The acceleration of a bullet that leaves a 28-in rifle barrel at 2700 ft/s is 1.562 x 10⁶ m/s² and its time in the barrel is 0.0017s
a) How to determine the acceleration of the bullet
Converting inches to a foot
1 inch = 1/12 foot
28-in = 1/12 x 28 ft
28-in = 2.33ft
Therefore, the distance, s, covered by the bullet is 2.333ft
Using one of the equations of motion
2as = u² - v²
The initial velocity is 0
Where a = acceleration of the bullet
s = distance
u = initial velocity
v = final velocity
a = v² - u² ÷ 2s
= (2700 ft/s)² - 0 ÷ 2 (2.333ft)
a = 1.562 x 10⁶ m/s²
b) How to determine the time of the bullet in the barrel
s = v² + u² ÷ 2
t = 2 ÷ (v² + u²)
t = 2 (2.333ft) ÷ (0 + 2700 ft/s)
t = 0.0017s
In summary, a bullet that leaves a 28-in rifle barrel at 2700 ft/s would have an acceleration of 1.562 x 10⁶ m/s², spending a time of 0.0017s in the barrel.
Learn more about the acceleration of a bullet leaving a rifle here at: https://brainly.com/question/18522474
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