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It is well known that bullets and other missiles fired at Superman simply bounce off his chest. Suppose that a gangster sprays Superman's chest with 5.6 g bullets at the rate of 110 bullets/min, and the speed of each bullet is 480 m/s. Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman's chest from the stream of bullets

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Answer:

4.93 N

Explanation:

We know that force F = dP/dt where dP/dt = rate of change of momentum and P = momentum = mv

Now, F = dP/dt = dmv/dt = vdm/dt + mdv/dt

Now since we have 5.6 g bullets, m = 5.6 g = 0.0056 kg, dm/dt = rate of change of mass of bullet hitting superman per second = mass of one bullet × rate of change of bullet = 0.0056 kg/bullet × 110 bullets/min = 0.0056 kg/bullet × 110 bullets/min × 1 min/60 s = 0.0103 kg/s, v = speed of bullet = 480 m/s and dv/dt = 0.

So, F = vdm/dt

= 0.01027 kg/s × 480 m/s

= 4.929 kgm/s²

= 4.929 N

≅ 4.93 N

So, the average force is 4.93 N

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