Question

A bullet is fired from a gun. the force on the bullet is given by : f = 600 – 2 × 105 t. where f is in newton and t in second. the force on the bullet becomes zero as soon as it leaves the barrel. what is the average impulse imparted to the bullet?

Answer 1

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SOLUTION--------

F = 600 - 2 * 10⁵t
0 = 600 - 2 * 10⁵ t
t = 600 / (2 * 10⁵)
t = 3 * 10⁻³ s
t = 3 milliseconds

Time taken by bullet to leave the barrel is 3 milliseconds

F.dt = (600 - 2* 10⁵t).dt
J = [600t - (2 * 10⁵ t²/2)]
J = [600*(3*10^-3) - (2 * 10^5 * (3 * 10^-3)^2 / 2)]
J = 0.9 Ns

Impulse imparted to bullet is 0.9 Ns

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Answer 2

HY

FRIEND.

THANKYOU YOU ASKING

QUESTION IN BRAINLY SITES

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HERE YOU ANSWER

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SOLUTION--------

F = 600 - 2 * 10⁵t

0 = 600 - 2 * 10⁵ t

t = 600 / (2 * 10⁵)

t = 3 * 10⁻³ s

t = 3 milliseconds

Time taken by bullet to leave the barrel is 3 milliseconds

F.dt = (600 - 2* 10⁵t).dt

J = [600t - (2 * 10⁵ t²/2)]

J = [600*(3*10^-3) - (2 * 10^5 * (3 * 10^-3)^2 / 2)]

J = 0.9 Ns