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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 is
0.9 ns have a chat with me I will send u the process bro
0.9 ns have a chat with me I will send u the process bro
how???
I will send u as a question but it is the answe
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