7. If the impact of the golf club on the ball in the previous question occurs over a time of 2 x 10 seconds, whatforce does the ball experience to accelerate from rest to 73 m/s?

Answer:

E=930.84 N/C

Explanation:

Given that

I = 1150 W/m²

μ = 4Π x 10⁻⁷

C = 2.999 x 10⁸ m/s

E= C B

C=speed of light

B=Magnetic filed  ,E=Electric filed

Power  P = I A

A=Area=4πr²  ,I=Intensity

E=930.84 N/C

Therefore answer is 930.84 N/C

Final answer:

To find the magnitude Em of the electromagnetic waves at the top of the earth's atmosphere, we use the intensity of electromagnetic wave and solving the equation Em = sqrt(2Icμo), we can find the magnitude of Em in units of N/C.

Explanation:

To find the magnitude Em of the electromagnetic waves at the top of the Earth's atmosphere, we use the fact that the power received per unit area is the intensity I of the electromagnetic wave. According to the given information, this intensity is 1150 W/m2. The relationship between the intensity and electromagnetic fields is given by the equation I = 0.5 * E²/c * μo. Solving for Em, we get Em = sqrt(2Icμo), where μo = 4π × 10-7 T N/A² is the permeability of free space and c = 2.99792 × 10⁸ m/s is the speed of light.

Subbing in the given values, we can compute Em as:

Em = sqrt[2 * 1150 W/m² * 2.99792 × 10⁸ m/s * 4π × 10-7 T N/A²]

This computation will give the strength of the electric field at the top of the earth’s atmosphere in units of N/C.

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To solve this problem we will start from the definition of Force, as the product between the electric field and the proton charge. Once the force is found, it will be possible to apply Newton's second law, and find the proton acceleration, knowing its mass. Finally, through the linear motion kinematic equation we will find the speed of the proton.

PART A ) For the electrostatic force we have that is equal to

Here

q= Charge

E = Electric Force

PART B) Rearrange the expression F=ma for the acceleration

Here,

a = Acceleration

F = Force

m = Mass

Replacing,

PART C) Acceleration can be described as the speed change in an instant of time,

There is not then

Rearranging to find the velocity,

Final answer:

The magnitude of the electric force felt by the proton is 4.4 x 10^-16 N. The proton's acceleration is 2.64 x 10^11 m/s^2. The proton's speed after 1.40 μs in the field is 3.70 x 10^5 m/s.

Explanation:

The charge of a proton is 1.6 x 10-19 coulombs and the electric field strength is 2750 N/C. Therefore, the magnitude of the electric force felt by the proton is (1.6 x 10-19 C)(2750 N/C) = 4.4 x 10-16 N. The mass of a proton is approximately 1.67 x 10-27 kilograms. Therefore, the proton's acceleration is (4.4 x 10-16 N)/(1.67 x 10-27 kg) = 2.64 x 1011 m/s2. Since the proton starts from rest, its initial velocity (u) is 0. Therefore, the proton's speed after 1.40 μs is v = (2.64 x 1011 m/s2)(1.40 x 10-6 s) = 3.70 x 105 m/s.

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Answer:

The temperature of silver at this given resistivity is 2971.1 ⁰C

Explanation:

The resistivity of silver is calculated as follows;

where;

Rt is the resistivity of silver at the given temperature

Ro is the resistivity of silver at room temperature

α is the temperature coefficient of resistance

To is the room temperature

T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature

Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m

When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;

Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C

Answer:

What displacement must the physics professor give the car

= 12.91 METERS

Explanation:

Check the attached file for explanation

Answer:

V = 1.69 * 10^6 V

Explanation:

Parameters given:

Electric field, E = 59V/m

Charge, q = 5.40C

We need to first find the distance between the electric charge and the point of consideration to be able to find the Electric potential difference.

Electric field is given as:

E = (kq/r^2)

k = Coulombs constant

=> r^2 = kq/E

=> r^2 = (9 * 10^9 * 5.4) / 59

r^2 = 8.2 * 10^8

r = 2.84 * 10^4 m

We can now find the Electric Potential by using:

V = kq/r

Hence,

V = (9 * 10^9 * 5.4) / (2.84 * 10^4)

V = 1.69 * 10^6 V

Answer:

-9Q

Explanation:

Electric field at origin is:

Electric field due to first charge at origin would be:

Electric field due to second charge would be:

If the second charge is Q', then should be:

compare the above two values to find the possible values of Q':

The net electric field at origin is greater than the one due to first charge. It means the second charge adds on to the electric field at the origin. Thus, it should be a negative charge.

Thus, Q' = -9Q

One value is possible as the location of the second charge is given to be on the positive x-axis.

Final answer:

The possible values for the unknown charge are 1/9 of the magnitude of the known charge.

Explanation:

To find the possible values for the unknown charge, we need to use the principle of superposition. The net electric field at the origin is given by the sum of the electric fields due to each charge. We know that the magnitude of the net electric field is 2keQ/a^2, so we can set up the equation:

2keQ/a^2 = keQ/(-a)^2 - keq/(3a)^2

By solving this equation, we can find the possible values for the unknown charge. Simplifying the equation, we get:

2 = 1 - 1/9

1/9 = 1

After solving the equation, we find that the possible value for the unknown charge is 1/9 of the magnitude of the known charge.

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