A 5.6 cm diameter parallel-plate capacitor has a 0.58 mm gap. What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000 V/s?

A heat engine that uses work to move heat

hope this helps :)

Answer:

D. A heat engine that uses heat to do work

Explanation:

Combustion = Heat

Combustion is used to create motion (work)

Answer:

= 3.869 × 10^-6 J/m³

Explanation:

Intensity is given as W/m^2 which is equivalent to J/ (s*m^2)

Speed of light is 2.99792 × 10^8 m/s.

Therefore;

Electromagnetic energy per cubic meter = Intensity/speed of light

      = 1160 J/ (s*m³)/ 2.99792 × 10^8 m/s.

      = 3.869 × 10^-6 J/m³

Answer:

255.4 N/m

Explanation:

We can consider the system eyeball-attached to the musculature as a mass-spring system in simple harmonic motion, whose frequency of oscillation is given by

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]

where in this case, we know:

f = 29 Hz is the frequency of oscillation

k is the spring constant, which is unknown

m = 7.7 g = 0.0077 kg is the mass of the eyeball

Solving the equation for k, we find the spring constant of the musculature attached to the eyeball:

[tex]k=(2\pi f)^2 m=(2 \pi (29 Hz))^2 (0.0077 kg)=255.4 N/m[/tex]

To find the effective spring constant of the musculature attached to the eyeball, use the formula k = (4π²m)/(frequency²)

To determine the effective spring constant of the musculature attached to the eyeball, we can use the formula for the resonant frequency of a mass-spring system:

frequency = 1 / (2π√(m/k))

where m is the mass of the eyeball and k is the spring constant. Rearranging the formula, we can solve for k:

k = (4π2m)/(frequency2)

Plugging in the values, we have m = 7.7 g and frequency = 29 Hz:

k = (4π2(7.7 g))/(29 Hz)2

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Answer:

612.2 m

Explanation:

The work needed to lift the object is equal to its increase in gravitational potential energy:

[tex]W=\Delta U=mg \Delta h[/tex] (1)

where

m = 1.0 kg is the mass

g = 9.8 m/s^2

[tex]\Delta h[/tex] is the vertical distance

The power provided is

[tex]P=0.10 kW = 100 W[/tex]

In one minute (t = 1 min = 60 s), the work provided is

[tex]W=Pt=(100 W)(60 s)=6000 J[/tex]

Substituting this into (1) and solving for [tex]\Delta h[/tex], we find

[tex]\Delta h=\frac{W}{mg}=\frac{6000 J}{(1.0 kg)(9.8 m/s^2)}=612.2 m[/tex]

The amount of energy needed to power a 0.10-kW bulb for one minute is equivalent to the work done in lifting a 1.0 kg object through a vertical distance. The distance is calculated using the formula Work = Force * Distance. The distance is found to be 0.61 meters.

The amount of energy needed to power a 0.10-kW bulb for one minute is 6 Joules. This energy is equivalent to the work done in lifting a 1.0 kg object through a vertical distance. To calculate the distance, we can use the formula:

Work = Force * Distance

In this case, the force is equal to the weight of the object, which is 1.0 kg * 9.8 m/s² (acceleration due to gravity). So, the distance is equal to:

Distance = Work / (Force * Acceleration due to gravity)

Substituting the values, we get:

Distance = 6 J / (1.0 kg * 9.8 m/s²) = 0.61 meters

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Answer:

10 m/s

Explanation:

The speed of a wave is given by

[tex]v=\lambda f[/tex]

where

[tex]\lambda[/tex] is the wavelength

f is the frequency

For the wave in this problem,

- The frequency is given by the number of oscillations per second, so

[tex]f=\frac{2 cycles}{1 s}=2 Hz[/tex]

- The wavelength is

[tex]\lambda=5 m[/tex]

So, the wave speed is

[tex]v=(5 m)(2 Hz)=10 m/s[/tex]

The speed of the water wave is calculated using the formula speed = frequency × wavelength. With a frequency of 2 Hz and a wavelength of 5 meters, the wave speed is 10 m/s.

The student's question is about calculating the speed of a water wave based on the given wavelength and the frequency of an object oscillating on the water's surface. To find the wave speed, we can use the formula speed = frequency × wavelength. Given that the object completes 2 cycles per second (which is the frequency), and the wavelength is 5 meters, we simply multiply the two values.

