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marishachu [46]
3 years ago
10

A particle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vec

tor is oriented 55.2° clockwise from the vertical axis, as shown. if the magnitude of the electric field is 4.82 n/c, how much work is done by the electric field as the particle is made to move a distance of d = 0.556 m straight up?
Physics
1 answer:
miss Akunina [59]3 years ago
3 0
The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:
W=F d cos \theta
where the force is F=qE, d=0.556 and \theta=55.2^{\circ}. Using the value of q and E given by the problem, we find
W=qEdcos\theta = 6.39\cdot10^{-5}J
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Answer:

0.001 s

Explanation:

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F=\frac{\Delta p}{\Delta t}

where

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\Delta t is the time interval

The change in momentum can be written as

\Delta p=m(v-u)

where

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So the original equation can be written as

F=\frac{m(v-u)}{\Delta t}

In this problem:

m = 5 kg is the mass of the fist

u = 9 m/s is the initial velocity

v = 0 is the final velocity

F = -45,000 N is the force applied (negative because its direction is opposite to the motion)

Therefore, we can re-arrange the equation to solve for the time:

\Delta t=\frac{m(v-u)}{F}=\frac{(5)(0-9)}{-45,000}=0.001 s

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Explanation:

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