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:
W=315 x 10⁵ J
Explanation:
Given that
F= 2.5 x 10⁵ N
d= 90 m
K.E.=5.4 x 10⁷ J
We know that work done by all force is equal to the change in kinetic energy
Lets take work done by catapult is W
W + F.d= K.E.
W= 5.4 x 10⁷ - 2.5 x 10⁵ x 90 J
W= (540 - 25 x 9) 10⁵ J
W=315 x 10⁵ J
The gravitational force between a mass and the Earth is the object'sweight. Mass is considered a measure of an object's inertia, and its weight is the force exerted on the object in a gravitational field. On the surface of the Earth, the two forces are related by the acceleration due to gravity: Fg = mg.
Hoped this helped!
In theory, yes. The 2 problems are the materials used for clinical thermometers, & the temperature capacity of the clinical thermometer. If anything, change the material & extend the measurement threshold. At that point, it wouldn´t be used for clinical garbage anymore.
Answer:
oxygen
Explanation:
as it is in the air it can't be depleted or used up