A spring with an mm-kg mass and a damping constant 9 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 2.5 newtons. If the spring is stretched 1 meters beyond its natural length and then released with zero velocity, find the mass that would produce critical damping.

4.05 kg

Explanation:

From Hooke's law, force required to stretch the spring is  represented as follows:

k (0.5) = 2.5

Spring Constant, k = 5

For critical damping, c² - 4mk =0

m = c² / 4 k

c= damping constant

m = Mass to produce critical damping

There fore, m = (9²/4*5)

= 81/20

= 4.05 kg

Therefore, the mass that would produce critical damping. = 4.05 kg

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.