Question #95ec3

Question

1587 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-09T02:36:22

K E of an object of mass m and velocity v is given by the formula

$E_k=1/2mv^2$

For first bullet $m=4.0 g=4xx10^-3kg and "velocity"= 100" m/s"$

Its KE $E_(k1)=1/2xx4xx10^-3xx100^2=20J=20xx10^7erg$

In Calorie KE $E_(k1)=20/4.2cal~~4.762cal$

For 2nd bullet $m=8.0 g=8xx10^-3kg and "velocity"= 100" m/s"$

Its KE $E_(k2)=1/2xx8xx10^-3xx100^2=40J=40xx10^7erg$

In Calorie KE $E_(k2)=40/4.2cal~~9.524cal$