Question

One of Lex Luthor's henchman attacks Superman, shooting a rapid-fire stream of 3.3 g bullets at him at a rate of 112/min. The speed of each bullet is 527 m/s. Because of his super-power, the bullets just stop and fall straight to the ground after striking his chest. What is the magnitude of the average force on Superman's chest?

Answer 1

Answer:

3.2451N

Explanation:

Mass of the bullet (m) $= 3.3g = 3.3*10^{-3}Kg$

Speed of the bullet (V)$= 527m/s$

Rate of bullet (r) $= 112/min = 1.866\sec$

We can calculate with this information the average acceleration of bullets

$a=V*r = 527\frac{m}{s}\frac{1.866}{s} = 983.38m/s^2$

The force is given by,

$F=ma\\F=(3.3*10^{-3})*983.38m/s^2 = 3.2451N$

That is just because he is Superman.