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# Laws of Motion Class 11 Notes NEET 2023 Physics

## Laws of Motion Class 11 Notes NEET 2023 Physics - Free PDF Download

Last updated date: 26th Mar 2023
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One of the crucial chapters of Physics Class 11 is Laws of Motion. This chapter holds immense importance for the preparation of the NEET exam as the conceptual questions are asked to test the aptitude of the aspirants. This chapter will introduce concepts that are the more advanced version of what you have studied in the previous classes. The laws of motion will be explained with proper scientific derivations to show their implications. To make it easier for your NEET preparation, refer to the concise Laws of Motion Class 11 notes prepared by the subject matter experts of Vedantu.

These notes have been prepared to offer a simpler explanation of all the basic and advanced concepts related to laws of motion. This chapter will prepare a strong foundation of kinematics and mechanics taught in the other chapters.

Competitive Exams after 12th Science

## Aristotle’s Fallacy:

The Aristotelian law says that external forces are necessary to keep a body in motion, but external forces are also necessary to overcome frictional forces in solids and viscous forces in fluids. These forces are always present in nature.

Newton studied the Ideas of Galileo regarding the motion of an object. He formulated three fundamental laws that govern the motion of objects. These three laws are known as Newton's laws of motion, which are as given below:

## Newton's First Law of Motion

Any object remains in the state of rest or in a uniform Motion along a straight line until it is compelled to change the state by applying an external force. This law as also known as Law of Inertia.

Tips: How to find causes the motion of a body which is initially in the state of rest.

• First apply the first law of motion.

• Now check the options by putting law.

• Now if only one option matches then that option is correct, if not then hop on to the other options.

## Applications of Newton's First Law of Motion

The following applications are based on Newton's first law of motion:

1. The passengers in a bus fall backward when it starts suddenly.

2. This is because the sudden start of the bus brings motion to the bus as well as to our feet in contact with the floor of the bus but the rest of our body opposes this motion because of inertia, so they fall backwards.

3. When a carpet or a blanket is beaten with a stick, then the dust particles separate out from it.

## Newton's Second Law of Motion

The second law of motion states that the rate of change of momentum of an object is proportional to the applied external force and takes place in the direction in which external force acts.

According to the second law of motion, force $F \propto \dfrac{\text { change in momentum }}{\text { time }}$

We know that, change in velocity with respect to time is called acceleration, $\left[ a=\dfrac{v-u}{t} \right]$

And also we know that momentum $(p)=m\times v$

Then, ${{p}_{2}} = mv$ and ${{p}_{1}} = mu$

$F = K\dfrac{{{p}_{2}}-{{p}_{1}}}{t}=K\dfrac{mv-mV}{t}=K\dfrac{m(v-u)}{t}=Kma\quad$

$K=\text { constant of proportionality. }$

## Application of Newton's Second Law of Motion

The following applications are based on Newton’s second law of motion:

1. A cricket player (or fielder) moves his hands backward on catching a fast cricket ball.

2. During athletics meet, a high jumping athlete is provided either a cushion or a heap of send on the ground to fall upon.

## Law of Conservation of Linear Momentum

When no external force acts on a system of several interacting particles, the total linear momentum of the system is conserved. The total linear momentum is the vector sum of the linear momentum of all the particles of the system.

Consider an isolated system (the system on which no external force acts) of particles. Suppose the m particles have masses ${{m}_{1}},{{m}_{2}},{{m}_{3}}$……${{m}_{n}}$ and are moving with velocities ${{v}_{1,}}{{v}_{2}},{{v}_{3,\ldots }}{{v}_{n}}$ respectively.

Then total linear momentum of the system is

$\text{p}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}+{{m}_{3}}{{v}_{3}}+\ldots +{{m}_{n}}{{v}_{n}}$

$\text{p}={{p}_{1}}+{{p}_{2}}+{{p}_{3}}+\ldots +{{p}_{n}}$

If $F$ is the external force acting on the system, then according to Newton’s second law $F=\dfrac{dp}{\text{ dt }}$

For an isolated system, $F=0$ or $\dfrac{dp}{dt}=0$

As the derivative of a constant is zero, so

${{\text{p}}_{1}}+{{\text{p}}_{2}}+{{\text{p}}_{3}}+\ldots {{\text{p}}_{n}}=\text{ constant }$

Thus in the ambiance of any external force, the total linear momentum of the system is constant. This is the law of conservatism of linear momentum.

## Newton's Third Law of Motion

When two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction. The third law is also known as the law of action and reaction.

## Applications of Newton's Third Law of Motion

Some application of Newton’s third law is as follows:

1. Walking of a person, a person can walk because of Newton's third law of motion. During walking a person pushes the ground in a backward direction and in the reaction the ground also pushes the person with an equal magnitude of force but in the opposite direction. This enables him to move in a forwarding direction against the push.

2. Recoil of the gun when the bullet is fired from a gun, the bullet also pushes the gun in the opposite direction with an equal magnitude of force. This results in the gunman feeling a backward push, i.e., recoil force from the butt of the gun.

### Equilibrium of a Particle

A particle is in equilibrium if it does not move when acted upon by two or more forces. Thus, the force resultant is zero, and the particle remains stationary.

i.e., $\Sigma \vec{F}=0$

Then $\Sigma \mathrm{F}_{\mathrm{x}}=0, \Sigma \mathrm{F}_{\mathrm{y}}=0$ and $\Sigma \mathrm{F}_{\mathrm{z}}=0$

### Stable Equilibrium

When a body is slightly displaced from equilibrium position, it tends to regain its original position, it is said to be in stable equilibrium. This means that P.E. is minimum  $\left(\dfrac{d^{2} u}{d r^{2}}=+v e\right)$

So the centre of gravity is at its lowest point.

### Unstable Equilibrium

A body is considered to be in unstable equilibrium if it travels in the direction of the displacement after being displaced from the equilibrium position.

P.E. is the maximum in this instance.$\left(\dfrac{d^{2} u}{d r^{2}}=-v e\right)$

As a result, the centre of gravity is highest.

## Impulse

A large force that acts on an object for a very short time and produces a large change in its momentum is called an impulsive force.

The impulse of a force acting on an object is equal to the product of the large force and small time interval for which it acts on the object.

Impulse $(I)=$ Force $\times$ Time

Its unit is Newton second.

## Friction

Friction is a force which opposes the relative motion between the two objects when one object moves or tries to move over the surface of another object.

### Types of Friction

There are three types of friction:

Types of Friction

Static Friction: It is an opposing force which comes into play when one object tends to move over the surface of another object.

Limiting Friction: The maximum static frictional force which comes into play when one object just begins to slide over the surface of another object.

Kinetic Friction: The opposing force that comes into play when one object moves over the Surface of another object.

Kinetic Friction is of Two Types:

Types of Kinetic Friction

Sliding Friction: It comes into play when one object slides over the surface of another object.

Rolling Friction: It comes into play when one object rolls over the surface of another object.

### Force and its Types

Force is the interaction that alters or seeks to alter the body's resting or moving condition after acting on a body. Six different types of forces rule nature.

### Conservative Force

It is the term for a force whose actions are unaffected by the direction taken. The conservative force's work depends solely on the starting and ending points. Therefore, a conservative force in a closed loop does not work. Examples include the forces of gravity, static electricity, and elastic force.

### Non-conservative Force

The term "nonconservative force" refers to a force whose work depends on the path a particle takes.

Example: Friction force, air drag force, etc.

### Constant Force

Constant force is the term used to describe a force that is constant over time and space. It always stays the same.

### Variable Force

Variable force is defined as a force that varies in time and distance. It depends on various factors, such as time, space, velocity, etc.

### Contact Force

Requires physical contact between the body and the force's application.

The Non-Contact Force doesn't need the body and the force's agent to make eye contact.

In mechanics, gravity is the sole non-contact force that we deal with. Other non-contact forces include magnetic force, electrostatic force, and others. They do not, however, fall under the mechanic's category. In the following articles, we will learn more about them. However, many exam questions focus primarily on the various Contact Forces. We will thus research them.

### Types of Contact Forces

1. Tension force

2. Friction force

3. Spring force

4. Drag force

5. Pseudo force

1. Tension Force

The mass that is connected to the string is balanced by the tension in the string.

2. Friction Force

"The frictional force occurs between the two surfaces when two items are in touch and one of the objects moves or desires to move."

Alternatively stated, "the force preventing the relative motion of solid surfaces is known as friction."

For instance, the car's tyres experience friction as it travels along the road. When we push a big stone and it doesn't move, friction force stops the motion.

The friction force is usually denoted by "f".

Friction force $(f)=\mu \mathrm{N}$

Here $\mathrm{N}$ is the normal force and is $\mu$ the coefficient of friction.

### Types of Friction

Friction is divided into the following categories according to its state of motion and its state of rest:

Static Friction:

• Static friction is the friction that exists between the surfaces of two bodies when they are not moving relative to one another but nevertheless have a tendency to move. When the body is at rest, there is friction.

• For example: when trying to move a block that is resting on a table while applying force, yet the block does not move. So, in this instance, static friction is acting. In this case, the static friction force balances the applied force.

Static Friction

• Friction is being adjusted via static friction. The amount of static friction grows along with an increase in the applied external force until it reaches a specified maximum value. The limiting value of static friction is referred to as this maximum value.

• When an external force is applied to the body, static friction only becomes a factor. Static friction won't exist if a body is held at rest and no force is used to move it.

