An air bubble of volume 20 cm3 is at the bottom of a lake 40 m deep, where the temperature is 4.00C. The bubble rises to the surface, which is at a temperature of 200C. Take the temperature of the bubble’s air to be the same as that of the surrounding water. Just as the bubble reaches the surface, what is its volume?

The circuit courts in Illinois have original jurisdiction as trial courts. There are 24 judicial circuits in the state, and each one includes one or more of the 102 counties that make up Illinois.

All cases filed in the State of Illinois other than those for which the Supreme Court has original jurisdiction are first heard by the Circuit Court, which is the trial court with wide jurisdiction.

A Supreme Court, an Appellate Court, and Circuit Courts are in charge of the legal system. (Source: Constitution of Illinois.) JUDICIAL DISTRICTS SECTION 2 For the purpose of choosing judges for the Supreme and Appellate Courts, the State is divided into five judicial districts.

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a) The speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

b) We cannot calculate the work done by the friction force.

c) The net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

(a) The work done on the 6.00 kg block by gravity can be calculated using the formula:

Work_gravity = force_gravity * displacement * cos(theta),

where force_gravity is the weight of the block, displacement is the downward displacement of the block, and theta is the angle between the force and displacement vectors (which is 0 degrees in this case).

The weight of the block is given by:

force_gravity = mass * acceleration_due_to_gravity = 6.00 kg * 9.8 m/s^2 = 58.8 N.

Plugging in the values, we get:

Work_gravity = 58.8 N * 0.800 m * cos(0) = 47.04 J.

The work done on the 6.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(180) = -29.6 J.

The negative sign indicates that the tension is in the opposite direction of the displacement.

Using the work-energy theorem, we can find the speed of the 6.00 kg block after descending 0.800 m:

Work_net = change_in_kinetic_energy.

Since the block starts from rest, its initial kinetic energy is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * mass * velocity^2.

Solving for velocity, we get:

velocity = sqrt(2 * Work_net / mass).

The net work done on the block is the sum of the work done by gravity and the tension:

Work_net = Work_gravity + Work_tension = 47.04 J - 29.6 J = 17.44 J.

Plugging in the values, we get:

velocity = sqrt(2 * 17.44 J / 6.00 kg) = 2.07 m/s.

Therefore, the speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

(b) The total work done on the 8.00 kg block during the 0.800 m displacement can be calculated using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the 8.00 kg block is not moving vertically, its initial and final kinetic energies are zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 0.

The work done on the 8.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(0) = 29.6 J.

The work done on the 8.00 kg block by the friction force can be calculated using the formula:

Work_friction = force_friction * displacement * cos(theta),

where force_friction is the frictional force on the block. However, the problem statement does not provide the value of the friction force. Therefore, we cannot calculate the work done by the friction force.

(c) The net work done on the system of two blocks during the 0.800 m displacement of the 6.00 kg block can be found using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the system starts from rest, the initial kinetic energy of the system is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * (6.00 kg + 8.00 kg) * velocity^2.

Simplifying, we get:

Work_net = 1/2 * 14.00 kg * velocity^2.

Using the value of velocity calculated in part (a), we get:

Work_net = 1/2 * 14.00 kg * (2.07 m/s)^2 = 29.13 J.

The work done on the system of two blocks by gravity is the sum of the work done on the individual blocks by gravity:

Work_gravity_system = Work_gravity_6kg + Work_gravity_8kg = 47.04 J + 0 J = 47.04 J.

The work done on the system of two blocks by the tension in the rope is the sum of the work done on the individual blocks by the tension:

Work_tension_system = Work_tension_6kg + Work_tension_8kg = -29.6 J + 29.6 J = 0 J.

Therefore, the net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

Note: The calculations for part (b) and (c) were based on the given information, but the value of the friction force was not provided in the problem statement.

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When people are interested in changing their personalities, they are usually most interested in more neurotic less assertive.

Personality makes up an individual, they are the features or characteristics that forms an individual.

Each individual is made of distinctive character which makes one of different from other.