Speed = Frequency × Wavelength = 2 Hz × 5 m = 10 m/s

Therefore, the speed of the water wave is 10 meters per second (m/s).

(a) [tex]4.26\cdot 10^5 V/m[/tex]

In a velocity selector, the speed of the beam is related to the magnitude of the electric field and of the magnetic field by the formula:

[tex]v=\frac{E}{B}[/tex]

where

E is the magnitude of the electric field

B is the magnitude of the magnetic field

In this problem, we have

[tex]B=0.115 T[/tex] (magnetic field)

[tex]v=3.7\cdot 10^6 m/s[/tex] (speed of the particles)

Solving the equation for E, we find the electric field:

[tex]E=vB=(3.7\cdot 10^6 m/s)(0.115 T)=4.26\cdot 10^5 V/m[/tex]

(b) 3.2 kV

The relationship between electric field and potential difference between the two plates is:

[tex]V=Ed[/tex]

where, in this problem:

[tex]E=4.26\cdot 10^5 V/m[/tex] is the magnitude of the electric field

[tex]d=0.75 cm=0.0075 m[/tex] is the separation between the plates

Substituting into the equation, we find the potential difference:

[tex]V=(4.26\cdot 10^5 V/m)(0.0075 m)=3195 V=3.2 kV[/tex]

Answer:

[tex]R_3 < R_1 < R_2[/tex]

Explanation:

The resistance of a wire is given by:

[tex]R=\frac{\rho L}{A}[/tex]

where

[tex]\rho[/tex] is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

1) The first wire has length L and cross-sectional area A. So, its resistance is:

[tex]R_1=\frac{\rho L}{A}[/tex]

2) The second wire has length twice the first one: 2L, and same thickness, A. So its resistance is

[tex]R_2=\frac{2\rho L}{A}[/tex]

3) The third wire has length L (as the first one), but twice cross sectional area, 2A. So, its resistance is

[tex]R_3=\frac{\rho L}{2A}[/tex]

By comparing the three expressions, we find

[tex]R_3 < R_1 < R_2[/tex]

So, this is the ranking of the wire from most current (least resistance) to least current (most resistance).

Answer:

E) its speed and wavelength decrease, but its frequency stays the same

Explanation:

First of all, the frequency of a light wave does not depend on the medium, while wavelength and speed do. Therefore, the frequency remains costant.

In particular, the speed of light in a medium is given by:

[tex]v=\frac{c}{n}[/tex]

where c is the speed of light in a vacuum and n is the index of refraction. From the formula, we see that v and n are inversely proportional: so, when the light moves into a material with higher index of refraction, its speed decreases.

Moreover, speed is related to wavelength by

[tex]v=\lambda f[/tex]

where [tex]\lambda[/tex] is the wavelength and f is the frequency. Since the two quantities are directly proportional, this means that since the speed decreases, the wavelength decreases as well.

So, the correct choice is

E) its speed and wavelength decrease, but its frequency stays the same

When light goes from one material into another material having a HIGHER index of refraction , E) its speed and wavelength decrease, but its frequency stays the same.

[tex]\texttt{ }[/tex]

Let's recall Snell's Law about Refraction as follows:

[tex]\boxed{n_1 \sin \theta_1 = n_2 \sin \theta_2}[/tex]

where:

n₁ = incident index

θ₁ = incident angle

n₂ = refracted index

θ₂ = refracted angle

[tex]\texttt{ }[/tex]

In this problem, we will use this following formula:

[tex]\boxed{ \frac{n_1}{n_2} = \frac{v_2}{v_1} }[/tex]

where:

n₁ = incident index

v₁ = incident speed

n₂ = refracted index

v₂ = refracted speed

[tex]\texttt{ }[/tex]

From above formula we could conclude that the speed of light is inversely proportional to the index of refraction. Therefore, when light goes from one material into another material having a HIGHER index of refraction , its speed will decrease.

[tex]\texttt{ }[/tex]

As we know that :

[tex]\boxed{v = \lambda f}[/tex]

Speed of light is directly proportional to the wavelength of light.

If the speed of light decreases , then wavelength of light will also decrease.

[tex]\texttt{ }[/tex]

When light goes from one material into another material having a HIGHER index of refraction , E) its speed and wavelength decrease, but its frequency stays the same

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Light

b is the correct answer

Answer:

c- interact with charged particles.

Explanation:

When a charged particle is placed in the field of the other charged particle then it will experience electromagnetic force on it.

As we know that force on a charged particle due to some other electric field is given by

F = qE

here we know

q = charge

E = electric field intensity

Similarly when a charged particle is placed in external magnetic field then the force on that moving charge is given by

[tex]F = q(v\times B)[/tex]

so here if charge is moving in the magnetic field due to some other system then the force on it is given by above equation.

So here we can say that electromagnetic force is present when electromagnetic fields

c- interact with charged particles.

ok interesting what is you question

The answer is B

Bimetallic strip is used to create a bimetallic coil for a thermometer which reacts to the heat from a lighter, by uncoiling and then coiling back up when the lighter is removed.

A Bimetallic Coil is formed from two pieces of different metals stuck together lengthwise.

Option A

Explanation:

It is formed with two metal pieces that are stuck together in proper length. It is also called bimetallic strip which is mainly used for converting temperature into mechanical displacement. This coil or strip is consist of two different metal that are "steel and copper" or "steel and brass". With "riveting", welding and brazing the strip in the metals are joined length wise together. The displacement in the sideways of strips is more than smaller length ways expansion. The effect produced is used is basically used in mechanical and electrical devices.

Given the data in the question;

Electricity used; [tex]E = 1000kWh = 36*10^8J[/tex]

Time; [tex]t = 1 month = 2.592 * 10^6s[/tex]

Voltage;  [tex]V = 120V[/tex]

Calculate the average rms current to the house.

First we determine the power

Power is the amount of energy transferred per unit time:

[tex]Power = \frac{E}{t}[/tex]

So we substitute in our values

[tex]Power = \frac{36*10^8J}{2.592*10^6s}\\\\Power = 1.388 * 10^3 J/s \\\\Power = 1.388 * 10^3 W[/tex]

Next we Calculate the Current

From Ohms Law:

[tex]P = I * V[/tex]

We substitute in our values

[tex]1.388*10^3W = I \ *\ 120V \\\\I = \frac{1.388*10^3W}{120V} \\\\I = 11.57 W/V\\\\I = 12A[/tex]

Therefore, the average rms current in the power line to the house is 12A

The resistance of a household

From Ohm's Law:

[tex]R = \frac{V}{I}[/tex]

We substitute in our values

[tex]R = \frac{120V}{12A}\\\\R = 10ohms[/tex]

Therefore, The average resistance of a household is 10Ω.

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Answer:

radio waves, microwaves, infrared, visible, UV, X-rays, gamma rays

Explanation:

Electromagnetic waves consist of oscillating electric and magnetic fields which vibrate in a direction perpendicular to the direction of motion of the wave (transverse wave). Electromagnetic waves travel in a vacuum with a speed of [tex]c=3.0\cdot 10^8 m/s[/tex] (speed of light), and they are classified into 7 different types according to their wavelength.

From longer to shorter wavelength, these types are:

Radio waves (wavelength > 30 cm)

Microwaves (30 cm - [tex]5 \mu m[/tex])

Infrared ([tex]5 \mu - 750 nm[/tex])

Visible light (750 nm - 380 nm)

UV radiation (380 nm - 8 nm)

X-rays (8 nm - 6 pm)

Gamam rays (< 6 pm)

A pure substance because it cannot be separated into parts

Answer: Absolute brightness is the actual amount of light produced by the star, whereas apparent brightness changes with distance from the observer.

Explanation:

Answer: d

Explanation:

1) D. The collisions between molecules are inelastic

Explanation:

The kinetic theory of the gases describe the property of the gases by looking at microscopic level. At these level, some assumptions are made on the motion/collisions of the molecules of the gas:

- Molecules move by random motion

- The volume of the molecules is negligible compared with the volume of the gas

- The molecules obey Newton's laws of motion

- The intermolecular forces between the molecules are negligible except during the collisions

- Collisions between molecules are elastic

Therefore, the following statement

D. The collisions between molecules are inelastic

is wrong.

2) [tex]\sqrt{2}[/tex]

The kinetic energy Ek of a gas is directly proportional to its absolute temperature T:

[tex]E_k = \frac{3}{2}kT[/tex]

where k is the Boltzmann's constant. However, the kinetic energy depends on the square of the average velocity of the particles, [tex]v^2[/tex]:

[tex]E_k = \frac{1}{2}mv^2=\frac{3}{2}kT[/tex]

where m is the mass of the particles. This means that the velocity is proportional to the square root of the temperature:

[tex]v \propto \sqrt{T}[/tex]

So, if the temperature of the gas is doubled, the average speed increases by a factor [tex]\sqrt{2}[/tex], and the ratio v2/v1 is

[tex]\frac{v_2}{v_1}=\sqrt{2}[/tex]

Answer:

The current halves

Explanation:

The relationship between voltage, current and resistance in a circuit is given by Ohm's law:

[tex]V=RI[/tex]

where

V is the voltage

R is the resistance

I is the current

We can rewrite the formula as

[tex]I=\frac{V}{R}[/tex]

we see that I is directly proportional to V and inversely proportional to R.  In this problem, V is held constant while R is doubled:

[tex]R'=2R[/tex]

so, the new current in the circuit will be

[tex]I'=\frac{V}{R'}=\frac{V}{2R}=\frac{1}{2}I[/tex]

So, the current halves.

A process occurs in which a system's potential energy increases while the environment does work on the system. The kinetic energy of a system decreases while its potential energy and thermal energy are unchanged.

Kinetic energy is the energy which is present in the body of an object which is under motion. Kinetic energy of an object transforms into potential energy and in the form of work when the object changes its state of motion to rest.

When a process occurs, the potential energy of the system increases while the environment does some work on the system. In this case, the kinetic energy of the object decreases while the potential energy and thermal energy of the system remains unchanged. This is because kinetic energy is used in doing work.

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Your question is incomplete, most probably the complete question is:

A process occurs in which a system's potential energy increases while the environment does work on the system. Does the system's KINETIC energy increase, decrease, or stay the same?

Answer:

[tex]8.4\cdot 10^{-13} N[/tex]

Explanation:

The magnitude of the magnetic force on the proton is given by:

[tex]F=qvB sin \theta[/tex]

where:

[tex]q=1.6\cdot 10^{-19} C[/tex] is the proton charge

[tex]v=4.2\cdot 10^6 m/s[/tex] is the proton velocity

[tex]B=2.5 T[/tex] is the magnetic field

[tex]\theta=30^{\circ}[/tex] is the angle between the direction of v and B

Substituting into the formula, we find

[tex]F=(1.6\cdot 10^{-19}C)(4.2\cdot 10^6 m/s)(2.5 T) sin 30^{\circ}=8.4\cdot 10^{-13} N[/tex]

(a) [tex]1.03\cdot 10^{-16} N[/tex], -k direction

First of all, let's find the magnetic field produced by the wire at the location of the electron:

[tex]B=\frac{\mu_0 I}{2 \pi r}[/tex]

where

I = 19.0 A is the current in the wire

r = 1.9 cm = 0.019 m is the distance of the electron from the wire

Substituting,

[tex]B=\frac{(1.256\cdot 10^{-6})(19.0A)}{2 \pi (0.019 m)}=2\cdot 10^{-4} T[/tex]

and the direction is +j direction (tangent to a circle around the wire)

Now we can find the force on the electron by using:

[tex]F=qvBsin \theta[/tex]

where

[tex]q=1.6\cdot 10^{-19}C[/tex] is the electron's charge

[tex]v=3.23\cdot 10^6 m/s[/tex] is the electron speed

[tex]B=2\cdot 10^{-4} T[/tex] is the magnetic field

[tex]\theta[/tex] is the angle between the direction of v and B

In this case, the electron is travelling away from the wire, while the magnetic field lines (B) form circular paths around the wire: this means that v and B are perpendicular, so [tex]\theta=90^{\circ}, sin \theta=1[/tex]. So, the force on the electron is

[tex]F=(1.6\cdot 10^{-19}C)(3.23\cdot 10^6 m/s)(2\cdot 10^{-4} T)(1)=1.03\cdot 10^{-16} N[/tex]

The direction is given by the right hand rule:

- Index finger: direction of motion of the electron, +i direction (away from the wire)

