Question

4. If a bullet leaves the muzzle of a rifle at 600.0 m/s,
and the barrel is 0.90 m long, what was the acceleration
of the bullet while in the barrel?​

Answer 1

Answer:

$u = 0m {s}^{ - 1}$

$v = 600m {s}^{ -1 }$

$s = 0.90m$

${v}^{2} - {u}^{2} = 2as$

${600}^{2} - {0 }^{2} = 2 \times 0.90 \times s$

$a = \frac{360000}{2 \times 0.90}$

$s = 200000m {s}^{ - 2}$

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Answer 2

The acceleration of the bullet is $2\times 10^5\ m/s^2$.

Explanation:

Let us assume that the initial speed of bullet, u = 0

Final speed of the bullet, v = 600 m/s

It covers a distance of 0.9 m. We need to find the acceleration of the bullet. Using third equation of motion as :

$v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(600)^2}{2\times 0.9}\\\\a=2\times 10^5\ m/s^2$

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