Therefore, when people are interested in changing their personalities, they are usually most interested in more neurotic less assertive

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The skater's final angular velocity is approximately 9.86 rad/s.

The skater's final angular velocity can be calculated using the principle of conservation of angular momentum. The equation for angular momentum is given by:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Initially, the skater has an angular momentum of:

L_initial = I_initial * ω_initial

Substituting the given values:

L_initial = 2.12 kg m² * 3.25 rad/s

The skater's final angular momentum remains the same, as angular momentum is conserved:

L_final = L_initial

The final moment of inertia is given as 0.699 kg m². Therefore, the final angular velocity can be calculated as:

L_final = I_final * ω_final

0.699 kg m² * ω_final = 2.12 kg m² * 3.25 rad/s

Solving for ω_final:

ω_final = (2.12 kg m² * 3.25 rad/s) / 0.699 kg m²

Hence, the skater's final angular velocity is approximately 9.86 rad/s.

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Aster's final position from her initial position is 63 m approximately. She will head north west direction to return to her initial position

Displacement is the distance travelled in a specific direction. It is a vector quantity.

Given that a person walks first 70 m in the direction 37° north of east, and then walks 82 m in the direction 20° south of east, and finally walks 28 m in the direction 30° west of north.

a) Let P be the Aster's final position from her initial position?

We can use bearing by using Cosine formula to solve this question.

P² = 70² + 82² - 2 × 70 × 82 cos 73

P² = 4900 + 6724 - 11480 cos 73

P² = 11624 - 3356.43

P² = 8267.57

P = √8267.57

P = 90.9 m

P = 90.9 - 28

P = 62.9 m

We can get the angle by using Sine rule

82/ sin Ф = 90.9 / sin 73

sin Ф = 0.8627

Ф = \(Sin^{-1}\) (0.8627)

Ф = 59.6°

Ф = 60°

b)  She will head north west direction to return to her initial position

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Given,

The spring constant of the spring, k=27.4 N/m

The stretch in the spring, x=2.1 m

The magnitude of the restoring force of the spring is given by,

Where F is the force required to stretch the given spring 2.1 m

On substituting the known values,

The force required to stretch the spring is 57.54 N

Answer:

t = 5 s

Explanation:

Data:

Use formula:

Replace:

Solve the subtraction of the numerator:

It divides:

How much time did it take the car to reach this final velocity?

It took a time of 5 seconds.

Answer:

\(\phi=4344.72Nm^2/c\)

Explanation:

From the question we are told that:

Dimension of Wall:

 \((L*B)=(8.7 m * 3.2 m)\)

Electric field \(B=210 N/C\)

Angle \(\theta =42 \textdegree North\)

Generally the equation for electric Flux is mathematically given by

 \(\phi=EAcos\theta\)

 \(\phi=210*(8.7*3.2)*cos 42\)

 \(\phi=4344.72Nm^2/c\)

Explanation:

you move 10 -³ up

so: 100* 10³ /5.5*10

100000/55

which is approximately 18182

P equals 625000 watts The engine provides power during this period, enabling anything to be done or accomplished. political or governmental power

influencing oneself, others, and circumstances This kind of influence comes from personal qualities rather than institutional authority. Personal power is more of a mindset or attitude. Self-efficacy and collaboration skills are important to those with high personal power.

Real power is energy, and as our awareness and self-awareness expand, it grows from within. Being powerful depends on having insight. A person with true authority does not affect the world without taking the larger picture into account.

Work done by the steam engine, W=4500*10³J

Time taken to do work, t = 7.2 seconds

                P=W/t

                  =4500*10³/7.2

                  =625000watt

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Answer:

m = 80[kg]

Explanation:

The kinetic energy can be calculated by means of the following equation.

\(E_{k}=\frac{1}{2}*m*v^{2}\)

where:

m = mass [kg]

v = velocity = 5 [m/s]

Ek = kinetic energy = 1000 [J]

Now replacing:

\(1000 = \frac{1}{2} *m*5^{2}\\2000 = 25*m\\m=80[kg]\)

The mass of a ball will be "80 kg".

According to the question,

Kinetic energy, \(E_k\) = 1000 Joules

Velocity, V = 5 m/s

As we know the formula,

→ Kinetic energy = \(\frac{1}{2}\) × mass × (velocity)²

→      \(E_k = \frac{1}{2}\) × m × v²

By substituting the values,

     \(1000= \frac{1}{2}\times m\times (5)^2\)

\(1000\times 2=25\times m\)      

     \(2000=25\times m\)

         \(m = \frac{2000}{25}\)

             \(= 80\) kg

Thus the above answer is correct.

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1. b. electrical potential energy.

2. a. True. For electric potential energy, a reference position must be defined.

3. a. -4 J. The potential energy between two charges is given by the equation U = k(q1q2)/r, where k is the electrostatic constant, q1 and q2 are the charges, and r is the separation between them. Since the potential energy is given as 2 J initially, and q2 is replaced by q3 (which is negative and twice the magnitude of q2), the potential energy becomes -4 J.

4. c. decreases. The potential energy between two charges decreases as they are moved closer together.

5. b. -3.24 J. The electric potential energy between two point charges is given by the equation U = k(q1q2)/r, where k is the electrostatic constant, q1 and q2 are the charges, and r is the separation between them. Substituting the values into the equation, we get U = (9x10^(-6) C)(-4x10^(-6) C)/(0.1 m) = -3.24 J.

6. a. B or D. Electric potential energy can never be negative if the charges repel each other.

7. c. measures energy per unit charge.

8. 12 J. The potential energy is given by the equation U = qV, where U is the potential energy, q is the charge, and V is the potential. Substituting the values into the equation, we get U = (6 C)(2 V) = 12 J.

9. b. 10 V. The electric potential is given by the equation V = U/q, where V is the potential, U is the potential energy, and q is the charge. Substituting the values into the equation, we get V = 50 J/5 C = 10 V.

10. c. 14 V. The electric potential is inversely proportional to the distance from a point charge. When the distance is doubled, the potential is halved. Therefore, the electric potential at the new position of q2 is 28 V/2 = 14 V.

11. c. increases. The potential at the position of q2 due to q1 increases when a negative test charge is substituted with twice the magnitude of the positive test charge.

12. By convention, the direction of a current is taken to be the direction of motion of negative charges. In reality, it is actually electrons that move in wires.

13. c. 0.25 A. The current is defined as the rate of flow of charge. Given that 15 C flows through the wire each minute, the current is 15 C/60 s = 0.25 A.

14. b. 1.2x10^6 C. The charge flowing through the wire is given by the equation Q = It, where Q is the charge, I is the current, and t is the time. Substituting the values into the equation, we get Q = (3x10^(-3) A)(1 hour)(3600 s/hour) = 1.2x10^6 C.

15. d. current. An ampere is a unit of electric current.

16. electrons. As current flows through a wire, it is actually electrons that move. The flow of electrons constitutes the electric current.

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Answer:

\(a)-3346.8\;N \\ b)-4937.04\;N-m\)

Explanation:

a) - In Free-body diagram :

At point D, the free body diagram of a man :

\(D = Fn\\Fn=T+mg\\put\;values\;in\;it\\ .\;\;\;\;=300+(80)(9.81)=1084.8\;N\)

\(Mg=200\times9.81=1962\;N\)

\(\sum Fx=0\; where\; Ax=0\)

\(\sum Fy=0\; where\; Ay-Mg-Fn-T=0\)

Then, put the value in the equation.

\(Ay=3346.8\;N\)

b)-

\(Ma=Mg(AE)+Fn(AD)+T=4937.04\;N-m\)

Based on the given figure, you have that the components of the forces on the direction of the incline are:

applied force:

Fa = 500N*cos(-30°) = 433.01N

friction force:

Fr = µk*N = µk*m*g*cos(30°) = (0.42)(40kg)(9.8 m/s^2)cos(30°) = 142.58N

gravitational force:

Fg = m*g*sin(30°) = (40kg)(9.8)sin(30°) = 196.0N

Next, consider that the work done by a force is equal to the product of the force and the distance at which the force is applied. In this case, such distance is 8m.

