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2 years ago

A 100 Kg man is diving off a 50 meter cliff. What is his kinetic energy when he is 20 meters from the water?

Physics

1 answer:

iren2701 [21]2 years ago

8 0

Answer:

$K.E=29.403125J$

Explanation:

From the question we are told that

Mass $M=100$

Height $50-20=30m$

Generally the equation for velocity before impact is is is mathematically given by

$v=\sqrt{2gh}$

$v=\sqrt{2*9.8*30}$

$v=24.25$

Generally the equation for Kinetic Energy is is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}*100*(24.25)^2\\$

$K.E=29403.125J$

$K.E=29.403125J$

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3 years ago

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.

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since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

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Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

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