Answer:
1.3m/s
Explanation:
Data given,
Mass,m=1.0kg,
Amplitude,A=0.10m,
Frequency,f=2.0Hz.
From the equation of a simple harmonic motion, the displacement of the object at a given time is define as
we can express the velocity by the derivative of the displacement,
Hence
at equilibrium, the velocity becomes
Hence if we substitute values we arrive at
Answer:
Maximum force, F = 1809.55 N
Explanation:
Given that,
Diameter of the anterior cruciate ligament, d = 4.8 mm
Radius, r = 2.4 mm
The tensile strength of the anterior cruciate ligament,
We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,
So, the maximum force that could be applied to anterior cruciate ligament is 1809.55 N
Answer:
d = 4.9 m
Explanation:
It is mentioned that a stone is dropped from a certain height. It is required to find the distance covered by it in one second.
The initial speed of the stone is equal to 0 as it was at rest. Let d is the distance covered by the stone.
Using second equation of motion :
Put u = 0 and a = g
So, the distance covered by it in one second is 4.9 m.