• Static friction has a limit value that is independent of the contact area. The only factors are the surface's composition and the normal force.$\left(f_{s}\right)_{\max }=\mu_{s} N$

• Here $\mu_{s}$ is the coefficient of static friction and $\mathrm{N}$ is the normal force acting on the block.

• The force of static friction is typically expressed by:$\left(f_{s}\right)_{\max } \leq \mu_{s} N$.

• If the external force $(\mathrm{F})$ exceeds this limiting friction $\left(f_{s}\right)_{\max }$ then the block starts moving on the surface.

• Rolling friction is the friction that occurs between a body's surface and a surface when a body is rolling on it.

• When a body is rolling, the point of contact with the surface is not moving relative to the surface that the body is rolling on.

• A finite area (not a point) of the body comes into contact with the surface as a body rolls because the surfaces in contact temporarily flex a little. The overall result is that motion is opposed by the contact force component that is parallel to the surface.

• Rolling friction is denoted by $\left(f_{R}\right)$.

• For example: The friction caused by the road on the wheel of a moving vehicle is known as rolling friction. friction on the bicycle wheel.

Rolling Friction

### Points to Remember

• There is a difference between the static and kinetic coefficients of friction. The type of surface and temperature have an impact on the coefficient of friction.

• Because any object must have enough force of friction between it and the surface in order to prevent sliding on it and conduct a pure rolling motion, rolling friction is always less than static friction. Because there is no relative motion at the point of contact, the rolling friction produces no effort.

### Pseudo Forces:

There is a force exerted on the object we are witnessing whenever we observe it in relation to an accelerating body. A pseudo force is what we refer to as here.

Imaginary force is known as pseudo force.

Pseudo force $\left(F_{P}\right)=$ ma

Where $a$ is the acceleration of the body with respect to which we are watching the object, and $m$ is the mass of the object that we are observing.

A pseudo force will exert itself on a body outside the car, for instance, if we see it while we are seated inside a speeding vehicle.

### Types of Non-Contact Forces

These forces act at a distance.

1. Gravitational Force: Gravitational force is the pull of gravity that one object experiences from another due to their respective masses. It is inversely proportional to the square and directly proportional to the product of the masses of those items.

$\mathrm{F}\propto\dfrac{\left(\mathrm{m}_{1}\mathrm{~m}_{2}\right)}{\mathrm{r}^{2}}$

Where,

$F$ is the gravitational force between two bodies

$m_{1}$ is the mass of the first body

$m_{2}$ is the mass of the second body

$r$ is the distance between the centres of two bodies

The force exerted by two bodies with mass $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$, whose centres is $\mathrm{r}$ units apart is given as:

$F=G \dfrac{\left(m_{1} m_{2}\right)}{r^{2}}$

2. Electrical Force: Electric force is the interaction of any two charged bodies that is either attractive or repulsive.

The magnitude of the electrostatic interaction between two point charges is inversely proportional to the square of the distance between them and directly proportional to the scalar multiplication of the charges.

Electric force formula can be obtained from Coulomb’s law as follows:

$\vec{F}=K \dfrac{q_{0} q_{1}}{r^{2}} \hat{r}$

Where,

- $\vec{F}$ is the electric force directed between two charged bodies

- $\mathrm{K}$ is the constant of proportionality

- $q_{0}$ and $q_{1}$ are the amounts of charge on each body

- $r$ is the distance between the charged bodies

- $\hat{r}$ is the variable unit vector

3. Magnetic Force: The attraction or repulsion that develops between electrically charged particles as a result of their motion is known as magnetic force.

We know that the formula of magnetic force is $F=q v B \sin \theta$

$\boldsymbol{F}=$ magnetic force

$q$ = charge of moving particle

$v=$ particle velocity

$B=$ magnetic field

$\theta=$ angle between velocity and magnetic field vectors

### The Expression for Velocity of Body Moving in a Vertical Circle:

Consider a little object with mass m that is strung up on one end and rotated in a vertical circle with radius r. In this instance, the body's acceleration rises while it moves up the vertical circle and falls when it moves down it. As a result, the body's speed is constantly changing. The vertical circle's bottom position is where it is highest, and its top position is where it is lowest. As a result, the body is not moving in a circular motion that is constant. The weight "mg," regardless of the particle's location on the circle, always works vertically downward.

Velocity of Body Moving in a Vertical Circle

Energy at point $P=$ Energy at point $L$

$\dfrac{1}{2} m v^{2}+m g h=\dfrac{1}{2} m u^{2}$

$\therefore v^{2}+2 g h=u^{2}$

$\therefore v^{2}=u^{2}-2 g h$

$\therefore v=\sqrt{u^{2}-2 g h} \ldots \ldots \ldots . .(1)$

This is an expression for the velocity of a particle at any point performing a circular motion in a vertical circle.

Consider the centripetal force at point $P$

$T-\mathrm{mg} \cos \theta=\dfrac{m v^{2}}{r} \ldots \ldots \ldots.(2)$

$\therefore T=\mathrm{mg} \cos \theta+\dfrac{m v^{2}}{r}$

From figure, $\cos \theta=\dfrac{r-h}{r}$

Substituting in equation (2)

$T=\mathrm{mg}\left(\dfrac{r-h}{r}\right)+\dfrac{m v^{2}}{r}$

$\therefore T=\dfrac{\mathrm{m}}{r}\left(\mathrm{~g}(r-h)+v^{2}\right)$

$\text { But } v^{2}=u^{2}-2 g h$

$T=\dfrac{\mathrm{m}}{r}\left(g r-g h+u^{2}-2 g h\right)$

$T=\dfrac{\mathrm{m}}{r}\left(u^{2}-3 g h+g r\right)$

This is the expression for the tension in the string.

Case - I: When the body is at the lowermost position i.e. body is at L)

$(\mathrm{h}=0)$

$\text { We have, } T=\dfrac{\mathrm{m}}{r}\left(u^{2}-3 g h+g r\right)$

$T_{L}=\dfrac{\mathrm{m}}{r}\left(u^{2}-3g(0)+gr\right)=\dfrac{\mathrm{m}}{r}\left(u^{2}+g r\right)$

Case - II: When the body is at the uppermost position i.e. body is at $\mathrm{H}(\mathrm{h}=2 \mathrm{r})$

$\text { We have, } T=\dfrac{\mathrm{m}}{r}\left(u^{2}-3 g h+g r\right)$

$T_{H}=\dfrac{\mathrm{m}}{r}\left(u^{2}-3g(2r)+gr\right)=\dfrac{\mathrm{m}}{r}\left(u^{2}-5 g r\right)$

Case - III: When the string is horizontal i.e. body is at $M(h=r)$

$\text { We have, } T=\dfrac{\mathrm{m}}{r}\left(u^{2}-3 g h+g r\right)$

$T_{M}=\dfrac{\mathrm{m}}{r}\left(u^{2}-3g(r)+gr\right)=\dfrac{\mathrm{m}}{r}\left(u^{2}-2 g r\right)$

### Relation Between Tension at the Highest point and at the Lowest Point:

$T_{L}-T_{H}=\dfrac{\mathrm{m}}{r}\left(u^{2}+g r\right)-\dfrac{\mathrm{m}}{r}\left(u^{2}-5 g r\right)$

$\therefore T_{L}-T_{H}=\dfrac{\mathrm{m}}{r} u^{2}+m g-\dfrac{\mathrm{m}}{r} u^{2}+5 m g$

$\therefore T_{L}-T_{H}=6 \mathrm{mg}$

### Lowest Point $\mathrm{L}(\mathrm{h}=0)$ :

This is the least speed at which the body must move in order for it to loop, or complete one full round of the circle.

Hence at the highest point $T_{H} \geq 0$

$\therefore \dfrac{\mathrm{m}}{r}\left(u^{2}-5 g r\right) \geq 0$

$\therefore u^{2}-5 g r \geq 0$

$\therefore u^{2} \geq 5 g r$

$\therefore u \geq \sqrt{5 g r}$

### When String is Horizontal $(\mathrm{h}=\mathrm{r})$ :

We have,

$v=\sqrt{u^{2}-2 g h}$

$\therefore v =\sqrt{(\sqrt{5 g r})^{2}-2 g(r)}$

$\therefore v =\sqrt{5 g r-2 g r}=\sqrt{3 g r}$

### Tension in String and Acceleration of Particle:

At the Highest Point:

$T_{H}=\dfrac{m v^{2}}{r}+m g$

$a=v^{2} / r$

Banking of roads is the practice of raising the outer borders of curving roads over the inner edge in order to provide cars the required centripetal force to make a turn safely. Let's review what centripetal force is now. It is the force that causes an item to move in an angular or circular fashion by pulling or pushing it in that direction as it moves. Let's talk about the angle of banking and the vocabulary used in road banking in the following parts.

Consider a vehicle of mass "m" travelling at speed "v" on a radius "r" banked road. Let the banking angle be and the amount of friction between the vehicle's tyres and the road be f.

Total force upwards equals Total force downwards.

$N \cos \Theta=m g+f \sin \Theta$

Where,

$N \cos \Theta$

one of the components of normal reaction along the vertical axis

mg: weight of the vehicle acting vertically downward

$f \sin \Theta$

one of the components of frictional force along the vertical axis

therefore,

$m g=N \cos \Theta-f \sin \Theta$

(eq.1)

$\dfrac{m v^{2}}{r}=N \sin \Theta+f \cos \Theta$

Where,

$N \sin \Theta$ one of the components of normal reaction along the horizontal axis

$f \cos \Theta$

one of the components of frictional force along the horizontal axis

$\dfrac{\dfrac{m v^{2}}{r}}{m g}=\dfrac{N \sin \Theta+f \cos \Theta}{N \cos \Theta-f \sin \Theta}$

(after diving eq.1 and eq.2)

therefore,

$\dfrac{v^{2}}{r g}=\dfrac{N \sin \Theta+f \cos \Theta}{N \cos \Theta-f \sin \Theta}$

Frictional force

$f=\mu_{s} N$

$\dfrac{v^{2}}{r g}=\dfrac{N \sin \Theta_{+} \mu_{s} N \cos \Theta}{N \cos \Theta-\mu_{s} N \sin \Theta}$

$\dfrac{v^{2}}{r g}=\dfrac{N\left(\sin \Theta+\mu_{s} \cos \Theta\right)}{N\left(\cos \Theta-\mu_{s} \sin \Theta\right)}$

$\dfrac{v^{2}}{r g}=\dfrac{\left(\sin \Theta+\mu_{s} \cos \Theta\right)}{\left(\cos \Theta-\mu_{s} \sin \theta\right)}$

$\dfrac{v^{2}}{r g}=\dfrac{\left(\tan \Theta+\mu_{s}\right)}{\left(1-\mu_{s} \tan \Theta\right)}$

therefore,

$v=\sqrt{\dfrac{r g\left(\tan \Theta+\mu_{s}\right)}{\left(1-\mu_{s} \tan \Theta\right.}}$

$v_{\max }=\sqrt{r g \tan \Theta}$

$\tan \Theta=\dfrac{v^{2}}{r g}$

$\Theta=\tan ^{-1} \dfrac{v^{2}}{r g}$

### Important Formula

 Name of the Concept Formula Static friction $\mathbf{f}_{\mathrm{s}} \leq \boldsymbol{\mu}_{\mathrm{s}} \mathbf{N}$ Kinetic friction $\mathbf{f}_{\mathbf{k}}=\boldsymbol{\mu}_{\mathrm{k}} \mathbf{N}$ Rolling friction $\mathbf{f}_{\mathbf{r}}=\boldsymbol{\mu}_{\mathbf{r}} \mathbf{N}$ Calculation of Necessary Force in Different Conditions: Minimum pulling force $P$ at an angle $\alpha$ from the horizontal. $P=\dfrac{W \sin \theta}{\cos (\alpha-\theta)}$ Minimum pushing force $P$ at an angle $\alpha$ from the horizontal. $P=\dfrac{W \sin \theta}{\cos (\alpha+\theta)}$ Minimum pulling force $P$ to move the body up an inclined plane. $P=\dfrac{W \sin (\theta+\lambda)}{\cos (\alpha-\theta)}$ Minimum force on body in downward direction along the surface of inclined plane to start its motion. $P=\dfrac{W \sin (\theta-\lambda)}{\cos (\alpha-\theta)}$ Minimum force to avoid sliding a body down an inclined plane. $P=W\left[\dfrac{\sin (\lambda-\theta)}{\cos (\theta+\alpha)}\right]$ Minimum force for motion and its direction. $P_{\min }=\dfrac{\mu m g}{\sqrt{1+\mu^{2}}}$ Co-Efficient of Friction $\boldsymbol{\mu}=\mathbf{F} / \mathbf{R}$

Example 1: A small block of mass $0.1 \mathrm{~kg}$ lies on a fixed inclined plane PQ which makes an angle $\theta$ with the horizontal. A horizontal force of $1 \mathrm{~N}$ acts on the block through its centre of mass as shown in the figure. The block remains stationary if (take $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )

Block

(A) $\theta=45^{\circ}$

(B) $\theta>45^{\circ}$ and a frictional force acts on the block towards $P$.

(c) $\theta>45^{\circ}$ and a frictional force acts on the block towards $Q$.

(D) $\theta<45^{\circ}$ and a frictional force acts on the block towards Q.

Hint: The opposing force that is produced when two surfaces attempt to move in the same direction or in opposite directions is known as frictional force. A frictional force is primarily intended to produce resistance to the mobility of one surface across another surface.

### Complete Solution:

The forces acting on the block are $F=1 \mathrm{~N}$ towards the left, weight $\mathrm{mg}=0.1 \times 10=1 \mathrm{~N}$ downwards, normal force N, and the frictional force f. Resolve F and mg along and perpendicular to the inclined plane.

Different Components of Forces Acting on a Block.

When $\theta=45^{\circ}$, the net force that brings the block down is

$F_{d}=m g \sin \theta-F \cos \theta=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}=0$

Thus, the block is stationary if $\theta=45^{\circ}$. When $\theta>45^{\circ}$, the force $F_{d}>0$, and hence the block tends to move down. Thus, the frictional force f acts on the block upwards i.e., towards Q.

Example 2: Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. The masses of A, B and C are m, 2m and 2m respectively. A moves towards B with a speed of 9 m/s and makes an elastic collision with it. Thereafter B makes a completely inelastic collision with C. All motions occur along the same straight line. The final speed of C is :

Objects

(A) $6 \mathrm{~m} / \mathrm{s}$

(B) $9 \mathrm{~m} / \mathrm{s}$

(C) $4 \mathrm{~m} / \mathrm{s}$

(D) $3 \mathrm{~m} / \mathrm{s}$

Hint:

Complete Solution:

Collision between A and B

Collision between A and B

$m \times 9=m v_{1}+2 m v_{2}($ from momentum conservation $)$

$e=1=\dfrac{v_{2}-v_{1}}{9}$

$\Rightarrow v_{2}=6 \mathrm{~m} / \mathrm{sec} \cdot \mathrm{v}_{1}=-3 \mathrm{~m} / \mathrm{sec} \text {. }$

collision between $\mathrm{B}$ and $\mathrm{C}$

Collision between B and C

$2 \mathrm{~m} \times 6=4 \mathrm{mv}$ (from momentum conservation)

$\mathrm{v}=3 \mathrm{~m} / \mathrm{s}$

Hence, the final speed of C is 3m/s.

## Importance of Class 11 Physics NEET Laws of Motion

This chapter is the advanced version of the laws of motion you have studied in the previous classes. In this chapter, you will be introduced to the three laws of motion framed and proved by Sir Isaac Newton. Get in-depth knowledge of how these laws are scientifically proved. Find out how the derivations of these laws result into the formulation of new terms related to kinematics and mechanics.

Refer to the Laws of Motion notes to find out how the mathematical tools of derivatives are used to explain all three laws. This chapter will also explain how the concepts of displacement, velocity, acceleration, force, work, and power can be derived from these derivations.

Find out the meaning of advanced terminologies such as impulse, mass, weight, and the above-mentioned terms and their correlations. You will also get a good idea of what inertia is and how it affects the motion of objects in different cases.

The chapter is very important as it describes the fundamental concepts of force, work, and power used in the conceptual chapters of advanced physics. Understand how the laws define such terms and how these physical quantities can be calculated using their formulae and units.

## Benefits of Vedantu’s Revision Notes on Laws of Motion

• These revision notes are compiled to offer a simpler version of all the fundamental concepts related to the motion of objects. Study these terms and find the interrelation between them. Understand how these terms are determined from the explanation of laws of motion and how to calculate them easily.

• Get to know how the experts have framed the topics easily and enhance your understanding of this chapter. Gaing in-depth knowledge and understanding of these topics of this chapter using the revision notes on Laws of Motion and make your study sessions more productive.

• Revise these notes to complete preparing this chapter before a crucial exam. Remember all you have studied due to the simpler and concise representation of all definitions and derivations of the scientific principles of laws of motion.

• Proceed to answer the sample questions to evaluate your preparation level. Check where you need to work more in this fundamental Physics chapter to make your preparation better.

Get the free PDF version of Laws of Motion NEET notes and complete your study material for this chapter. Make your study sessions more productive and take a step ahead. Utilise the easier description of all the concepts related to laws of motion and become better at solving questions asked in the NEET exam and perform excellently.

## FAQs on Laws of Motion Class 11 Notes NEET 2023 Physics

1. What do you mean by motion?

When a body is moving in terms of time from one point to the other, this state is called motion.

2. What is inertia?

The tendency of a body to remain at rest or in motion even when an external force is applied is called inertia. It is directly proportional to the mass of an object.

3. What is conservation of momentum?

A body tries to remain in its state of rest or motion in an isolated system. It is called the conservation of momentum.

4. What is a force?

According to the Laws of Motion Class 11 notes for NEET, a physical quantity that tries to cause a change in motion or rest is called force.


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## kaykim95 2 years ago True or false?? This is a convergent geometric series

• This Question is Closed
1. kaykim95

2. estebananaya

i think its convergent, because you can put the argument of the sum like (-3/5)^(i+1) over -5

3. Lumenaire

It is convergent. It's common ratio is -3/5, the absolute value of which is smaller than one. So it does converge.

4. kaykim95

so true?

5. Lumenaire

yes

6. ZeHanz

Rewrite it a little:$\sum_{i=1}^{\infty}\frac{ 3^{i+1} }{ (-5)^{i+2} }=\sum_{i=1}^{\infty}\frac{ 3^{i+1} }{ (-5)(-5)^{i+1} }=-\frac{ 1 }{ 5 }\sum_{i=1}^{\infty}\left( -\frac{ 3 }{ 5 } \right)^{i+1}$Now read @Lumenaire 's answer (again)!

