It is well known that bullets fired at superman simply bounce off his chest. suppose that a gangster sprays superman's chest with bullets of mass m = 1.7 g at a rate of r = 165 bullets/min. the speed of each bullet is v = 380 m/s. suppose the bullets rebound with no change in speed. what is the average force exerted by the stream of bullets on superman's chest?

1.77 N

Explanation:

Mass of bullet m = 1.7 x 10⁻³ kg

velocity v = 380 m /s .

Momentum of one bullet

= 1.7 x 10⁻³ x 380

= 646 x 10⁻³ kg m/s

momentum of 165 bullets

= 165 x 646 x 10⁻³ kg m/s

= 106.59 kg m/s

Final momentum after bouncing

= 0

change in momentum

= 106.59  - 0

= 106.59 kg m/s

This change occurs in one minute

so rate of change in momentum

= 106.59 / 60 kg m/s per second

= 1.77 kg m/s per second

rate of change in momentum = force

This is force on superman's chest

Force required = 1.77 N .

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever