Question

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Answer 1

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Question

A bullet of mass 30 g moving with a speed of 500 ms penetrates 10 cm into a fixed target. Calculate the average force exerted by target on the bullet.

Solution

Mass of bullet,

m = $\frac{30}{1000} g$ = 0.03kg

Speed (v) = ${500 \: ms}^{ - 1}$

Initial kinetic energy = $\frac{1}{2} {mv}^{2}$

$= \frac{1}{2} \times 0.03 \times ( {500})^{2}$

$= \frac{1}{2} \times 0.03 \times 250000 = 3750J$

Final Kinetic Energy $= \frac{1}{2} {mv}^{2}$ = 0

Loss in kinetic energy = 3750 J

Suppose:

F = average force applied by block on bullet

S = displacement = 10 cm = 0.10m

Applying work energy principle,

W = Δk

Fs = Δk

F × 10 = 3750

$F \: = \: \frac{3750}{0.10} = 3.5 \times {10}^{4} N$

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Answer 2

Thanks for asking.........