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Solnce55 [7]
2 years ago
8

In a ballistics test, a 24 g bullet traveling horizontally at 1300 m/s goes through a 25-cm-thick 360 kg stationary target and e

merges with a speed of 930 m/s. The target is free to slide on a smooth horizontal surface. How long is the bullet in the target? What average force does it exert on the target? What is the target's speed just after the bullet emerges?
Physics
1 answer:
MaRussiya [10]2 years ago
8 0

Answer:

A)0.00022s b)40363.6N c) 0.025m/s

Explanation:

Mass = 24g = 0.024kg, distance though the target = thickness of the target = 25cm = 0.25m

Initial speed of the bullet = 1300m/s, final speed = 930m/s

Using equation of motion

Distance = 1/2(vf+vi)*t (time in seconds)

t = 0.25*2/(1300+930) = 0.00022s

B) force exerted on the body

F = ma = m* (vf-vi)/t = 0.024*(930-1300)/0.00022

F = -40363N, it is negative because the body decelerated during this motion

C) using law of conservation of momentum,

M1*U1+ M2*U2(M2and U1 are the mass and initial speed of the body) = M1V1+ M2V2

The target was at rest so initial speed U2 = 0

0.024*1300 + 360*0 = 0.024*930 + 360*V2

31.2 = 22.32+360*V2

31.2-22.33 = 360*V2

V2 = 8.88/360 = 0.025m/s

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A raindrop of mass 0.5 * 10^-4 kg is falling verctically under the influence of gravity. The air drag on the raindrop is fdrag =
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Answer:

The displacement of the air drop after 3 second is 18.27 m.

Explanation:

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Weight of the rain drop = W

Duration of time = t = 3 seconds

W=m\times g

Drag force on rain drop = D=0.2\times 10^{-5} v^2

W=0.5\times 10^{-4} kg\time 10 m/s^2=0.5\times 10^{-3} N

Motion of the rain drop:

F=m\times a

Net force on the rain drop , F=  W - D

W-D=m\times a

0.5\times 10^{-3} N-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times a

0.5\times 10^{-3} kg m/s^2-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times \frac{v}{t}

0.006v^2+0.05v-1.5=0

v = 12.18 m/s

Initial velocity of the rain drop = u = 0 (since, it is starting from rest)

v=u+at (First equation of motion)

12.18 m/s=0m/s+a\times 3 s

a=4.06 m/s^2

s=ut+\frac{1}{2}at^2 (second equation of motion)

s=0\times 3s+\frac{1}{2}\times 4.06m/s^2\times (3 s)^2

s = 18.27 m

The displacement of the air drop after 3 second is 18.27 m.

6 0
2 years ago
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