Answer:

Explanation:
Given that,
Mass of the bowling ball, m = 5 kg
Radius of the ball, r = 11 cm = 0.11 m
Angular velocity with which the ball rolls, 
To find,
The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.
Solution,
The translational kinetic energy of the ball is :



The rotational kinetic energy of the ball is :



Ratio of translational to the rotational kinetic energy as :

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2
Well the chocolate bar may melt at the heat of the machine but why is there a chocolate bar there in the first place is my question xD
Answer:
69.68 N
Explanation:
Work done is equal to change in kinetic energy
W = ΔK = Kf - Ki = 
W = 
where m = mass of the sprinter
vf = final velocity
vi = initial velocity
W = workdone
kf = final kinetic energy
ki = initial kinetic energy
d = distance traveled
Ftotal = total force
vf = 8m/s
vi= 2m/s
d = 25m
m = 60kg
inserting parameters to get:
W = ΔK = Kf - Ki = 



= 39.7
we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

-- pick a planet from the table
-- take it's mass and radius from the table, and plug them into the big ugly formula above the table
-- do the arithmetic with your pencil or your calculator. The answer is the acceleration of gravity on the planet you picked. Write it down so you don't lose it.
-- do the same for the other 3 planets in the table
Answer:
The difference in the decibel corresponses to a constant difference in the loudness perceived.
The refore the sound intensity from the orchestra is like 100 times that of the violin.
Explanation: