The answer is c.
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Answer:
0.76
Explanation:
we are given:
radius (r) =5.7 m
speed (s) = 1 revolution in 5.5 seconds
acceleration due to gravity (g) = 9.8 m/s^{2}
coefficient of friction (Uk) = ?
we can get the minimum coefficient of friction from the equation below
centrifugal force = frictional force
m x r x ω^{2} = Uk x m x g
r x ω^{2} = Uk x g
Uk = 
where ω (angular velocity) = 
= 
= 1.14
Uk = 
= 0.76
Yes , increased tension suggests increased molecular attraction between the molecules of the ropes which affect the increase in the speed of wave.
Answer: To increase the rigidity of the system you could hold the ruler at its midpoint so that the part of the ruler that oscillates is half as long as in the original experiment.
Explanation:
When a rule is displaced from its vertical position, it oscillates back and forth because of the restoring force opposing the displacement. That is, when the rule is on the left there is a force to the right.
By holding a ruler with one hand and deforming it with the other a force is generated in the opposite direction which is known as the restoring force. The restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. The momentum gained causes the ruler to move to the right leading to opposite deformation. This moves the ruler again to the left. The whole process is repeated until dissipative forces reduce the motion causing the ruler to come to rest.
The relationship between restoring force and displacement was described by Hooke's law. This states that displacement or deformation is directly proportional to the deforming force applied.
F= -kx, where,
F= restoring force
x= displacement or deformation
k= constant related to the rigidity of the system.
Therefore, the larger the force constant, the greater the restoring force, and the stiffer the system.
is the acceleration of the box.
<u>Explanation:</u>
Given data:
Mass of the box = 3.74 kg
Flat friction-less ground is pulled forward by a 4.20 N force at a 50.0 degree angle and pulled back by a 2.25 N force at a 122 degree angle.
First, we need to find the net horizontal force acting on the box. With the given data, the equation can be formed as below. Net horizontal force acting on the box (F) is given by
F = 2.699676 – 1.192275 = 1.507 N
Next, find acceleration of the box using Newton's second law of motion. This states that the link between mass (m) of an objects and the force (F) required to accelerate it. The equation can be given as
