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?

Respuesta :

Answer:

3.2451N

Explanation:

Mass of the bullet (m) [tex]= 3.3g = 3.3*10^{-3}Kg[/tex]

Speed of the bullet (V)[tex]= 527m/s[/tex]

Rate of bullet (r) [tex]= 112/min = 1.866\sec[/tex]

We can calculate with this information the average acceleration of bullets

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

The force is given by,

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

That is just because he is Superman.

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