Two 3.4g bullets are fired with speeds of 42.9m/s and 77.4m/s, respectively a) if the kinetic energy of the first bullet is 3.128697 in units of...

Please not that the kinetic energy information provided in the question is not essential for answering either part a) or part b) of the question. Kinetic energy of first bullet can also be calculated the same way as that that of second bullet.

Given:

Mass of the bullets = m 3.4 g = 3.4/1000 kg = 0.0034 kg

Speed of first bullet = v1 = 42.9 m/s

Speed of second bullet = v2 = 77.4 m/s

Kinetic energy of first bullet = K1 = 3.12869 J

b) Kinetic energy of second Bullet

Kinetic energy (K) of a moving object is given by the formula:

K = (m*v^2)/2

Therefore kinetic energy of the second bullet (K2) is given by:

K2 = (m*v^2)/2 = (0.0034*77.4^2)/2 = 10.184292 J

c) Ratio of kinetic energies

Ratio of kinetic Energies = K2/K1 = 3.12869/10.184292 = 3.255122

Please note that it is not necessary to calculate calculate or know the kinetic energies to calculate the ratio of kinetic energies. It can also be worked out as:

K2/K1 = (v2^2)/(v1^2) = (77.4^2)/(42.9^2) = 3.255122