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Properties

 Label 3380.2.a.f Level $3380$ Weight $2$ Character orbit 3380.a Self dual yes Analytic conductor $26.989$ Analytic rank $0$ Dimension $1$ CM no Inner twists $1$

Related objects

Newspace parameters

 Level: $$N$$ $$=$$ $$3380 = 2^{2} \cdot 5 \cdot 13^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 3380.a (trivial)

Newform invariants

 Self dual: yes Analytic conductor: $$26.9894358832$$ Analytic rank: $$0$$ Dimension: $$1$$ Coefficient field: $$\mathbb{Q}$$ Coefficient ring: $$\mathbb{Z}$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 260) Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$

$q$-expansion

 $$f(q)$$ $$=$$ $$q + q^{3} - q^{5} - q^{7} - 2 q^{9} + O(q^{10})$$ $$q + q^{3} - q^{5} - q^{7} - 2 q^{9} + 3 q^{11} - q^{15} - 3 q^{17} + 5 q^{19} - q^{21} + 9 q^{23} + q^{25} - 5 q^{27} - 9 q^{29} + 8 q^{31} + 3 q^{33} + q^{35} - 7 q^{37} + 3 q^{41} - q^{43} + 2 q^{45} - 6 q^{49} - 3 q^{51} + 6 q^{53} - 3 q^{55} + 5 q^{57} + 9 q^{59} - q^{61} + 2 q^{63} + 5 q^{67} + 9 q^{69} + 9 q^{71} + 2 q^{73} + q^{75} - 3 q^{77} + 8 q^{79} + q^{81} + 3 q^{85} - 9 q^{87} + 3 q^{89} + 8 q^{93} - 5 q^{95} + 17 q^{97} - 6 q^{99} + O(q^{100})$$

Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
 0
0 1.00000 0 −1.00000 0 −1.00000 0 −2.00000 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

Atkin-Lehner signs

$$p$$ Sign
$$2$$ $$-1$$
$$5$$ $$1$$
$$13$$ $$1$$

Inner twists

This newform does not admit any (nontrivial) inner twists.

Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 3380.2.a.f 1
13.b even 2 1 3380.2.a.i 1
13.c even 3 2 260.2.i.a 2
13.d odd 4 2 3380.2.f.d 2
39.i odd 6 2 2340.2.q.f 2
52.j odd 6 2 1040.2.q.i 2
65.n even 6 2 1300.2.i.d 2
65.q odd 12 4 1300.2.bb.b 4

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
260.2.i.a 2 13.c even 3 2
1040.2.q.i 2 52.j odd 6 2
1300.2.i.d 2 65.n even 6 2
1300.2.bb.b 4 65.q odd 12 4
2340.2.q.f 2 39.i odd 6 2
3380.2.a.f 1 1.a even 1 1 trivial
3380.2.a.i 1 13.b even 2 1
3380.2.f.d 2 13.d odd 4 2

Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(3380))$$:

 $$T_{3} - 1$$ $$T_{7} + 1$$ $$T_{19} - 5$$

Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$T$$
$3$ $$-1 + T$$
$5$ $$1 + T$$
$7$ $$1 + T$$
$11$ $$-3 + T$$
$13$ $$T$$
$17$ $$3 + T$$
$19$ $$-5 + T$$
$23$ $$-9 + T$$
$29$ $$9 + T$$
$31$ $$-8 + T$$
$37$ $$7 + T$$
$41$ $$-3 + T$$
$43$ $$1 + T$$
$47$ $$T$$
$53$ $$-6 + T$$
$59$ $$-9 + T$$
$61$ $$1 + T$$
$67$ $$-5 + T$$
$71$ $$-9 + T$$
$73$ $$-2 + T$$
$79$ $$-8 + T$$
$83$ $$T$$
$89$ $$-3 + T$$
$97$ $$-17 + T$$


text:
# LOG#116. Basic Neutrinology(I).

This new post ignites a new thread.

Subject: the Science of Neutrinos. Something I usually call Neutrinology.

I am sure you will enjoy it, since I will keep it elementary (even if I discuss some more advanced topics at some moments). Personally, I believe that the neutrinos are the coolest particles in the Standard Model, and their applications in Science (Physics and related areas) or even Technology in the future ( I will share my thoughts on this issue in a forthcoming post) will be even greater than those we have at current time.

Let me begin…

The existence of the phantasmagoric neutrinos ( light, electrically neutral and feebly -very weakly- interacting fermions) was first proposed by W. Pauli in 1930 to save the principle of energy conservation in the theory of nuclear beta decay. The idea was promptly adopted by the physics community but the detection of that particle remained elusive: how could we detect a particle that is electrically neutral and that interact very,very weakly with normal matter? In 1933, E. Fermi takes the neutrino hypothesis, gives the neutrino its name (meaning “little neutron”, since it was realized than neutrinos were not Chadwick’s neutrons) and builds his theory of beta decay and weak interactions. With respect to its mass, Pauli initially expected the mass of the neutrino to be small, but necessarily zero. Pauli believed (originally) that the neutrino should not be much more massive than the electron itself. In 1934, F. Perrin showed that its mass had to be less than that of the electron.

By the other hand, it was firstly proposed to detect neutrinos exploding nuclear bombs! However, it was only in 1956 that C. Cowan and F. Reines (in what today is known as the Reines-Cowan experiment) were able to detect and discover the neutrino (or more precisely, the antineutrino). In 1962, Leon M. Lederman, M. Schwartz, J. Steinberger and Danby et al. showed that more than one type of neutrino species $\nu_e,\nu_\mu$ should exist by first detecting interactions of the muon neutrino. They won the Nobel Prize in 1988.

When we discovered the third lepton, the tau particle (or tauon), in 1975 at the Stanford Linear Accelerator Center, it too was expected to have an associated neutrino particle. The first evidence for this 3rd neutrino “flavor” came from the observation of missing energy and momentum in tau decays. These decays were analogue to the beta decay behaviour leading to the discovery of the neutrino particle.

In 1989, the study of the Z boson lifetime allows us to show with great experimental confidence that only 3 light neutrino species (or flavors) do exist. In 2000, the first detection of tau neutrino ($\nu_\tau$ in addition to $\nu_e,\nu_\mu$) interactions was announced by the DONUT collaboration at Fermilab, making it the latest particle of the Standard Model to have been discovered until the recent Higgs particle discovery (circa 2012, about one year ago).

In 1998, research results at the Super-Kamiokande neutrino detector in Japan (and later, independently, from SNO, Canada) determined for the first time that neutrinos do indeed experiment “neutrino oscillations” (I usually call NOCILLA, or NO for short, this phenomenon), i.e., neutrinos flavor “oscillate” and change their flavor when they travel  “short/long” distances. SNO and Super-Kamiokande tested and confirmed this hypothesis using “solar neutrinos”. this (quantum) phenomenon implies that:

1st. Neutrinos do have a mass. If they were massless, they could not oscillate. Then, the old debate of massless vs. massive neutrinos was finally ended.

2nd. The solar neutrino problem is solved. Some solar neutrinos scape to the detection in Super-Kamiokande and SNO, since they could not detect all the neutrino species. It also solved the old issue of “solar neutrinos”. The flux of (detected) solar neutrinos was lesser than expected (generally speaking by a factor 2). The neutrino oscillation hypothesis solved it since it was imply the fact that some neutrinos have been “transformed” into a type we can not detect.

3rd. New physics does exist. There is new physics at some energy scale beyond the electroweak scale (the electroweak symmetry breaking and typical energy scale is about 100GeV). The SM is not complete. The SM does (indeed) “predict” that the neutrinos are massless. Or, at least, it can be made simpler if you make neutrinos to be massless neutrinos described by Weyl spinors. It shows that, after the discovery of neutrino oscillations, it is not the case. Neutrinos are massive particles. However, they could be Dirac spinors (as all the known spinors in the Standard Model, SM) or they could also be Majorana particles, neutral fermions described by “Majorana” spinors and that makes them to be their own antiparticles! Dirac particles are different to their antiparticles. Majorana particles ARE the same that their own antiparticles.

In the period 2001-2005, neutrino oscillations (NO)/neutrino mixing phenomena(NEMIX) were observed for the first time at a reactor experiment (this type of experiment are usually referred as short baseline experiment in the neutrino community) called KamLAND. They give a good estimate (by the first time) of the difference in the squares of the neutrino masses. In May 2010, it was reported that physicists from CERN and the Italian National Institute for Nuclear Physics, in Gran Sasso National Laboratory, had observed for the first time a transformation between neutrino flavors during an accelerator experiment (also called neutrino beam experiment, a class of neutrino experiment belonging to “long range” or “long” baseline experiments with neutrino particles). It was a new solid evidence that at least one neutrino species or flavor does have mass. In 2012, the Daya Bay Reactor experiment in China, and later RENO in South Korea measured the so called $\theta_{13}$ mixing angle, the last neutrino mixing angle remained to be measured from the neutrino mass matrix. It showed to be larger than expected and it was consistent with earlier, but less significant results by the experiments T2K (another neutrino beam experiment), MINOS (other neutrino beam experiment) and Double Chooz (a reactor neutrino experiment).

With the known value of $\theta_{13}$ there are some probabilities that the $NO\nu A$ experiment at USA can find the neutrino mass hierarchy. In fact, beyond to determine the spinorial character (Dirac or Majorana) of the neutrino particles, and to determine their masses (yeah, we have not been able to “weight” the neutrinos, but we are close to it: they are the only particle in the SM with no “precise” value of mass), the remaining problem with neutrinos is to determine what kind of spectrum they have and to measure the so called CP violating processes. There are generally 3 types of neutrino spectra usually discussed in the literature:

A) Normal Hierarchy (NH): $m_1<. This spectrum follows the same pattern in the observed charged leptons, i.e., $m(e)<. The electron is about 0.511MeV, muon is about 106 MeV and the tau particle is 1777MeV.

