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: A pendulum is an object hung from a fixed point that swings back and forth under the action of gravity.
Example: playground swing
Answer:
Factors Affecting the Strength of the Magnetic Field of an Electromagnet: Factors that affect the strength of electromagnets are the nature of the core material, strength of the current passing through the core, the number of turns of wire on the core and the shape and size of the core.
Explanation:
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Answer:
<em>The PE of the mass increased by 6,972.95 J</em>
Explanation:
<u>Gravitational Potential Energy</u>
It's the energy stored in an object because of its vertical position or height in a gravitational field.
It can be calculated with the equation:
U=m.g.h
Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or 9.8 m/s^2.
We are given the mass of m=16 slug raised by a height h=10 ft. Both units will be converted to SI standard:
1 slug = 14.59 Kg, thus
16 slug = 16*14.59 Kg=233.44 Kg
1 ft = 0.3048 m, thus:
10 ft = 10*0.3048 m = 3.048 m
Thus, the PE of the mass increased by:
U = 233.44 * 9.8 * 3.048 = 6,972.95 J
the PE of the mass increased by 6,972.95 J
