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
The answer would be 40 cause 10*4=40
You do the net force by subtracting the sides. The direction of the box is moving forward to the right by 10 N.
Without the fig, we know nothing about AB, C_1, C_2, V_a, or V_b.
The probability of getting any kind of answer from anybody who hasn't
seen or tasted the fig is vanishingly small.
