Abox of mass 22 kg is at rest on a flat floor. the coefficient of static friction between the box and floor is 0.37. what is the minimum force needed to set the box in motion across the floor?

The answer is 80 N because

Force of friction = friction coeff. times weight in newtons

weight = 9.81 (on earth) times mass

.37*9.81*22 = 79.8534

round up to 80

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.

b.

explanation: