"Neon signs need 12,000 V to operate. If a transformer operates off a 240 V source and has 1000 turns in its primary coil, how may turns must the secondary coil have

Answer:

0.0081T

Explanation:

The magnetic field B in the toroid is proportional to the applied current I and the number of turns N per unit length L of the toroid. i.e

B ∝ I [tex]\frac{N}{L}[/tex]

B = μ₀ I [tex]\frac{N}{L}[/tex]                   ----------------(i)

Where;

μ₀ = constant of proportionality called the magnetic constant = 4π x 10⁻⁷N/A²

Since the radius (r = 4.2cm = 0.042m) of the toroid is given, the length L is the circumference of the toroid given by

L = 2π r

L = 2π (0.042)

L = 0.084π

The number of turns N = 1000

The current in the toroid = 1.7A

Substitute these values into equation (i) to get the magnetic field as follows;

B = 4π x 10⁻⁷ x  1.7 x  [tex]\frac{1000}{0.084\pi }[/tex]        [cancel out the πs and solve]

B = 0.0081T

The magnetic field along the central radius is 0.0081T

Answer:

The value of the inductance is 0.955 mH

Explanation:

Given;

frequency of the oscillator, f = 18 kHz = 18,000 Hz

the peak current, I₀ = 70 mA = 0.07 A

the root mean square voltage, [tex]V_{rms}[/tex] = 5.4 V

The root mean square current is given as;

[tex]I_{rms}= \frac{I_o}{\sqrt{2} }[/tex]

[tex]I_{rms} = \frac{0.07}{\sqrt{2} } \\\\I_{rms} = 0.05 \ A[/tex]

Inductive reactance is given by;

[tex]X_L =\frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.05} \\\\X_L = 108 \ ohms[/tex]

Inductance is given by;

[tex]L = \frac{X_L}{2\pi f} \\\\L = \frac{108}{2\pi *18,000} \\\\L = 9.55 *10^{-4} \ H[/tex]

L = 0.955 mH

Therefore, the value of the inductance is 0.955 mH

The value of the inductance (L) for this oscillating circuit is equal to [tex]9.55 \times 10^{-4}[/tex] Henry.

Given the following data:

To determine the value of the inductance (L):

First of all, we would find the root mean square (rms) current by using the formula:

[tex]I_{rms} = \frac{I_o}{\sqrt{2} }\\\\I_{rms} = \frac{70 \times 10^{-3}}{1.4142} \\\\I_{rms} = 0.050 \;A[/tex]

Next, we would calculate the inductive reactance of the oscillator by using the formula:

[tex]X_L = \frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.050} \\\\X_L = 108 \; Ohms[/tex]

Now, we can solve for the value of the inductance (L):

[tex]L = \frac{X_L}{2\pi f}[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]L = \frac{108}{2 \times 3.142 \times 18 \times 10^3} \\\\L = \frac{108}{113112}[/tex]

L = [tex]9.55 \times 10^{-4}[/tex] Henry.

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Answer:

Explanation:

The whole wave front may be divided into two halves , the upper half and the lower half . Waves coming from top of the slit or top of upper half and top of lower half or from the mid point of slit can form minima at given point only when there is phase difference of π radian or path difference of λ or one wavelength. Every other point in upper half and corresponding point in lower half will interfere destructively at that point and will form dark spot at the given point . In this way minima will be formed at that point

Hence the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit at first minima  is π radian .

Answer:

The time rate of change of the electric field between the plates is  [tex]\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}[/tex]  

Explanation:

From the question we are told that

    The  radius is  [tex]r = 2.8 \ cm = 0.028 \ m[/tex]

     The distance of separation is  [tex]d = 1.1 \ mm = 0.0011 \ m[/tex]

      The  current is  [tex]I = 5 \ A[/tex]

Generally the electric field generated is mathematically represented as

         [tex]E = \frac{q }{ \pi * r^2 \epsilon_o }[/tex]

Where [tex]\epsilon_o[/tex] is the permitivity of free space with a value

          [tex]\epsilon_o = 8.85*10^{-12 }\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

So the time rate of change of the electric field between the plates is mathematically represented as

        [tex]\frac{E }{t} = \frac{q}{t} * \frac{1 }{ \pi * r^2 \epsilon_o }[/tex]

But [tex]\frac{q}{t } = I[/tex]

So  

       [tex]\frac{E }{t} = * \frac{I }{ \pi * r^2 \epsilon_o }[/tex]

substituting values  

        [tex]\frac{E }{t} = * \frac{5 }{3.142 * (0.028)^2 * 8.85 *10^{-12} }[/tex]

        [tex]\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}[/tex]

Answer:

B. only when the current in the first coil changes

Explanation:

This is because for current to be induced in the second coil they must be an change in current inyhe first coil in line with Faraday's first law of electromagnetic induction. Which state that whenever their is a change in magnetic lines of flux an emf is induced

Answer:

B

Explanation:

Protons have a positive electrical charge of +1,

Electrons have a negative charge of -1,

Neutrons have a neutral charge of about 0.

Answer:

The maximum emf induced in the loop is 0.132 Volts

Explanation:

Given;

radius of the circular loop, r = 9.5 cm

intensity of the wave, I = 0.0295 W/m²

wavelength, λ = 6.40 m

The intensity of the wave is given as;

[tex]I = \frac{B_o^2*c }{2\mu_o}[/tex]

where;

B₀ is the amplitude of the field

c is the speed of light = 3 x 10⁸ m/s

μ₀ is permeability of free space = 4π x 10⁻⁵ m/A

[tex]I = \frac{B_o^2*c }{2\mu_o}\\\\B_o^2 = \frac{I*2\mu_o}{c} \\\\B_o^2 = \frac{0.0295*2*4\pi*10^{-7}}{3*10^8} \\\\B_o^2 = 2.472 *10^{-16}\\\\B_o = \sqrt{2.472 *10^{-16}}\\\\ B_o = 1.572*10^{-8} \ T[/tex]

Area of the circular loop;

A = πr²

A = π(0.095)²

A = 0.0284 m²

Frequency of the wave;

f = c / λ

f = (3 x 10⁸) / (6.4)

f = 46875000 Hz

Angular velocity of the wave;

ω = 2πf

ω = 2π(46875000)

ω = 294562500 rad/s

The maximum induced emf is calculated as;

emf = B₀Aω

       = (1.572 x 10⁻⁸)(0.0284)(294562500)

       = 0.132 Volts

Therefore, the maximum emf induced in the loop is 0.132 Volts

Answer:

Explanation:

Using equations of motion:

(a)

v=u+at

∴0=19.4−8.57t

∴t=19.4/8.57

=2.3s

B. Using s= ut + 1/2 at²

19.4(2.3)-1/28.57(2.3)²

= 21.92rad

The amount of charge that passes per unit time is called electric current .

Current has dimensions of [Charge] / [Time] .

It's measured and described in units of ' Ampere ' .

1 Ampere means 1 Coulomb of charge passing a point every second.

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, [tex]q_1=-3\ nC[/tex]

It is placed at a distance of 9 cm at x axis

Charge, [tex]q_2=+4\ nC[/tex]

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

[tex]\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0[/tex]

Here,

[tex]r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}[/tex]

So,

[tex]\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}[/tex]

Squaring both sides,

[tex]3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m[/tex]

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

According to the question,

Charge,

Now,

→ [tex]\frac{kq_1}{r_1} +\frac{kq_2}{r_2} =0[/tex]

or,

→ [tex]\frac{kq_1}{r_1} =-\frac{kq_2}{r_2}[/tex]

→  [tex]\frac{q_1}{r_1} = \frac{q_2}{r_2}[/tex]

here,

[tex]r_1 = \sqrt{y^2+81}[/tex]

[tex]r^2 = \sqrt{y^2+225}[/tex]

By substituting the values,

→      [tex]\frac{-3 }{\sqrt{y^2+225} } = -\frac{4}{\sqrt{y^2+225} }[/tex]

By applying cross-multiplication,

[tex]3\times \sqrt{y^2+225} = 4\times \sqrt{y^2+81}[/tex]

By squaring both sides, we get

→  [tex]9(y^2+225) = 16(y^2+81)[/tex]

    [tex]9y^2+2025 = 16 y^2+1296[/tex]

   [tex]2025-1296=7y^2[/tex]

               [tex]7y^2=729[/tex]

                  [tex]y = 10.2 \ m[/tex]

Thus the solution above is correct.

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Answer:

sphere #1 carries positive charge and #2 carries negative charge

This is because from the laws of static electricity, disconnecting the copper wire makes #1 to be positively charged and #2 to be negatively charged

Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

M₀ = r x F

From the question,

r = i + j + k

F = 1i + 0j +  5k

Therefore,

M₀ = (i + j + k) x (1i + 0j +  5k)

M₀ = [tex]\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right][/tex]

M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)

M₀ = i(5) - j(4) + k(-1)

M₀ = 5i - 4j - k

Therefore, the moment about the origin O of the force F is

M₀ = 5i - 4j - k

Answer:

34.58 N

Explanation:

From the question,

Tension in the cord =  Weight of the solid metal- Upthrust

T = W-U................... Equation 1

But,

W = mg

Where m = mass of solid, g = acceleration due to gravity.

