Energy in a meteor

We know the kinetic energy formula...

$\large {E_k=\frac {1}{2} mv^2}$

Here, we have $\large {m=10~\textrm{tons}=(10 \times 907)~\textrm{kg}=9070~\textrm{kg}}$

And, $\large{v=14600~\textrm{m/s}}$

Plugging in, we get...

$\large {E_k=\frac{1}{2} \times 9070~\textrm{kg} \times (14600~\textrm{m/s})^2}$

$\large{=(\frac{1}{2} \times 9070 \times 14600 \times 14600)~\textrm{J}}$

$\large{=(9070 \times 7300 \times 14600)~\textrm{J}}$

$\large{=(9.07 \times 7.3 \times 1.46 \times 10^3 \times 10^3 \times 10^4)~\textrm{J}}$

$\large{=(96.66806 \times 10^{10})~\textrm{J}}$

$\large{=(9.67 \times 10^{11})~\textrm{J}}~~~\textrm{(approx.)}$

Therefore, the answer is... $\Large{\fbox{9.67E+11}}$

The formula for kinetic energy is $\frac{1}{2}mv^{2}$ . We know that the mass of the meteor is 9070 kilograms and the magnitude of velocity is 14600 m/s. If we plug that information into the formula. we get $9.67 \times 10^{11}$ J.

mass of the meteor = 9070 kg velocity = 14600 m/s K.E.= 1/2 mv2 So putting the values in the formula we get , K.E. =966680600000 J

use formula Ek=1/2 (mass,m)(velocity,v)^2

Ek=1/2(907*10)(14600)^2

Ek = 9.67E+11