- Middle finger: direction of magnetic field, +j direction (tangent to a circle around the wire)

- Thumb: direction of the force --> since the charge is negative, the sign must be reversed, so it means -k direction (anti-parallel to the current in the wire)

(b) [tex]1.03\cdot 10^{-16} N[/tex], +i direction

The calculation of the magnetic field and of the force on the electron are exactly identical as before. The only thing that changes this time is the direction of the force. In fact we have:

- Index finger: direction of motion of the electron, +k direction (parallel to the current in the wire)

- Middle finger: direction of magnetic field, +j direction (tangent to a circle around the wire)

- Thumb: direction of the force --> since the charge is negative, the sign must be reversed, so it means +i direction (away from the wire)

(c) 0

In this case, the electron is moving tangent to a circle around the wire, in the +j direction. But this is exactly the same direction of the magnetic field: this means that v and B are parallel, so [tex]\theta=0, sin \theta=0[/tex], therefore the force on the electron is zero.

A large elliptical galaxy

Answer:

A Super Massive black hole

Explanation:

C. How much sleep each student gets

It says directly that you are trying to test how the amount of sleep a student gets affects their performance. If that is the case, it would make sense to change the hours or amount of sleep each student gets before taking the test.

I shall draw this out as these tend to be answered best visually

Answer:

KE = 0.16 J

Explanation:

As we know that total energy is always remains conserved

so here we can say that initial potential energy stored in the spring is equal to the kinetic energy of the object when it comes to relaxed state

So here we have

[tex]U = \frac{1}{2}kx^2[/tex]

here we know that

[tex]k = 125 N/m[/tex]

x = 0.05 m

now from above equation

[tex]U = \frac{1}{2}(125)(0.05)^2[/tex]

[tex]U = 0.16 J[/tex]

so total potential energy here is same as the final kinetic energy which is 0.16 J

Answer:

13 Hz

Explanation:

The frequency of a wave is given by the equation:

[tex]f=\frac{v}{\lambda}[/tex]

where

f is the frequency

v is the speed of the wave

[tex]\lambda[/tex] is the wavelength

In this problem, we have:

v = 5.2 m/s is the speed of the wave

[tex]\lambda=0.40 m[/tex] is the wavelength (distance between two adjacent crests)

Substituting into the formula, we find the frequency of the wave:

[tex]f=\frac{5.2 m/s}{0.40 m}=13 Hz[/tex]

Answer:

D.13 Hz

Explanation:

A

p

e

x

Answer:Which statements describe Rutherford’s model of the atom? Select all that apply.

The atom is mostly empty space.

The atom cannot be divided into smaller particles.

Electrons orbit around the center of the atom.

The atom’s positive charge is located in the atom’s nucleus.

The electrons are located within the positive material of the nucleus.

Electron clouds are regions where electrons are likely to be found.

Explanation:

Sample Response: The one major change that occurred was the placement and organization of the electron. Rutherford’s model identified that the electrons were at a distance from the nucleus, Bohr’s model identified that the electrons occurred at levels that related to their available energy, and the modern atomic model shows that electrons are located in a predicted area but cannot be identified in a specific point.

This is chemical energy the substance that makes it glow are chemicals.

Answer:

Chemical .

Explanation:

On Halloween, you take a glow stick, crack the capsule inside and shake it until it glows. This is an example of light energy being created from ___________________ energy

From the principle of the conservation of energy which states that energy can neither be created nor destroyed but can converted from one form to another

Phenyl oxalate ester is responsible for the luminescence in a glow stick.When it reacts with hydrogen peroxide, the liquid inside a glow stick to glow.

therefore we can say that chemical energy is been converted to light energy

the glow lights are by no means radioactive. it can be made to glow again by putting it in a freezer.

Answer:

[tex]4.7\cdot 10^6 Pa[/tex]

Explanation:

The total force exerted by the man on the chair is equal to his weight:

[tex]F=W=mg[/tex]

where m=95.0 kg is the man's mass and g=9.8 m/s^2. Substituting,

[tex]F=(95.0 kg)(9.8 m/s^2)=931 N[/tex]

Since there are 4 legs, we can assume that the force is equally distributed over the 4 legs; so the force supported by each leg is

[tex]F=\frac{931 N}{4}=232.8 N[/tex]

The radius of each leg is [tex]r=0.400 cm=0.004 m^2[/tex], so the area of each leg is

[tex]A=\pi r^2 = \pi (0.004 m)^2=5\cdot 10^{-5} m^2[/tex]

And the pressure exerted on each leg is equal to the ratio between the force supported by each leg and the area:

[tex]p=\frac{F}{A}=\frac{232.8 N}{5\cdot 10^{-5} m^2}=4.7\cdot 10^6 Pa[/tex]

The pressure on each leg is 4655kNm-2.

