Atrap-jaw ant snaps its mandibles shut at very high speed, a good trait for catching small prey. but an ant can also slam its mandibles into the ground; the resulting force can launch the ant into the air for a quick escape. a 12 mg ant hits the ground with an average force of 47 mn for a time of 0.13 ms; these are all typical values. at what speed will it leave the ground?

v = 0.509 m/s

Explanation:

Given that,

Mass of the ant,

Force exerted,

Time,

To find,

Speed with which it leave the ground.

Solution,

The change in momentum is equal to the product of force and time as :

Since, u = 0

v = 0.509 m/s

So, the speed with which it will leave the ground is 0.509 m/s.

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm