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3 years ago

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A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 5.70 x 10^5 m/s^2 for 9.

60 x 10^−4 s. What is its muzzle velocity (in m/s) (that is, its final velocity)? (Enter the magnitude.)

1 answer:

s2008m [1.1K]3 years ago

3 0

Answer:

Final speed, v = 547.2 m/s

Explanation:

Initial velocity of the bullet, u = 0

Acceleration of the bullet, $a=5.7\times 10^5\ m/s^2$

Time taken, $t=9.6\times 10^{-4}\ s$

Let v is the final velocity of the muzzle. It can be calculated using the first equation of motion as :

$v=u+at$

$v=at$

$v=5.7\times 10^5\ m/s^2\times 9.6\times 10^{-4}\ s$

v = 547.2 m/s

So, the final speed of the muzzle is 547.2 m/s. Hence, this is the required solution.

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When the spring is in its equilibrium position, that is $x=0$ , the object speed its maximum. So, we have:

$\frac{k(0)^2}{2}+\frac{mv_{max}^2}{2}=\frac{kA^2}{2}\\A^2=\frac{mv_{max}^2}{k}\\A=\sqrt{\frac{mv_{max}^2}{k}}$

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Answer:

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