Then, you have:

Work done by the applied force:

WFa = (433.01N)(8.0m) = 3464.08 J

Work done by the friction force:

WFr = (142.58N)(8.0m) = 1140.64 J

Work done by the gravitational force:

WFg = (196.0N)(8.0m) = 1568.0 J

Finally, the net work, based on the direction of the forces, is:

Net work = WFa - WFr - WFg = 3464.08J - 1140.64J - 1568.0J = 755.44J

Hence, the net work is 755.44 J

Answer: Wfr = -1883.25 J

Explanation:

1) In the x direction (direction of motion of the crate):

Fcos(30^o) - f_fr - mgsin(30^o ) = 0 (because a_x = 0)

f_fr = Fcos(30^o ) - mgsin(30^o)

f_fr = (498)(√3/2) - (40)(9.81) (1/2) = 235.28 N

⇒ W_fr = f_fr.s.cos(180^o) = -(235.28)(8.0) = -1882.25 J (1)

2) In the y direction (perpendicular to the motion of the crate):

n - Fsin(30^o) - mgcos(30^o) = 0 (because a_y = 0)

⇒ n = Fsin(30^o) + mgcos(30^o) = (498)(1/2) + (40)(9.81)(√3/2) = 588.83 N

⇒ f_fr = μ_k.n = (0.40)(588.83) = 235.53 N

⇒ W_fr = f_fr.s.cos(180^o ) = -(235.53)(8.0) = -1884.24 J (2)

Average of calculations (1) & (2): W_fr = -1883.25 J

The centripetal force affecting the body has to be doubled.

1. The centripetal force acting on a body moving in a circular path of radius r with speed v is given by F = mv²/r, where m is the mass of the body.

2. If the speed of the body increases to √2v while moving in the same circular path, the new centripetal force acting on the body can be calculated as follows:

  F' = m(√2v)²/r = 2mv²/r

3. Comparing the new centripetal force F' with the initial centripetal force F, we get:

  F' = 2F

4. As a result, the centripetal force acting on the body must be twice in order for the body to proceed in the same circular direction at 2v.

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The repulsive force is 2.4435 X 10 ^ -4 N between the two pith balls with charge equale to -30.5 nC and distance of 18.5 cm.

What is Coulomb's Law?

According to this law, charges repel one another and attract one another with a force that is inversely proportional to the square of the distance between them and proportional to the product of the charges.

According to Coulomb's law, the force of attraction or repulsion between two charged bodies should be directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The image shows the calculation of force between the two pith balls using Coulomb's law.

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The force between the two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Hence, if distance between charges is halved (charges remaining kept constant), the force between the two charges is quadrupled.

Answer:

Walking –We can walk only if we apply frictional force. Friction is what holds your shoe to the ground. The friction present on the ice is very little, this is the reason why it is hard to walk on the slippery surface of the ice.

Writing – A frictional force is created when the tip of the pen comes in contact with the surface of the paper. Rolling friction is what comes into play while writing with a ballpoint pen while sliding friction arises when one writes with a pencil.

Skating – A thin film of water under the blade is necessary to make the skate slide. The heat generated by the skate blade rubbing against the surface of ice causes some of the ice to melt right below the blade where the skater glides over the ice. This water acts as a lubricant reducing friction.

Lighting a matchstick – When the head of the matchstick is rubbed against a rough surface, heat is generated and this heat converts red phosphorous to white phosphorous. White phosphorous is highly inflammable and the match stick ignites. Sometimes, matchsticks fail to ignite due to the presence of water. Water lowers friction.

Driving of the vehicle on a surface- While driving a vehicle, the engine generates a force on the driving wheels. This force initiates the vehicle to move forwards. Friction is the force that opposes the tyre rubber from sliding on the road surface. This friction avoids skidding of vehicles.

Applications of breaks in the vehicle to stop it- Friction braking is the most widely used braking method in vehicles. This process involves the conversion of kinetic energy to thermal energy by applying friction to the moving parts of a vehicle. The friction force resists the motion and in turn, generates heat. This conversion of energy eventually bringing the velocity to zero.

Flight of aeroplanes- Drag is the force that opposes the forward motion of the aeroplane. The friction which resists the motion of an object moving through a fluid or immobile in a moving fluid, as occurs when we fly a kite. The friction of the air is created as it meets and passes over an aeroplane and its components. Drag is generated by air impact force, skin friction and displacement of the air.

Drilling a nail into the wall- Friction is responsible for fixing of nails in a wall. As the nail is driven into the wall, the nearby material to the nail of the wall gets compressed. This exerts a force on the nail. This force is the friction that converts the normal force exerted by the compressed layers of the wall into the resisting shear force. In this manner the friction cause nails and screws to hold on to walls.

The dusting of the carpet by beating it with a stick- When the carpet is beaten with the stick, the dust comes out. The dust is carried off by the wind or falls on the floor. The carpet exhibits a little static friction that holds the dust to the carpet.  When the carpet is beaten, it will overcome the friction and the carpet will move away from the dust making the carpet free from dust.

Sliding on a garden slide- We know that friction is a force that is present whenever two objects rub against each other. In case of a slide in the garden such as a slide and a person’s backside rub each other’s surface. Without friction, a slide would accelerate the rider too quickly, resulting in possible injury due to the fall. The friction reduces the velocity of the sliding person and makes him stop.

Hope that helps! :D Sorry if it's too lengthy...

Explanation:

Answer:

If 2N = 30mm

Then ? =35mm

2N×(35mm÷30mm)

2N×1.167

= 2.33N

To get the additional force required, subtract 2N from 2.33N

2.33N - 2N

= 0.33N

∴ The additional force needed is 0.33N

Explanation:

With this question, we establish the principles of proportion. That is, if more less divides and vice versa. If 2N is stretching the material to 30mm, then how  more force will be needed to stretch the material to 35mm?

Answer: I think its zero-

I dont really know about this...

Answer:

Breaking an atom refers to a process called nuclear fission, which involves splitting the nucleus of an atom into smaller nuclei. This is typically accomplished by bombarding the atom with a neutron, which causes the nucleus to become unstable and split apart, releasing a large amount of energy in the process. This energy is what is harnessed in nuclear power plants to generate electricity. However, it should be noted that nuclear fission can also have potentially harmful effects, such as the release of radioactive material and the potential for nuclear accidents.

A rope pulls a Tesla out of mud. The guy pulls a force F⊥ of 300N, and theta = 4.2°. The tension force T is 298.44__ Newton.

The problem describes a Tesla that is stuck in the mud and needs to be pulled out using a rope. the guy pulls a force F⊥ of 300N and that the angle between the rope and the horizontal plane is θ = 4.2°. The goal is to find the tension force T exerted by the rope.To solve for T, we'll need to use trigonometry. We can break the force vector into its horizontal and vertical components as follows:

Fx = F⊥ cosθ and Fy = F⊥ sinθ.

Since the rope is pulling the Tesla horizontally, the horizontal component of the force will be the tension force T. So we have:

T = Fx = F⊥ cosθ = (300 N) cos(4.2°) ≈ 298.44 N

Taking the cosine of the angle is necessary since it's the adjacent side that we're interested in, which is the horizontal component of the force. Therefore, the tension force exerted by the rope is approximately 298.44 N.

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Since no work is done, the power exerted is zero. Therefore, the man exerts no power on the wall.

In physics, force is defined as any action that can change the motion of an object or cause an object to accelerate. Force is a vector quantity, meaning that it has both magnitude (size or strength) and direction. The unit of force in the International System of Units (SI) is the Newton (N), which is defined as the amount of force required to accelerate a mass of one kilogram at a rate of one meter per second squared (1 N = 1 kg × 1 m/s^2). Force can be measured using a variety of instruments, such as spring scales, strain gauges, or force plates. Some common types of forces include gravitational force, electromagnetic force, frictional force, and normal force. The study of forces and their effects on the motion of objects is known as mechanics and is a fundamental concept in physics.

Here,

a. The man does not do any work on the wall because the wall does not move. Work is only done when there is a displacement in the direction of the force applied.

b. Since no work is done, no energy is used or transferred.

c. The power exerted by the man can be calculated using the formula:

Power = Work / Time

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Answer:

The weight of the planet is 29083.5 N .

Explanation:

mass of satellite, m = 6463 kg

height of orbit, h = 4.82 x 10^5 m

period, T = 2 h

radius of planet, R = 4.29 x 10^6 m

Let the acceleration due to gravity at the planet is g.

\(T = 2\pi\sqrt\frac{(R+h)^3}{gR^2}\\\\2\times 3600 = 2\times3.14\sqrt\frac{(4.29+0.482)^3\times10^{18}}{g\times 4.29\times 4.29\times 10^{12} }\\\\24.2 g =108.67\\\\g = 4.5 m/s^2\)

The weight of the satellite at the surface of the planet is

W = m g = 6463 x 4.5 = 29083.5 N

Answer:

a = ω^2 A      formula for max acceleration (ignoring sign)

V = ω A         formula for max velocity

V^2 = ω^2 A^2 = a A   from first equation

E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J

(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule

Answer:

Below

Explanation:

0-1 sec descends at constant rate from 10 to 6 m

1-2 sec stops at 6m

2-3 sec descends at constant rate to 2 m

3-4 sec stops at 2 m

4-5 sec descends at another constant rate to 0 m

The total distance covered by the car is 300 miles.

The total displacement covered by the car is zero.

The average speed of the car is 17.88 m/s.

The average velocity of the car is also zero.

Distance between the points A and B, d = 150 miles

Time taken by the car to travel from A to B, t₁ = 3 hours

Time taken by the car to travel from B to A, t₂ = 5 hours

a) Given that the car travelled from A to B and then back to A.

Therefore, the total distance covered by the car is,

Distance = 2 x d

Distance = 2 x 150

Distance = 300 miles

Since the car is travelling from A to B and then returning back to the initial point A, the total displacement covered by the car is zero.

b) The speed with which the car travelled from A to B is,

v₁ = d/t₁

v₁ = 150/3

v₁ = 50 miles/hr

v₁ = 22.35 m/s

The speed with which the car travelled from B to A is,

v₂ = d/t₂

v₂ = 150/5

v₂ = 30 miles/hr

v₂ = 13.41 m/s

Therefore, the average speed of the car is,

v = (v₁ + v₂)/2

v = (22.35 + 13.41)/2

v = 17.88 m/s

As, the total displacement of the car is zero, the average velocity of the car is also zero.

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Answer:

F₁ = 4 F₀

Explanation:

The force applied on the string by the ball attached to it, while in circular motion will be equal to the centripetal force. Therefore, at time t₀, the force on ball F₀ is given as:

F₀ = mv₀²/r   --------------- equation (1)

where,

F₀ = Force on string at t₀

m = mass of ball

v₀ = speed of ball at t₀

r = radius of circular path

Now, at time t₁:

v₁ = 2v₀

F₁ = mv₁²/r

F₁ = m(2v₀)²/r

F₁ = 4 mv₀²/r

using equation (1):

F₁ = 4 F₀

Two microscopic bags each contain two protons. When separated by distance d, electric force on one due to the other is F. When you transfer a proton from one to another without changing anything else. The electric force now is ¾F.

Electric force is described as the repulsive or attractive interaction between any two charged bodies.

The electric force is one of the various forces that act on objects and has its impact and effects on the given body which is described by Newton’s laws of motion.

The electric force between two electrons is equal to the electric force between two protons when placed at equal distances and this describes that the electric force is not based on the mass of the object, but depends on the quantity known as the electric charge.

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