Acoin is placed 32 cm from the center of a horizontal turntable, initially at rest. the turntable then begins to rotate. when the speed of the coin is 110 cm/s (rotating at a constant rate), the coin just begins to slip. the acceleration of gravity is 980 cm/s^2 . what is the coefficient of static friction between the coin and the turntable?

0.39

Explanation:

distance from the center (r) = 32 cm = 0.32 m

speed of the coin (v) = 110 cm/s = 1.1 m/s

acceleration due to gravity (g) = 980 m/s^{2} = 9.8 m/s^{2}

find the coefficient of static friction (k) between the coin and the turn table

frictional force = kmg

before the table begins to move, the frictional force balances the centripetal force ()

therefore

frictional force = centripetal force

kmg =

kg =

k = ÷ g

k = ÷ 9.8 = 0.39

c, 1a hope this

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.