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

3.2451N

Step-by-step 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(m)/(s)(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.