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
Answer:
The tension is 
Explanation:
The free body diagram of the question is shown on the first uploaded image From the question we are told that
The distance between the two poles is 
The mass tied between the two cloth line is 
The distance it sags is 
The objective of this solution is to obtain the magnitude of the tension on the ends of the clothesline
Now the sum of the forces on the y-axis is zero assuming that the whole system is at equilibrium
And this can be mathematically represented as

To obtain
we apply SOHCAHTOH Rule
So 
![\theta = tan^{-1} [\frac{opp}{adj} ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%5B%5Cfrac%7Bopp%7D%7Badj%7D%20%5D)
![= tan^{-1} [\frac{1}{7}]](https://tex.z-dn.net/?f=%3D%20tan%5E%7B-1%7D%20%5B%5Cfrac%7B1%7D%7B7%7D%5D)






Answer:
See the explanation below
Explanation:
This happens because the coffee was at rest, at the time when the car accelerates to move forward the container containing the coffee also accelerates in the same direction, the liquid when moving suddenly spills. The same will happen if the car slows down, any sudden movement changes will cause the coffee to spill, understood by sudden movements when it accelerates or slows down.
Answer:
You're four sentences should include about how the roller coaster has the most potential energy at the top of the track, and the opposing energy, "kinetic" has the most kinetic energy when going down the hill.
Explanation:
Kinetic - In-Motion.
Potential - Gathering Energy to go into Motion.
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