The term pressure is defined as the ratio of force per unit area. Firts we have to find the force acting on the chair and the area covered by the force.

The force acting on the ground is the weight of the man and the chair.

F = W = mg = 95.0 kg * 9.8 ms-2 = 931 N.

Since this force is evenly distributed, the force on each leg = 931/4 = 232.75 N

The area of each leg = πr^2 = 3.142 * (0.4 * 10^-2)^2 = 5 * 10^-5 m^2

Pressure = force/area = 232.75 N/ 5 * 10^-5 m^2 = 4655kNm-2

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Answer:

1. 5.12068 × 1011 N/C away from the proton

Explanation:

The electric field produced by a single point charge is given by:

[tex]E=k\frac{q}{r^2}[/tex]

where

k is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

[tex]q=1.6\cdot 10^{-19}C[/tex] is the charge of the proton

[tex]r=5.3\cdot 10^{-11} m[/tex] is the distance at which we want to calculate the field

[tex]k=8.99\cdot 10^9 Nm^2C^{-2}[/tex] is the Coulomb's constant

Substituting into the formula,

[tex]E=(8.99\cdot 10^9 Nm^2C^{-2})\frac{1.6\cdot 10^{-19}C}{(5.3\cdot 10^{-11}m)^2}=5.12068\cdot 10^{11} N/C[/tex]

And the direction of the electric field produced by a positive charge is away from the charge, so the correct answer is

1. 5.12068 × 1011 N/C away from the proton

The magnitude of the electric field set up by the proton at the position of the electron in a hydrogen atom is approximately 5.14 x 10¹¹ N/C. This is based on the formula for the electric field E = kQ/r². The direction of the electric field is away from the proton.

The electric field created by a charge is given by the formula E = kQ/r², where Q is the charge, r is the distance, and k is the Coulomb constant. In this case the magnitude of the charge of a proton (Q) is +1.602 x 10⁻¹⁹ C, the average distance (r) between a proton and an electron in a hydrogen atom is 5.3 x 10⁻¹¹ m, and the Coulomb constant (k) is 8.99 x 10⁹ N.m²/C².

Using these values in the formula gives E = (8.99 x 10⁹ N.m²/C²)(1.602 x 10⁻¹⁹ C)/(5.3 x 10⁻¹¹ m)² = 5.14 x 10¹¹ N/C. Therefore, the magnitude of the electric field set up by the proton at the position of the electron is approximately 5.14 x 10¹¹ N/C.

As for the direction of the electric field, electric field lines point in the direction that a positive test charge would move if placed in the field. Since the proton has a positive charge, the field lines (and hence the direction of the electric field) point away from the proton. So, the correct answer is ~5.14 x 10¹¹ N/C away from the proton.

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Answer:

2639 N

Explanation:

The impulse on the baseball is given by:

[tex]I=F \Delta t[/tex] (1)

where F is the force exerted on the ball and [tex]\Delta t[/tex] the contact time between the bat and the ball.

The impulse is also equal to the change in momentum of the ball:

[tex]I=\Delta p = m (v-u)[/tex] (2)

where m is the ball's mass, v is its final velocity, u is its initial velocity.

By using (1) and (2) simultaneously we can write an expression for F, the force exerted on the ball:

[tex]F=\frac{m(v-u)}{\Delta t}[/tex]

where:

m = 0.145 kg

u = 35.0 m/s

v = -56.0 m/s

[tex]\Delta t=5.00 \cdot 10^{-3}s[/tex]

Substituting,

[tex]F=\frac{(0.145 kg)((-56.0 m/s)-35.0 m/s)}{5.00\cdot 10^{-3} s}=-2639 N[/tex]

and the negative sign means that the force is simply in the opposite direction to the ball's initial direction.