $$a = 8$$
• $$n^\rm{th}$$ term, $$l = {a_n} = 62$$
• Sum up to nth terms, $${S_n} = 210$$

As $${S_n} = \frac{n}{2}\left[ {a + l} \right]$$

\begin{align}210& = \frac{n}{2}\left[ {8 + 62} \right]\\210 &= \frac{n}{2} \times 70\\n &= 6\end{align}

We know that $$n^\rm{th}$$ term of the AP series, $${a_n} = a + \left( {n - 1} \right)d$$

\begin{align}62 &= 8 + \left( {6 - 1} \right)d\\62 - 8 &= 5d\\54 &= 5d\\d &= \frac{{54}}{5}\end{align}

(viii) Given $${a_n} = 4,{\rm{ }}d = 2,{\rm{ }}{S_n} = - {\rm{ }}14,$$ find $$n$$ and $$a$$.

What is Known?

$${a_n} = 4, d = 2, {S_n} = - 14$$

What is Unknown?

$$n$$ and $$a$$

Reasoning:

Sum of the first $$n$$ terms of an AP is Given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$ or $${S_n} = \frac{n}{2}\left[ {a + l} \right]$$, and $$nth$$ term of an AP is $$\,{a_n} = a + \left( {n - 1} \right)d$$

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms and $$l$$ is the last term.

Steps:

Given,

• Common difference, $$d = 2$$
• $$n^\rm{th}$$ term, $$l = {a_n} = 4$$
• Sum up to nth terms, $${S_n} = - 14$$

We know that nth term of AP series,$$\,{a_n} = a + \left( {n - 1} \right)d$$

\begin{align}4 &= a + (n - 1)2\\4 &= a + 2n - 2\\a &= 6 - 2n \qquad \dots(1)\end{align}

\begin{align}{S_n} &= \frac{n}{2}\left[ {a + l} \right]\\ - 14 &= \frac{n}{2}\left[ {6 - 2n + 4} \right].............[\text{from(1)}]\\ - 14& = n\left( {5 - n} \right)\\ - 14 &= 5n - {n^2}\\{n^2} - 5n - 14 &= 0\\{n^2} - 7n + 2n - 14 &= 0\\n\left( {n - 7} \right) + 2\left( {n - 7} \right) &= 0\\\left( {n - 7} \right)\left( {n + 2} \right) &= 0\end{align}

Either $$n - 7 = 0$$ or $$n + 2 = 0$$

$$n=7$$ or $$n = - 2$$

However, $$n$$ can neither be negative nor fractional.

Therefore, $$n=7$$

From equation (1), we obtain

\begin{align}a&= 6 - 2n\\a& = 6 - 2 \times 7\\a& = 6-14\\a& = - 8\end{align}

(ix) Given $$a = 3, n = 8, S = 192$$, find $$d$$.

What is Known?

$$a = 3, n = 8, S = 192$$

What is Unknown?

$$d$$

Reasoning:

Sum of the first $$n$$ terms of an AP is Given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$is the number of terms.

Steps:

Given,

• First term, $$a = 3$$
• Number of terms, $$n = 8$$
• Sum up to nth terms, $${S_n} = 192$$

\begin{align}{S_n}& = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\192 &= \frac{8}{2}\left[ {2 \times 3 + \left( {8 - 1} \right)d} \right]\\192 &= 4\left[ {6 + 7d} \right]\\48 &= 6 + 7d\\42 &= 7d\\d &= 6\end{align}

(x) Given $$l = 28, S = 144$$ and there are total $$9$$ terms. Find $$a$$.

What is Known?

$$l = 28, S = 144, n = 9$$

What is Unknown?

$$a$$

Reasoning:

Sum of the first $$n$$ terms of an AP is Given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$ or $${S_n} = \frac{n}{2}\left[ {a + l} \right]$$

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$is the number of terms and $$l$$ is the last term.

Steps:

Given,

• Last term, $$l = {a_n} = 28$$
• Number of terms, $$n = 9$$
• Sum up to nth terms, $${S_n} = 192$$

\begin{align}{S_n} &= \frac{n}{2}\left( {a + l} \right)\\144 &= \frac{9}{2}\left( {a + 28} \right)\\32 &= a + 28\\a &= 4\end{align}


text:
# Why does the cross section for Compton scattered photons decrease with increasing energy of incident photon?

The cross section is independent of energy for energies less than electron's mass, but for greater energies the cross section of compton scattered electrons decreases, why?

Think about what happens as you increase the photon energy, you begin to move away from merely dislodging an electron to an energy regime in which pair production becomes more predominant.

As you probably know, in the pair-production reaction, we need a third body for momentum conservation. Not surprisingly, that heavy body is one of the nuclei in the surface of the metal. It recoils just a little, due to it's relatively high mass, so that leaves the majority of the incident photon's energy available for pair production.

We can also discover pair production in the region of an atomic, as opposed to free, electron within the metal. Obviously in this situation, much more energy is absorbed by recoil effects. In fact, the normal cutoff turns out to be $4mc^2$.

So the combined cross section required for pair production is the sum of the cross sections associated with the above two reaction pathways. The total pair-production cross section is the sum of the two components, nuclear and electronic. These cross sections depend on the energy of the incoming photon. At high energies, approximately equal to or greater than 100 MeV, pair production is the dominant mechanism of radiation interaction with matter.

So as you increase the incident energy, you begin to see less of the Compton effect and more pair production, Compton scattering is still a factor, at low energy it can be related to the photoelectric effect and when you increase the energy, it competes with pair production.

Energy Dependent Photoelectric Effect, Compton Scattering & Pair Production

A good example of this is incident photon energy on lead. Below 0.1 MeV, we find almost totally photoelectric processes occurring , then between between 0.1 MeV and 2.5 MeV we can detect both photoelectric and Compton processes and finally, as the energy level rises above to between 2.5 MeV and 100 MeV, we can observe both Compton scattering and pair production.


text:
URL linebreak in footnote

I use a footnote with a too long url, so that I'd like to see a linebreak. The url package does linebreaks automatically as far as i know, but in the footnote it won't do it. Anyone have a solution to this?

\footnote{\url{http://www.this-is-a-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-long.url/}}

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In general \url does a good job in breaking the URL; this particular example can't be broken, because the hyphen is not a permissible point. Can you show the real URL? –  egreg Jul 18 '11 at 14:49
Welcome to TeX.sx! Your question was migrated here from another stackexchange site. Please register on this site, too, and make sure that both accounts are associated with each other, otherwise you won't be able to comment on or accept answers or edit your question. –  Martin Scharrer Jul 18 '11 at 16:14

migrated from stackoverflow.comJul 18 '11 at 14:40

This question came from our site for professional and enthusiast programmers.

The url package documentation suggests the following:

Ordinarily, breaks are not allowed after "-" characters because this leads to confusion. (Is the "-" part of the address or just a hyphen?) The package option [hyphens] allows breaks after explicit hyphen characters. The \url command will never ever hyphenate words.

If your original example included a URL that has that specific structure, then \usepackage[hyphens]{url} will solve your problem.

\documentclass{article}
\usepackage[hyphens]{url}
\begin{document}
This is text\footnote{\url{http://www.this-is-a-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-very-long.url/}}.
\end{document}

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