Question

A scientist at a space research center is doing research on near-Earth objects and determining the kinetic energy of a theoretical object. A meteorite with a mass of 12 grams falls to Earth. If the meteorite’s velocity is 35 kilometers/second, what is its kinetic energy?

Answer 1

(Taking an online class with this exact question, Answer given by program)
KE= 1/2 mv^2
     = 1/2(0.012)(35,000)^2
     = 7.35x10^6
The kinetic energy of the meterorite is 7.35 x 10^6 Joules

Answer 2

its kinetic energy is 7350kJ 


Kinetic energy is given as = $\frac{1}{2} mv^{2}$

Now, m = 12 gms = 0.012 kg
 And, velocity = 35 kilometers/second = 35000 m/sec

Kinetic energy is given as = $\frac{1}{2} 0.012 kg * 35000*35000 m/[tex] s^{2}$
                        = 6 $10^{-2}$×1225 ×$10^{6}$ m/$s^{2}$ 
                        = 7350 kJ