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svetlana [45]
3 years ago
12

A charge of 0.14 C is moved from a position where the electric potential is 20 V to a position where the electric potential is 5

0 V. What is the change in potential energy of the charge associated with the change in position?
Physics
1 answer:
alukav5142 [94]3 years ago
7 0

The change in electric potential energy is given by:

ΔU = ΔVq

ΔU = change in PE, ΔV = potential difference, q = charge

Given values:

ΔV = 50V - 20V = 30V, q = 0.14C

Plug in and solve for ΔU:

ΔU = 30(0.14)

ΔU = 4.2J

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Explanation:

8 0
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Part A
7nadin3 [17]

Answer:

2.5 m/s²

Explanation:

Using the formula, v = u + at ( v = Final velocity; u = Initial velocity; t = Time; a = Acceleration)

25 = 0 + 10a

a = 25/10 = 2.5 m/s²

8 0
3 years ago
A 50kg meteorite moving at 1000 m/s strikes Earth. Assume the velocity is along the line joining Earth's center of mass and the
zysi [14]

As per the question, the mass of meteorite [ m]= 50 kg

                       The velocity of the meteorite [v] = 1000 m/s

When the meteorite falls on the ground, it will give whole of its kinetic energy to earth.

We are asked to calculate the gain in kinetic energy of earth.

The kinetic energy of meteorite is calculated as -

                                       Kinetic\ energy\ [K.E]\ =\frac{1}{2} mv^2

                                                             =\frac{1}{2}50kg*[1000\ m/s]^2

                                                               =\frac{1}{2}50* 10^{6}\ J

                                                               =25*10^6\ J    

Here, J stands for Joule which is the S.I unit of energy.

Hence,\ the\ kinetic\ energy\ gained\ by\ earth\ is\ 25*10^6\ J

4 0
3 years ago
Read 2 more answers
Select the correct answer.
Effectus [21]
A. logic, would be your answer i believe!
4 0
2 years ago
A girl and a boy are riding on a merry go round that is turning at a constant rate. The girl is near the outer edge, and the boy
galina1969 [7]

Answer:

The girl has greater tangential acceleration

Explanation:

The angular acceleration (\alpha) of the merry go round is equal to the rate of the change of the angular velocity, \omega:

\alpha = \frac{d\omega}{dt}

Since all the points of the merry go round complete 1 circle in the same time, the angular velocity of each point of the merry go round is the same, and so all the points also have the same angular acceleration.

The tangential acceleration instead is given by

a_t = \alpha r

where

\alpha is the angular acceleration

r is the distance from the centre of the merry go round

Since the girl is near the outer edge and the boy is closer to the centre, the value of r for the girl is larger than for the boy, so the girl has greater tangential acceleration.

5 0
3 years ago
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