Given: m = 4.7 kg, g = 9.8 m/s²

W = 4.7(9.8) = 46.1 N.

U = (Density of water×Volume of water displaced.)gravity

U = D×V×g.

But,

V = Mass of solid/density of solid

V = 4.7/4000

V = 1.175×10⁻³ m³.

Therefore,

U = 1.175×10⁻³(1000)(9.8)

U = 11.52 N

Therefore,

T = 46.1-11.52

T = 34.58 N

Answer:

1000j

Explanation:

work done = force x distance

w = 5 x 10 x 20 = 1000joules

Answer:

352 m/s

Explanation:

The velocity of sound in air is approximated as:

v ≈ 331 + 0.6 T

where v is the velocity in m/s and T is the temperature in Celsius.

At T = 35:

v = 331 + 0.6 (35)

v = 352 m/s

Answer:

c.) Their eligibility for social education services depends on whether their conditions adversely affect their educational functioning.

Explanation:

Chronic Illness is a human health condition in which a particular (or number of) illness is persistent in the body and the effects on the body are long-lasting and are often resistant to treatment. The word chronic is usually used when the disease/illness/sickness and its effects stay in the body for more than three months.

The likeliest answer from the options given is option C because before social education services are given, it has to be decided if their health condition adversely affects their education.

Answer:

408N at 89.89°

Explanation:

This problem requires that we resolve the force vectors into

x- and y

-componentsOnce this is done, we can add the components easily, as the one 2-dimensional problem will be two 1-dimensional problems.

Finally, we will convert the resultant force into standard form and find the equilibrant.

Resolve into components:

F1x =F1cos 180°= 232(−1)=−232N

F1y=F1sin180°=0N

F2x=F2cos(−140°)=194(−0.766)=−148.6N

F1y=F1sin(−140°)=232(−0.643)=−149.17N

Note the change of the angle used to give the direction of

F2. Standard angles (rotation from thex

-axis; counterclockwise is +) should be used to avoid sign errors in the results.

Now, we add the components:

Fx=F1x+F2x=−380.6N

Fy=F1y+F1y=−148.17N

Technically, this is the resultant force. However, it should be changed back into standard form. Here's how:

F=√(Fx)2(Fy)2=√(−380.6)^2(−148.17)^2=408N

θ=tan−1(−148.17−380.6)

=89.89°

Answer:

A. the inertia of shelly

Explanation:

Answer:

A.  the inertia of Shelly

Explanation:

newtons first law states that an object at rest stays at rest unless on opposite force act against it, so when the buss started to move it acted on shelly forcing her to move back. Shelly's inertia was at a halt and was forced back by the motion of the bus.

Answer:

(a) 0.59 ms

(b) 0.15 ms

(c) The significance is that the speed of sound in different media determines the time interval of perception by the ears, which are at constant distance apart.

Explanation:

(a) distance between ears = 20 cm = 0.2 m

speed of sound generated = 340 m/s

time = ?

speed = [tex]\frac{distance covered}{time taken}[/tex]

⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]

                        = [tex]\frac{0.2}{340}[/tex]

                        = 5.8824 × [tex]10^{-4}[/tex]

                        = 0.59 ms

The time interval of the arrival of the sound at the right ear to the left ear is 0.59 ms.

(b) distance between ears = 20 cm = 0.2 m

speed of sound in water = 1530 m/s

time = ?

speed = [tex]\frac{distance covered}{time taken}[/tex]

⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]

                         = [tex]\frac{0.2}{1530}[/tex]

                         = 1.4815 × [tex]10^{-4}[/tex]

                         = 0.15 ms

The sound heard by the right ear of the diver would arrive at the left 0.15 ms latter.

(c) The significance is that the speed of sound in different media, determines the time interval of perception by the ears, which are at constant distance apart.

A) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear is; t = 0.588 × 10⁻³ seconds

B) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s is; t = 0.131 × 10⁻⁵ seconds

C) The significance about the time interval of the two situations is that;

Transmission of sound varies with different mediums.

A) We are given;

Distance between the two ears; d = 20 cm = 0.2 m

Speed of sound; v = 340 m/s

Since the sound source is placed at the right ear, the time interval for it to get to the left ear is;

t = d/v

t = 0.2/340

t = 0.588 × 10⁻³ seconds

B) We are now told that the speed of sound in water is 1530 m/s. Thus;

t = 0.2/1530

t = 0.131 × 10⁻⁵ seconds

C) We can see that in answer A and B, the time interval is different even when the distance remained the same. This means that, the time interval of hearing a sound changes with respect to the medium of transmission.

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Answer:

processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.

Explanation:

The temperature of planet Earth is due to two main types of process, internal and external.

Internal processes are all chemical processes that occur that release heat into the environment or due to gases that trap heat on the planet, greenhouse effect

External processes is heating due to energy coming from the Sun. This includes direct heating of the surface by the absorption of energy and reflects of energy in different atmospheric layers.

These are the two terms that heat the Earth

In addition there are several processes so the planet loses energy,

* energy radiation to outer space that is a few degrees kelvin, for which there is a permanent emission

* endothermic processes that need to absorb heat to perform, this lowers the temperature of the system

* liquid (water) system that absorbs large amounts of heat to change state and temperature.

These processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.

Therefore it is impossible for the temperature to increase indefinitely since the emission would increase by decreasing the value

Answer:

(b) Increases

Explanation:

The potential energy between two point charges is given as;

[tex]U = F*r = \frac{kq_1q_2}{r}[/tex]

Where;

k is the coulomb's constant

q₁  ans q₂ are the two point charges

r is the distance between the two point charges

Since the two charges have opposite sign;

let q₁ be negative and q₂ be positive

Substitute in these charges we will have

[tex]U = \frac{k(-q_1)(q_2)}{r} \\\\U = - \frac{kq_1q_2}{r}[/tex]

The negative sign in the above equation shows that as the distance between the two charges increases, the potential energy increases as well.

Therefore, as you move the point charges farther and farther apart, the potential energy of this system relative to infinity Increases.

Answer:

1) 14 kW

2) 4.67 x 10^-5 N

Explanation:

Area of solar panel = 10 m^2

Intensity of sun's radiation incident on earth = 1.4 kW/m^2

Solar power absorbed = ?

We know that the intensity of radiation on a given area is

[tex]I[/tex] = [tex]\frac{P}{A}[/tex]

where I is the intensity of the radiation

P is the power absorbed due to this intensity on a given area

A is the area on which this radiation is incident

From the equation, we have

P = IA

P = 1.4 kW/m^2  x  10 m^2 = 14 kW

b) For a perfect absorbing surface, the radiation pressure is given as

p = I/c

where p is the radiation pressure

I is the incident light intensity = 1.4 kW/m^2 = 1.4 x 10^3 kW/m^2

c is the speed of light = 3 x 10^8 m/s

substituting values, we have

p = (1.4 x 10^3)/(3 x 10^8) = 4.67 x 10^-6 Pa

we know that Force = pressure x area

therefore force on the solar panels is

F = 4.67 x 10^-6 x 10 = 4.67 x 10^-5 N

Answer:

a) The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second, b) The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second, c) D. The magnitude of the change in the ball's momentum.

Explanation:

a) This phenomenon can be modelled by means of the Principle of Momentum Conservation and the Impact Theorem, whose vectorial form is:

[tex]\vec p_{o} + Imp = \vec p_{f}[/tex]

Where:

[tex]\vec p_{o}[/tex], [tex]\vec p_{f}[/tex] - Initial and final momentums, measured in kilogram-meters per second.

[tex]Imp[/tex] - Impact due to collision, measured in kilogram-meters per second.

The impact experimented by the ball due to collision is:

[tex]Imp = \vec p_{f} - \vec p_{o}[/tex]

By using the definition of momentum, the expression is therefore expanded:

[tex]Imp = m \cdot (\vec v_{f}-\vec v_{o})[/tex]

Where:

[tex]m[/tex] - Mass of the ball, measured in kilograms.

[tex]\vec v_{o}[/tex], [tex]\vec v_{f}[/tex] - Initial and final velocities, measured in meters per second.

If [tex]m = 0.275\,kg[/tex], [tex]\vec v_{o} = -2.10\,j\,\left [\frac{m}{s} \right][/tex] and [tex]\vec v_{f} = 1.90\,j\,\left [\frac{m}{s} \right][/tex], the vectorial change of the linear momentum is:

[tex]Imp = (0.275\,kg)\cdot \left[1.90\,j+2.10\,j\right]\,\left[\frac{m}{s} \right][/tex]

[tex]Imp = 1.1\,j\,\left[\frac{kg\cdot m}{s} \right][/tex]

The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second.

b) The magnitudes of initial and final momentums of the ball are, respectively:

[tex]p_{o} = (0.275\,kg)\cdot \left(2.10\,\frac{m}{s} \right)[/tex]

[tex]p_{o} = 0.578\,\frac{kg\cdot m}{s}[/tex]

[tex]p_{f} = (0.275\,kg)\cdot \left(1.90\,\frac{m}{s} \right)[/tex]

[tex]p_{o} = 0.523\,\frac{kg\cdot m}{s}[/tex]

The change in the magnitude of the ball's momentum is:

[tex]\Delta p = p_{f}-p_{o}[/tex]

[tex]\Delta p = 0.523\,\frac{kg\cdot m}{s} - 0.578\,\frac{kg\cdot m}{s}[/tex]

[tex]\Delta p = -0.055\,\frac{kg\cdot m}{s}[/tex]

The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second.

c) The quantity calculated in part a) is more related to the net force acting on the ball during its collision with the floor, since impact is the product of net force, a vector, and time, a scalar, and net force is the product of the ball's mass and net acceleration, which creates a change on velocity.

In a nutshell, the right choice is option D.

Answer:

3.26m

Explanation:

See attached file

Explanation:

It is given that, An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. We need to find the ratio of their masses.

As they are in equilibrium, the force of gravity due to each other is same. So,

[tex]\dfrac{Gm_xM}{r^2}=\dfrac{Gm_yM}{r^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{m_y}=(\dfrac{r_x^2}{r_y^2})\\\\\dfrac{m_x}{m_y}=(\dfrac{140^2}{481^2})\\\\\dfrac{m_x}{m_y}=0.0847[/tex]

So, the ratio of masses X/Y is 0.0847

Answer:

9.42*10^-4 m^3/s

Explanation:

d=3/100 m

=0.03 m

A=3.14*0.03^2/6

=4.71*10^-4 m^2

Volume flow rate V = A * s

= 4.71*10^-4 * 2

= 9.42*10^-4

So, the volume flw rate of fluid is 9.42*10^-4 m^3/s

Answer:

A. Attractive

B. ( μ₀I² ) / ( 2πd )

Explanation:

A. We know that currents in the same direction attract, and currents in the opposite direction repel, according to ampere's law. In this case the current in the two wires are flowing in the same direction, and hence the force between the two wires are attractive.

B. Suppose that two wires of length [tex]l_1[/tex] and [tex]l_2[/tex] both carry the current [tex]I[/tex] in the same direction ( given ). In the presence of a magnetic field produced by wire 1, a force of magnitude m say, is experienced by wire 2. The magnitude of the magnetic field produced by wire 1 at distance say d, from it's axis, should thus be the following -

[tex]B_1[/tex] = μ₀I / 2πd

The force experienced by wire 2 should thus be -

[tex]F_2[/tex] = I( [tex]l_2[/tex] [tex]*[/tex] [tex]B_1[/tex] )

= I [tex]*[/tex] [tex]l_2 * B_1 *[/tex] Sin( 90 )

= I [tex]*[/tex] [tex]l_2[/tex] ( μ₀I / 2πd )

Therefore the force per unit length experienced by wire 2 toward wire 1 should be ...

( [tex]F_2[/tex] / [tex]l_2[/tex] ) = ( μ₀I² ) / ( 2πd ) ... which is our solution

Answer:

Three halves of a wavelength I.e 7lambda/2

Explanation:

See attached file pls

Answer:

Electric potential at position, x = -20 m, = -100 V

Electric potential at position, x = 0 m, = 0

Electric potential at position, x = 10 m, = 50 V

Electric potential at position, x = 11 m, = 55 V

Electric potential at position, x = 99 m, 495 V

Explanation:

Given;

magnitude of electric field, E = 5.0 V/m

at position x = 10 m, electric potential = 50 V

Electric potential at position, x = -20 m

V = Ex

V = 5 (-20)

V = -100 V

Electric potential at position, x = 0 m

V = Ex

V = 5(0)

V = 0

Electric potential at position, x = 10 m

V = Ex

V = 5(10)

V = 50 V

Electric potential at position, x = 11 m

V = Ex

V = 5(11)

V = 55 V

Electric potential at position, x = 99 m

V = Ex

V = 5(99)

V = 495 V

Answer:

The speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s

Explanation:

Given;

speed of the ball thrown inside boxcar, [tex]V_B[/tex] = 23.2 m/s

speed of the boxcar moving along the tracks, [tex]V_T[/tex] = 34.9 m/s

Determine the speed of the ball relative to the ground if the ball is thrown forward.

If the ball is thrown forward, the speed of the ball relative to the ground will be sum of the ball's speed plus speed of the boxcar.

[tex]V_{relative \ speed} = V_B + V_T\\\\V_{relative \ speed} = 23.2 + 34.9\\\\V_{relative \ speed} = 58.1 \ m/s[/tex]

Therefore,  the speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s.