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
Explanation:
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The formula for the torque is
<span>τf = p F
where
</span><span>τf is the torque
p is the distance where the force is applied by the tendon
F is force applied by the tendon
If there are given values, substitute in the equation and solve for the torque.</span>
Answer : 
Explanation :
It is given that,
Mass of the engine, m = 30 kg
Thrust is equivalent to the force acting perpendicularly and it is F = 300 N
According to Newton's second law of motion :
a is the acceleration of the engine.
So, the acceleration of the engine is 
.
Hence, this is the required solution.
