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fomenos
3 years ago
6

A charge of 50 µC is pushed by a force of 25 µN a distance of 15 m in an electric field. What is the electric potential differen

ce? Recall that Fe = qE.
Physics
1 answer:
sineoko [7]3 years ago
6 0

The electric potential difference is 7.5 V.

Electric potential difference between two points is defined as the work done in moving unit positive charge between the two points. Electric potential difference can also be defined as the work done W per unit charge q in an electric field.

\Delta V=\frac{W}{q}......(1)

Work done by a force is the product of the force F and the displacement s made in the direction of the force. Assuming that the displacement is parallel to the line of action of the force,

W=F.s......(2)

Rewrite equation (1) using equation (2).

\Delta V=\frac{Fs}{q}......(3)

Substitute the given values of force, displacement and charge in the equation (3).

\Delta V=\frac{Fs}{q}\\ =\frac{(25*10^-^6N)(15m)}{(50*10^-^6C)} \\ =7.5V

The potential difference between the two points is 7.5 V


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the two balls will hit the ground at the same time.

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s = gt^2/2

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show how three identical 6 resistors must be connected tho have the following effective resistance values 9 and 4 ohms​
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Hen a gfci receptacle device is installed on a 20-ampere branch circuit (12 awg copper), what is the minimum volume allowance (i
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2.25in³

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3 years ago
You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between
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Answer:

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Explanation:

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u_1 = Velocity of ball before collision

u_2 = Velocity of the person before collision

v_1 = velocity of ball afer collision

v_2= velocity of the person after collision

We know that after the collision, as the person as the ball have both the same velocity, then,

v_1 = v_2

m_1u_1 + m_2u_2 = (m_1+m_2)v_2

Re-arrenge to find v_2,

v_2 = \frac{m_1u_1+m_2u_2}{m_1+m_2}

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Substituting,

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<em />

<em>The speed of the person would be 0.074m/s after the collision between him/her and the ball</em>

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a

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