Asize-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m above the base of the hill. we can model this ball as a thin-walled hollow sphere. (a) at what rate was it rotating at the base of the hill

W = 0.678 rad/s  

Explanation:

Using the conservation of energy:

Roll up and hill without slipping is the sumatory of two energys, rotational and translational, so:

where I is the moment of inertia, W the angular velocity at the base of the hill, m the mass of the ball, V the velocity at the base of the hill, g the gravity and h the altitude.

First, we will find the moment of inertia as:

I =

where m is the mass and R the radius, so:

I =

I = 36.26 Kg*m^2

Then, replacing values on the initial equation, we get:

also we know that:

V =WR

so:

Finally, solving for W, we get:

W = 0.678 rad/s

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.