B) Inverted Hierarchy (IH): $m_1<. This spectrum follows a pattern similar to the electron shells in atoms. Every “new” shell is closer in energy (“mass”) to the previous “level”.

C) Quasidegenerated (or degenerated) hierarchy/spectrum (QD): $m_1\sim m_2\sim m_3$.

While the above experiments show that neutrinos do have mass, the absolute neutrino mass scale is still not known. There are reasons to believe that its mass scale is in the range of some milielectron-volts (meV) up to the electron-volt scale (eV) if some extra neutrino degree of freedom (sterile neutrinos) do appear. In fact, the Neutrino OScillation EXperiments (NOSEX) are sensitive only to the difference in the square of the neutrino masses. There are some strongest upper limits on the masses of neutrinos that come from Cosmology:

1) The Big Bang model states that there is a fixed ratio between the number of neutrino species and the number of photons in the cosmic microwave background (CMB). If the total energy of all the neutrino species exceeded an upper bound about

$m_\nu\leq 50eV$

per neutrino, then, there would be so much mass in the Universe that it would collapse. It does not (apparently) happen.

2) Cosmological data, such as the cosmic microwave background radiation, the galaxy surveys, or the technique of the Lyman-alpha forest indicate that the sum of the neutrino masses should be less than 0.3 eV (if we don’t include sterile neutrinos, new neutrino species uncharged under the SM gauge group, that could increase that upper bound a little bit).

3) Some early measurements coming from lensing data of a galaxy cluster were analyzed in 2009. They suggest that the neutrino mass upper bound is about 1.5eV. This result is compatible with all the above results.

Today, some measurements in controlled experiments have given us some data about the squared mass differences (from both, solar neutrinos, atmospheric neutrinos produced by cosmic rays and accelerator/reactor experiments):

1) From KamLAND (2005), we get

$\Delta m_{21}^2=0\mbox{.}000079eV^2$

2) From MINOS (2006), we get

$\Delta m_{32}^2=0\mbox{.}0027eV^2$

There are some increasing efforts to directly determine the absolute neutrino mass scale in different laboratory experiments (LEX), mainly:

1) Nuclear beta decay (KATRIN, MARE,…).

2) Neutrinoless double beta decay (e.g., GERDA; CUORE, Cuoricino, NEMO3,…). If the neutrino is a Majorana particle, a new kind of beta decay becomes possible: the double beta decay without neutrinos (i.e., two electrons emitted and no neutrino after this kind of decay).

Neutrinos have a unique place among all the SM elementary particles. Their role in the cosmic evolution and the fundamental asymmetries in the SM (like CP violating reactions, or the C, T, and P single violations) make them the most fascinating and interesting particle that we know today (well, maybe, today, the Higgs particle is also as mysterious as the neutrino itself). We believe that neutrinos play an important role in Beyond Standard Model (BSM) Physics. Specially, I would like to highlight two aspects:

1) Baryogenesis from leptogenesis. Neutrinos can allow us to understand how could the Universe end in such an state that it contains (essentially) baryons and no antibaryons (i.e., the apparent matter-antimatter asymmetry of the Universe can be “explained”, with some unsolved problems we have not completely understood, if massive neutrinos are present).

2) Asymmetric mass generation mechanisms or the seesaw. Neutrinos allow us to build an asymmetric mass mechanism known as “seesaw” that makes “some neutrino species/states” very light and other states become “superheavy”. This mechanism is unique and, from some  non-subjective viewpoint, “simple”.

After nearly a century, the question of the neutrino mass and its origin is still an open question and a hot topic in high energy physics, particle physics, astrophysics, cosmology and theoretical physics in general.

If we want to understand the fermion masses, a detailed determination of the neutrino mass is necessary. The question why the neutrino masses are much smaller than their charged partners could be important! The little hierarchy problem is the problem of why the neutrino mass scale is smaller than the other fermionic masses and the electroweak scale. Moreover, neutrinos are a powerful probe of new physics at scales larger than the electroweak scale. Why? It is simple. (Massive) Neutrinos only interact under weak interactions and gravity! At least from the SM perspective, neutrinos are uncharged under electromagnetism or the color group, so they can only interact via intermediate weak bosons AND gravity (via the undiscovered gravitons!).

If neutrino are massive particles, as they show to be with the neutrino oscillation phenomena, the superposition postulates of quantum theory state that neutrinos, particles with identical quantum numbers, could oscillate in flavor space since they are electrically neutral particles. If the absolute difference of masses among them is small, then these oscillations or neutrino (flavor) mixing could have important phenomenological consequences in Astrophysics or Cosmology. Furthermore, neutrinos are basic ingredients of these two fields (Astrophysics and Cosmology). There may be a hot dark matter component (HDM) in the Universe: simulations of structure formation fit the observations only when some significant quantity of HDM is included. If so, neutrinos would be there, at least by weight, and they would be one of the most important ingredients in the composition of the Universe.

Regardless the issue of mass and neutrino oscillations/mixing, astrophysical interests in the neutrino interactions and their properties arise from the fact that it is produced in high temperature/high density environment, such as collapsing stars and/or supernovae or related physical processes. Neutrino physics dominates the physics of those astrophysical objects. Indeed, the neutrino interactions with matter is so weak, that it passes generally unnoticed and travels freely through any ordinary matter existing in the Universe. Thus, neutrinos can travel millions of light years before they interact (in general) with some piece of matter! Neutrinos are a very efficient carrier of energy drain from optically thick objects and they can serve as very good probes for studying the interior of such objects. Neutrino astronomy is just being born in recent years. IceCube and future neutrino “telescopes” will be able to see the Universe in a range of wavelengths and frequencies we have not ever seen till now. Electromagnetic radiation becomes “opaque” at some very high energies that neutrinos are likely been able to explore! Isn’t it wonderful? Neutrinos are high energy “telescopes”!

By the other hand, the solar neutrino flux is, together with heliosysmology and the field of geoneutrinos (neutrinos coming from the inner shells of Earth), some of the known probes of solar core and the Earth core. A similar statement applies to objects like type-II supernovae. Indeed, the most interesting questions around supernovae and the explosion dynamics itself with the shock revival (and the synthesis of the heaviest elements by the so-called r-processes) could be positively affected by changes in the observed neutrino fluxes (via some processes called resonant conversion, and active-sterile conversions).

Finally, ultra high energy neutrinos are likely to be useful probes of diverse distant astrophysical objects. Active Galactic Nuclei (AGN) should be copious emitters of neutrinos, providing detectable point sources and and observable “diffuse” background which is larger in fact that the atmospheric neutrino background in the very high energy range. Relic cosmic neutrinos, their thermal background, known as the cosmic neutrino background, and their detection about 1.9K are one of the most important lacking missing pieces in the Standard Cosmological Model (LCDM).

Do you understand why neutrinos are my favorite particles? I will devote this basic thread to them. I will make some advanced topics in the future. I promise.

May the Neutrinos be with you!

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# LOG#031. Entropic Gravity (II).

We will generalize the entropic gravity approach to include higher dimensions in this post. The keypoint from this theory of entropic gravity, according to Erik Verlinde, is that gravity does not exist as “fundamental” force and it is a derived concept. Entropy is the fundamental object somehow. And it can be generalized to a a d-dimensional world as follows.

The entropic force is defined as:

$\boxed{F=-\dfrac{\Delta U}{\Delta x}=-T\dfrac{\Delta S}{\Delta x}}$

The entropic force is a force resulting from the tendency of a system to increase its entropy. Since $\Delta S>0$ the sign of the force (whether repulsive or attractive) is determined by how we take the definition of $\Delta x$ as it is related to the system in question.

An arbitrary mass distribution M induces a holographic screen $\Sigma$  at some distance R that has encoded on it gravitational information. Today, we will consider the situation in d spatial dimensions.Using the holographic principle, the screen owns all physical information contained within its volume in bits on the screen whose number N is given by:

$\boxed{N=\dfrac{A_\Sigma (R)}{l_p^{d-1}}}$

This condition implies the quantization of the hyperspherical surface, where the hyperarea (from the hypersphere) is defined as:

$\boxed{A_\Sigma=\dfrac{2\pi^{d/2}}{\Gamma \left(\frac{d}{2}\right)}R^{d-1}}$

By the equipartition principle:

$\boxed{E=Mc^2=\dfrac{N}{2}k_BT}$

Therefore,

$k_BT=\dfrac{2Mc^2l_p^{d-1}}{A_\Sigma}$

The entropy shift due to some displacement is:

$\Delta S=2\pi k_B \dfrac{\Delta x}{\bar{\lambda}}=2\pi k_B mc\dfrac{\Delta x}{\hbar}$

Plugging the expression for the temperature and the entropy into the entropic force equation, we get:

$F=-T\dfrac{\Delta S}{\Delta x}=-\dfrac{2Mc^2}{k_B}\dfrac{2\pi k_B mc}{\hbar}\dfrac{l_p^{d-1}}{A_\Sigma}=-\dfrac{4\pi Mmc^3l_p^{d-1}}{\hbar A_\Sigma}$

and thus we finally get

$F=-\dfrac{2\pi^{1-\frac{d}{2}}\Gamma \left(\frac{d}{2}\right) l_p^{d-1}Mmc^3}{\hbar R^{d-1}}=-G_d\dfrac{Mm}{R^{d-1}}$

i.e.,

$\boxed{F=-G_d\dfrac{Mm}{R^{d-1}}}\leftrightarrow \boxed{F=-\dfrac{2\pi^{1-\frac{d}{2}}\Gamma \left(\frac{d}{2}\right) l_p^{d-1}Mmc^3}{\hbar R^{d-1}}}$

where we have defined the gravitational constant in d dimensions to be

$\boxed{G_d\equiv \dfrac{2\pi^{1-\frac{d}{2}}\Gamma \left(\frac{d}{2}\right) l_p^{d-1}c^3}{\hbar }}$


text:
# Why adding Gaussian noise increased MSE on the test set?

I have a multi-variable linear model. More precisely, a set of variables to explain a target variable. After adding Gaussian noise ($\sim N(0,1))$, the MSE increased from roughly $21$ to roughly $22$.

How can it be explained? Is that what's called overfitting?

• To what do you add the noise? To the dependent variables in the training set? Or in the test set? Or to independent variables? If so, in which set? – Stephan Kolassa Jun 23 '17 at 9:53
• I added the noise to the test set's y. I'm using the well-known data-set: cs.toronto.edu/~delve/data/boston/bostonDetail.html – Covvar Jun 23 '17 at 9:55
• the dependent variable is medv – Covvar Jun 23 '17 at 9:58


text:
# Overrandomized: Approximations with Integrals

Consider the problem Overrandomized. Intuitively, one can see something like Benford's law. Indeed, counting the leading digit works:

#include <algorithm>
#include <iostream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;

string Decode() {
unordered_map<char, int> char_counts; unordered_set<char> chars;
for (int i = 0; i < 10000; ++i) {
long long Q; string R; cin >> Q >> R;
char_counts[R[0]]++;
for (char c : R) chars.insert(c);
}
vector<pair<int, char>> count_chars;
for (const pair<char, int>& char_count : char_counts) {
count_chars.emplace_back(char_count.second, char_count.first);
}
sort(count_chars.begin(), count_chars.end());
string code;
for (const pair<int, char>& count_char : count_chars) {
code += count_char.second;
chars.erase(count_char.second);
}
code += *chars.begin();
reverse(code.begin(), code.end());
return code;
}

int main(int argc, char *argv[]) {
ios::sync_with_stdio(false); cin.tie(NULL);
int T; cin >> T;
for (int t = 1; t <= T; ++t) {
int U; cin >> U;
cout << "Case #" << t << ": " << Decode() << '\n';
}
cout << flush;
return 0;
}

Take care to read Q as a long long because it can be large.

It occurred to me that there's no reason the logarithms of the randomly generated numbers should be uniformly distributed, so I decided to look into this probability distribution closer. Let $R$ be the random variable representing the return value of a query.

\begin{align*} P(R = r) &= \sum_{m = r}^{10^U - 1} P(M = m, R = r) \\ &= \sum_{m = r}^{10^U - 1} P(R = r \mid M = m)P(M = m) \\ &= \frac{1}{10^U - 1}\sum_{m = r}^{10^U - 1} \frac{1}{m}. \end{align*} since $P(M = m) = 1/(10^U - 1)$ for all $m$.

The probability that we get a $k$ digit number that starts with a digit $d$ is then \begin{align*} P(d \times 10^{k-1} \leq R < (d + 1) \times 10^{k-1}) &= \frac{1}{10^U - 1} \sum_{r = d \times 10^{k-1}}^{(d + 1) \times 10^{k-1} - 1} \sum_{m = r}^{10^U - 1} \frac{1}{m}. \end{align*}

Here, you can already see that for a fixed $k$, smaller $d$s will have more terms, so they should occur as leading digits with higher probability. It's interesting to try to figure out how much more frequently this should happen, though. To get rid of the summation, we can use integrals! This will make the computation tractable for large $k$ and $U$. Here, I start dropping the $-1$s in the approximations.

\begin{align*} P\left(d \times 10^{k-1} \leq R < (d + 1) \times 10^{k-1}\right) &= \frac{1}{10^U - 1} \sum_{r = d \times 10^{k-1}}^{(d + 1) \times 10^{k-1} - 1} \sum_{m = r}^{10^U - 1} \frac{1}{m} \\ &\approx \frac{1}{10^U} \sum_{r = d \times 10^{k-1}}^{(d + 1) \times 10^{k-1} - 1} \left[\log 10^U - \log r \right] \\ &=\frac{10^{k - 1}}{10^{U}}\left[ U\log 10 - \frac{1}{10^{k - 1}}\sum_{r = d \times 10^{k-1}}^{(d + 1) \times 10^{k-1} - 1} \log r \right]. \end{align*}

Again, we can apply integration. Using integration by parts, we have $\int_a^b x \log x \,dx = b\log b - b - \left(a\log a - a\right)$, so \begin{align*} \sum_{r = d \times 10^{k-1}}^{(d + 1) \times 10^{k-1} - 1} \log r &\approx 10^{k-1}\left[ (k - 1)\log 10 + (d + 1) \log (d + 1) - d \log d - 1 \right]. \end{align*}

Substituting, we end up with \begin{align*} P&\left(d \times 10^{k-1} \leq R < (d + 1) \times 10^{k-1}\right) \approx \\ &\frac{1}{10^{U - k + 1}}\left[ 1 + (U - k + 1)\log 10 - \left[(d + 1) \log(d+1) - d\log d\right] \right]. \end{align*}

We can make a few observations. Numbers with lots of digits are more likely to occur since for larger $k$, the denominator is much smaller. This makes sense: there are many more large numbers than small numbers. Independent of $k$, if $d$ is larger, the quantity inside the inner brackets is larger since $x \log x$ is convex, so the probability decreases with $d$. Thus, smaller digits occur more frequently. While the formula follows the spirit of Benford's law, the formula is not quite the same.

This was the first time I had to use integrals for a competitive programming problem!


text:
Square Root Process Similar to Long Division
09-17-2021, 11:58 PM
Post: #2
 Albert Chan Senior Member Posts: 1,676 Joined: Jul 2018
RE: Square Root Process Similar to Long Division
(09-17-2021 11:00 PM)jeejohn Wrote:  I saw the thread "Third Order Convergence for Square Roots..."

I thought that I would offer up a process that allows calculations of square roots, similar to how long division is done

Nice. By the way, you can cut/paste table inside code block (like mine)

I just deleted a post about old-school square root from Third Order Convergence thread.
It really should belong here. Here goes.

For sqrt of value very close to 1, it is is even closer to 1. √(1+2ε) ≈ 1+ε
We can even *assume* result is 1, then adjust for it.

Example, √(0.9996)

Here is old school method for square root.
With nice divisor = 1, we do 3-digits at a time.
Numbers are scaled so that we work mostly with integer.

0.9996E6 = 1E3^2 - 400       // X = 1E3

Newtons correction (S-X*X)/2/X, for 2nd 3-digits.

-400E3/2 = -200E3 = -200X + 0

Negative digits OK, we will fix later. It just meant √(0.9996) ≈ 0.999800

Other digits followed below patterns:
1st column = quotient
2nd column = remainder, then corrected (shell-like cross multiply, except center)
Code:
1000 -200    0E3 - (-200)^2/2 = -20E3 = -20X + 0 -20     0E3 - (-20)*(-200) = -4E3 = -4X + 0 -4      0E3 - (-4)*(-200)-(-20)^2/2 = -1E3 = -X + 0 -1      0E3 - (-1)*(-200)-(-4)*(-20) = -280 = -X + 720 -1      720E3 - (-1)*(-200)-(-1)*(-20)-(-4)^2/2 = 719772 = 719X + 772 719     772E3 - 719*(-200)-(-1)*(-20)-(-1)*(-4) = 915776 = 915X + 776 915     776E3 - 915*(-200)-719*(-20)-(-1)*(-4)-(-1)^2/2 = 973375.5 = 973X + 375.5 973     375.5E3 - 973*(-200)-915*(-20)-719*(-4)-(-1)*(-1) = 591275 = 591X + 275 591     275E3 - ...

Normalize base-1000 digits:

√(.9996) = 0.999 799 979 995 998 999 719 915 973 591 ...

Ref: book "Dead Reckoning: Calculating without instruments", chapter on roots
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 Messages In This Thread Square Root Process Similar to Long Division - jeejohn - 09-17-2021, 11:00 PM RE: Square Root Process Similar to Long Division - Albert Chan - 09-17-2021 11:58 PM RE: Square Root Process Similar to Long Division - Albert Chan - 09-18-2021, 01:51 AM RE: Square Root Process Similar to Long Division - Albert Chan - 09-18-2021, 12:46 PM RE: Square Root Process Similar to Long Division - Albert Chan - 09-18-2021, 01:07 PM RE: Square Root Process Similar to Long Division - jeejohn - 09-18-2021, 08:55 PM

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Magyar Information Contest Journal Articles

# Problem P. 4096. (October 2008)

P. 4096. An object is 20 cm away from a concave mirror, and it is within the focal length of the mirror. If the mirror is changed to a plane mirror the image moves 15 cm closer to the mirror.

a) What is the focal length of the concave mirror?

b) Calculate the enlargement.

(3 pont)

Deadline expired on 11 November 2008.

### Statistics:

 161 students sent a solution. 3 points: 81 students. 2 points: 68 students. 1 point: 9 students. 0 point: 3 students.

 Our web pages are supported by: Morgan Stanley


text:
C. Bear and String Distance
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Limak is a little polar bear. He likes nice strings — strings of length n, consisting of lowercase English letters only.

The distance between two letters is defined as the difference between their positions in the alphabet. For example, , and .

Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example, , and .

