Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

v is the speed of cat, 

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.
Answer:
An object's acceleration depends on its mass and on the net force acting on it.
Explanation:
Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.
Answer:
0.0031792338 rad/s
Explanation:
= Angle of elevation
y = Height of balloon
Using trigonometry

Differentiating with respect to t we get

Now, with the base at 200 ft and height at 2500 ft
The hypotenuse is

Now y = 2500 ft


The angle is changing at 0.0031792338 rad/s
Answer:
S = 122.5m
Explanation:
Given the following data;
Acceleration due to gravity = 9.8m/s²
Time, t = 5 seconds
Since it's a free fall, initial velocity, u = 0
To find the displacement, we would use the second equation of motion given by the formula;

Where;
- S represents the displacement or height measured in meters.
- u represents the initial velocity measured in meters per seconds.
- t represents the time measured in seconds.
- a represents acceleration measured in meters per seconds square.
Substituting into the equation, we have;

S = 122.5m.