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2 years ago

8

A uniform electric field of 2 kNC-1 is in the x-direction. A point charge of 3 μC initially at rest at the origin is released. W

hat is the kinetic energy of this charge at x = 4m?

1 answer:

ANEK [815]2 years ago

8 0

The electric force acting on the charge is given by the charge multiplied by the electric field intensity:
$F=qE$
where in our problem $q=3 \mu C= 3 \cdot 10^{-6} C$ and $E=2 kN/C=2000 N/C$ , so the force is
$F=(3 \cdot 10^{-6} C)(2000 N/C)=0.006 N$

The initial kinetic energy of the particle is zero (because it is at rest), so its final kinetic energy corresponds to the work done by the electric force for a distance of x=4 m:
$K(4 m)=W=Fd=(0.006 N)(4 m)=0.024 J$

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2 years ago

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Answer:

The shell will land 10.18m away from the buoy.

Explanation:

In order to solve this problem, we must first do a sketch of what the problem looks like (see attached picture).

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Since it takes 3.19s for the shell to hit the water, that's the amount of time it spends flying horizontally.

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We can now find the distance between the ship to point B, which is the point the ball falls due to the tailwind. Since the movement will be accelerated in this scenario, we can find the distance by using the following formula:

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2 years ago