Limak gives you a nice string s and an integer k. He challenges you to find any nice string s' that . Find any s' satisfying the given conditions, or print "-1" if it's impossible to do so.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 106).

The second line contains a string s of length n, consisting of lowercase English letters.

Output

If there is no string satisfying the given conditions then print "-1" (without the quotes).

Otherwise, print any nice string s' that .

Examples
Input
4 26bear
Output
roar
Input
2 7af
Output
db
Input
3 1000hey
Output
-1


text:
# Can one always find a bigger global resolution

Let $$X$$ be a scheme. Let $$E$$ be a perfect complex of coherent sheaves on $$X$$ and suppose it admits two global resolutions $$F$$ and $$F'$$. By global resolution I mean that both $$F$$ and $$F'$$ are quasi-isomorphic to $$E$$ and both are complexes of vector bundles.

$$\bf{Question:}$$ Is it possible to find a global resolution $$H$$ of $$E$$ such that $$F, F'$$ map mono-morphically into $$H$$?

I know how to do this when $$E$$ has perfect amplitude $$[0, 1]$$ but I don't see how to generalize to say $$[0, n]$$ for arbitrary $$n$$.

• Why do you want this? – Sasha Jan 12 at 7:21
• Do you want the mono-morphism to be also a quasi-isomorphism? – Sándor Kovács Jan 28 at 20:41
• @SándorKovács no I don't think I need the mono to be a quasi-isomorphism :). – Anette Jan 28 at 21:07

Lemma Let $$A$$ and $$B$$ be complexes of sheaves. Then there exists a complex of sheaves $$C$$ such that both $$A$$ and $$B$$ admit a monomorphism into $$C$$ and the monomorphism $$A\hookrightarrow C$$ is a quasi-isomorphism. Furthermore, if both $$A$$ and $$B$$ are complexes of locally free sheaves, then $$C$$ can be chosen to be a complex of locally free sheaves.
Proof: Let $$D$$ denote the mapping cone of the identity morphism $$B\to B$$. Note that then by the definition of the mapping cone $$B$$ admits a monomorphism into $$D$$. Further note that $$D$$ is quasi-isomorphic to the zero complex. Now let $$C=A\oplus D$$. This has the required properties. $$\square$$
OK, so now take $$A=F$$ and $$B=F'$$. Note that this did not require $$F'$$ to be also quasi-isomorphic to $$E$$, only that it consists of locally free sheaves.


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# Inelastic Collision of a football tackle

1. Mar 16, 2009

### llauren84

1. The problem statement, all variables and given/known data
In an American football game, a 90.0-kg fullback running east with a speed of 5.00 m/s is tackled by a 95.0-kg opponent running north with a speed of 3.00 m/s. (a) Explain why the successful tackle constitutes a perfectly inelastic collision. (b) Calculate the velocity of the players immediately after the tackle. (c) Determine the mechanical energy that disappears as a result of the collision. Account for the missing energy.

2. Relevant equations
m1v1=m1+m2*v2
v2= m1/(m1+m2)*v1

3. The attempt at a solution
(a) Because they are stuck together, it is a perfectly inelastic collision.

(b)Theta= 40degrees = tan-1(3/5)
m1=90.0kg
m2=95.0kg
v1=(5.00i+3.00j)m/s= $$\sqrt{5.00^2+3.00^2}$$m/s= 5.83m/s

v2= m1/(m1+m2)*v1
=(90kg/185kg)*5.83 m/s =2.84 m/s
v2x= 2.84*sin40degrees=1.82i
v2y= 2.84*cos40degrees=2.16j
So, the vector for the velocity after impact is (1.82i+2.16j)m/s

(c)w=$$\Delta$$K=(1/2)m*$$\Delta$$v2
w=(1/2)(m1+m2)(v22-v12)
w=-2.405kJ
This is the heat generated from the collision.

I really worked out the problem, but I don't know if it's correct. Please check it for me. Thanks.

2. Mar 17, 2009

### rl.bhat

(b)Theta= 40degrees = tan-1(3/5)
m1=90.0kg
m2=95.0kg
v1=(5.00i+3.00j)m/s= LaTeX Code: \\sqrt{5.00^2+3.00^2} m/s= 5.83m/s

This is not correct.
Find the initial momentum of each player and find the resultant momentum of the tackled players.

3. Mar 17, 2009

### LowlyPion

Your equations don't accurately reflect the collision. You are dealing with 2 dimensions here. Perhaps if you separate out the components and deal with them independently it will become clearer?

For instance can treat them in x as though one was stationary. And again in y as though the other is stationary, yielding your 2 velocities, and then you can use Pythagoras to determine the resultant from the x,y as you were attempting.

4. Mar 17, 2009

### llauren84

So part a and c are correct?

5. Mar 17, 2009

### LowlyPion

A is ok. Perhaps you should correct b to determine whether c is?

6. Mar 17, 2009

### llauren84

Right.

So, I googled the Inelastic Collision to get the formulas.

Last edited: Mar 17, 2009
7. Mar 17, 2009

### llauren84

Part B:

m1v1i+m2v2i=(m1+m2)vf
vf=[m1v1i+m2v2i/(m1+m2)
I did the x direction and then the y direction:
vfx=1.56m/s
vfy=1.86m/s
Used pythagorean theroum.
vf=2.43m/s
tan-1(vfy/vfx)=50.0degrees
So, it is 50 degrees from east.

For part C:

w=1/2 m1+m2 (vf2-vi2)
w=(1/2)(185kg)(2.432-5.832)m2/s2
w=-2.60kJ

Is this correct? Do I need to break the work into componants too? I am so confused.

8. Mar 17, 2009

### LowlyPion

You may be confused, but it looks OK.

9. Mar 17, 2009

### llauren84

Thanks so much.


text:
Granny's Programming Pearls "inside of every large program is a small program struggling to get out" JavaRanch.com/granny.jsp
Big Moose Saloon
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# Hi , Iam ananya new to java tech,

ananya sa
Greenhorn

Joined: Sep 30, 2009
Posts: 3
I wrote simple java pgm with just one sysout., when i try to run it,,, iam getting problem like
Usage: javac <options> <source files>
use -help for a list of possible options

I put java path in my environment variables like this:
C:\Program Files\Java\jdk1.6.0\bin;C:\Program Files\Java\jre1.6.0\bin.

thanks,

Ananaya.
Bear Bibeault
Author and ninkuma
Marshal

Joined: Jan 10, 2002
Posts: 58817

59

Henry Wong
author
Sheriff

Joined: Sep 28, 2004
Posts: 17609

33

The Java compiler is complaining that it can't find the file named -- testpgm1.java.

Henry

Muhammed Patel
Greenhorn

Joined: Feb 26, 2009
Posts: 5
Hi ananya ,

Firstly on your path does not need to have both the C:\Program Files\Java\jdk1.6.0\bin and C:\Program Files\Java\jre1.6.0\bin. the C:\Program Files\Java\jdk1.6.0\bin is fine.
Also you should set C:\Program Files\Java\jdk1.6.0 to a new environment variable called JAVA_HOME and then in your Path environment variable you can add the following to the text : %JAVA_HOME%\bin .

That will just sort out your java paths.

Now with the problem that you are having!

1. Have you changed directory to the right directory. So if testpgm1.java is in c:\javafiles have you changed directory to it.
2. Make sure that you have the correct file name or you could also make sure that the name of the file is the same as the class it contains.

Here is a simple class. try saving it as Blah.java putting it in your C:\ and then compiling and running it with the following commands

C:\>javac Blah.java

C:\>java Blah
BLAH

Hope this helps.

And good luck.

In my world they are all ponies and eat rainbows and poop butterflies
ananya sa
Greenhorn

Joined: Sep 30, 2009
Posts: 3

Thanks,

It works now, i given the wrong path while compiling. that's the problem.

Muhammed Patel wrote:Hi ananya ,

Firstly on your path does not need to have both the C:\Program Files\Java\jdk1.6.0\bin and C:\Program Files\Java\jre1.6.0\bin. the C:\Program Files\Java\jdk1.6.0\bin is fine.
Also you should set C:\Program Files\Java\jdk1.6.0 to a new environment variable called JAVA_HOME and then in your Path environment variable you can add the following to the text : %JAVA_HOME%\bin .

That will just sort out your java paths.

Now with the problem that you are having!

1. Have you changed directory to the right directory. So if testpgm1.java is in c:\javafiles have you changed directory to it.
2. Make sure that you have the correct file name or you could also make sure that the name of the file is the same as the class it contains.

Here is a simple class. try saving it as Blah.java putting it in your C:\ and then compiling and running it with the following commands

C:\>javac Blah.java

C:\>java Blah
BLAH

Hope this helps.

And good luck.
ananya sa
Greenhorn

Joined: Sep 30, 2009
Posts: 3
hi , it compiled properly, but trying to run, it is saying :
C:\Documents and Settings\sri\Desktop>java testpgm1
Exception in thread "main" java.lang.UnsupportedClassVersionError: testpgm1 (Uns
upported major.minor version 50.0)
3)
at java.security.AccessController.doPrivileged(Native Method)

As you said i just kept C:\Program Files\Java\jre1.6.0\bin; in CLASS_PATH of environmntal variables .

What needs to do now?
Henry Wong
author
Sheriff

Joined: Sep 28, 2004
Posts: 17609

33

What needs to do now?

This is caused when you compile your program with one version of Java, but then run it with an earlier version of Java.

Basically, Java is backward compatible. It is not forward compatible. So, you need to run your program with a version that is at least the same or later than your compiler.

I am guessing that you have an older version of Java already installed (but no compiler already installed). So, when you compiled, it uses the compiler you installed, but when you run it, it uses the older previously installed JVM.

Henry

I agree. Here's the link: http://aspose.com/file-tools

subject: Hi , Iam ananya new to java tech,


text:
Synopsis

# Material Size Offers Check on Constants

Physics 12, s48
Searching for variations in fundamental constants—which are predicted by some theories of dark matter—might potentially be done by monitoring the sizes of solid crystals.

Fundamental constants may betray their name and vary over time or space. Certain dark matter theories, for example, predict that the fine-structure constant $𝛼$ and the proton-to-electron mass ratio $𝜇$ oscillate in time. Now, Lukáš Pašteka of Comenius University in Slovakia and colleagues have calculated that such variations might be detected by looking for subtle changes in the sizes of crystalline materials.

The idea that fundamental constants vary is not new, and several experiments have already explored the possibility in natural phenomena that depend on the constants. Astronomers have searched for variations in the spectra of distant objects (see 31 March 2004 Focus story), and physicists have monitored the ticks in atomic clocks to high precision (see 17 November 2014 Viewpoint). No variations in $𝛼$ or $𝜇$ have been detected, with the current upper limit on the rate of change for each constant being around 1 part in $1{0}^{17}$ per year.

Pašteka and colleagues considered the way materials might respond to fundamental constant variations. Specifically, they calculated how the atomic spacings in a crystal would be affected by a change in $𝛼$ or $𝜇$. For example, a fractional increase in $𝛼$ at the current limit of $1{0}^{-17}$/yr would cause a gold bar to shrink by a factor of $1{0}^{-18}$/yr, whereas a germanium crystal would grow by a factor of $1{0}^{-21}$/yr. The team says that such size changes could potentially be detected by monitoring these materials over several years with optical cavity techniques or with a resonant mass setup similar to those used in early attempts at gravitational-wave detection.

This research is published in Physical Review Letters.

–Michael Schirber

Michael Schirber is a Corresponding Editor for Physics based in Lyon, France.

## Subject Areas

Particles and Fields

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Long-baseline neutrino experiments are paving the way for the solution of two outstanding puzzles in neutrino physics—mass ordering and charge-parity violation. Read More »


text:
# Trig

How do we show (sin 10)^2 + (sin 50)^2 + (sin 70)^2 = 3/2 ???
I can't even think of a way to start this..

1. 0
asked by Ashley
1. since sin^2θ = (1-cos2θ)/2, we have

(1-cos20)/2 + (1-cos100)/2 + (1-cos140)/2 = 3/2
cos20+cos100+cos140 = 0

Using the sum-to-product rules,

cos20 + cos100+cos140
= cos20 + 2cos120cos20
= cos20 - cos20
= 0

posted by Steve
2. Thank you very much Steve

posted by Ashley

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5. ### algebra

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# General Bayer Data

Navigation: Models ➔ Alumina Models ➔ Bayer3 Species Model ➔ General Bayer Data

## General Description

All of the Bayer Properties Species Models implemented in SysCAD are used to calculate the properties of liquors within an Alumina project.

The models use equations defined in the public domain. The model equations can be selected in the Global Method. For information on the equations used, please see Model Theory.

The Bayer property models also generally include

The public domain bayer property model included with SysCAD is:

All of the properties that are not explicitly calculated by the Bayer species model are calculated using the Standard species model, which uses the mass weighted mean of the liquor (see Standard Species Model).

Refinery specific equations for various thermodynamic properties are often used to create client specific Alumina add-ons.

## Implications of using a Bayer Properties model

### Mixing Streams with different Properties Models

When multiple streams are combined in a unit, and some streams use the Standard Species Model and others use a Bayer Species model, then SysCAD selects the higher level Bayer Properties model for the combined mixture.

### Enthalpy

Hf@0 values will be the same as if using the Standard Species Model. For species other than water and steam, this means that the Hf@25 values entered in the species database will be changed to Hf@0 values using the heat capacities (Cp) values entered in the species database (rather than the bayer liquor heat capacities).
Due to the liquid heat capacity (Cp) being different for a bayer stream, the liquid Hs, Hz and Hf@T values will be different (remember that Hf@T = Hf@0 + Hs and Hs = f(Cp)).

### Entropy

For Builds up to 136, entropy values are not calculated by the Bayer model and hence will be the same as calculated by the Standard Species Model.
In Build 137 we have implemented optional calculation of entropy based on the LM1985 specific heat.
$S(T) = S^{25} + \int_{298.15}^T Cp/T dT$

### Caution

• The $S^{25}$ value is the standard species model evaluation of entropy, based on $S^{25}$ values for the individual species in Bayer liquor. There is little or no data for these, in particular for the pseudo-organic species and others; these usually just default to zero.
• Without accurate reference data, the absolute value of entropy as calculated is suspect and should be used with caution.
• The entropy model will give correct change in entropy for a fixed composition liquor stream at different temperatures.
• If however the stream has a change in composition - for example by Gibbsite precipitation, or flashing vapor - the resulting entropy change may not be accurate without having $S^{25}$ data - use with caution.
• If modelling requires entropy change for streams of varying composition, use heat of reaction or flash enthalpy data and the Clausius relation $dS = dQ/T$

### Heats of Reaction

The Bayer Properties model does not include calculations for individual species Cp's and Heats of Formation, it only has Cp correlations for a Bayer liquor. A reaction (Reaction Block) works with individual species and therefore relies on Hf and Cp data entered in the species database to calculate a heat of reaction. However Hf and Cp data are not required if the user specifies a Heat of Reaction (HoR).

When a user specifies a reaction in a unit, the heat of reaction is calculated from the difference in the sum of the heats of formation of the products and reactants, as shown below:

A + B -> C + D

Then the HoR = Heats of formation (Hf) for C + D - (A + B) (See Reaction Block Heat of Reaction)

• Now if Hf is known for A, B and C but not D and we enter real data for A, B, C and just put 0 for D we may get a HOR that is grossly incorrect. Heats of formation are generally an order of magnitude or so larger than heats of reaction (heat of reaction is just the difference between some large numbers) so that error in estimating Hf will result in large errors in HOR.
• One recommended approach is not to enter Hf values (ie leave these blank) in the species database for all the relevant Bayer Liquor species. Then in the reaction (RCT) file, a user defined HOR value is required for each reaction that includes any Bayer Liquor species. By leaving Hf blank in the species database, the SysCAD reaction block will know a user HOR override is required, so if the user forgets to specify a HOR the user is given a warning.
• Now in a set of reactions, the usual approach is to then enter a HOR of 0 (at 0°C) for almost all reactions. The exception is for dissolution and crystallization of Gibbsite and Boehmite which are very large sinks or sources for energy (they are the major ones for the process), so for these do not enter 0 for HOR. Many refineries have carried out extensive test work to determine the actual heats of reaction for these reactions occurring on their plants and these values are used in the RCT file. The HOR may be specified on a molar or mass basis (see Heat of Reaction).

The enthalpy of the product stream is calculated as follows:

   Enthalpy of Product Stream at Product Temperature = (Enthalpy of Feed Stream at Feed Temperature) - (Heat of Reaction at 0°C)

The HOR is typically specified at 0°C. When SysCAD displays a HOR at a different temperature, it calculates this value using the Cp data for the species involved. Often the Cp data is not known, which is why we recommend that the HOR is specified at 0°C and why the values shown for HOR at other temperatures are inaccurate. If you enter the HOR at a temperature other than 0°C then you do need to enter the Cp data for the species involved in the reaction, as SysCAD will use the Cp's to correct for HOR at varying temperatures.

Notes

1. Where no thermodynamic data (Hf) exists in the SysCAD species database for species involved in a reaction, the user MUST specify the Heat of Reaction.
2. For reactions like kaolin going into solution and re-precipitating as DSP, users often leave the HOR as zero. This will be slightly inaccurate, but it is a lot more accurate than deriving it from incorrect Hf's.
3. The HOR should be entered at a reference temperature of 0°C if the correct Cp correlations for all of the species involved in a reaction are not known.

Using the above method gives energy balances around digestors that are within +/- 5% of measured energy input. Obviously having more accurate data for the full set of species would be preferable.

Example:

Al[OH]3(s) + NaOH(aq) = NaAl[OH]4(aq)
Extent: Fraction Al[OH]3(s) = 1.0
HeatOfReaction: msFixed = 252.3 / Al[OH]3(s) At 0

In this example the reaction is endothermic with an energy requirement of 252.3 kJ/kg of Al[OH]3(s). The user may also specify the heat of reaction on a molar basis

### Volume Display of Individual Species

The volume flows or fractions on an individual basis as displayed in the access window give "confusing" values for Bayer liquids, see the volume fraction example below:

There is a simple explanation - the error is due to the fact that in a Bayer properties model, there is a defined density equation for a mixture but not for individual elements.

• In the access window shown above the individual species volume flows (or fractions) are calculated from the density of the species as defined in the database.
• If density correction information is present in the species database, the fields will show "*".
• These densities have NO correlation with the density of the Bayer mixture. Therefore when looking at numbers in isolation, the numbers are not correct.
• The various totals on the same page may also therefore "incorrect" as they are simply sums of the individual species.

Therefore, to make sure the right numbers are used, only use the volumes calculated on the Bayer/Props page [eg P_xx.Qo.LVolFlow25 (m^3/h)] or preceding Qo pages [eg P_xx.Qo.LQv (m^3/h)].

### Bayer Properties Display

The Bayer Properties are displayed on a separate Tab "Props", the example below shows some of the properties calculated.