3. A 64 lb weight stretches a spring 4 ft in equilibrium. The weight is initially displaced 6 inches above equilibrium and given
a downward velocity of 6 ft/sec. Find its displacement for t > 0 if the medium resists the motion with a force equal to 3 times the speed in ft/sec (that is, the damping constant c = 3). Note that there is no external force in this case.
1 answer:
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Answer:
Spring constant, k = 24.1 N/m
Explanation:
Given that,
Weight of the object, W = 2.45 N
Time period of oscillation of simple harmonic motion, T = 0.64 s
To find,
Spring constant of the spring.
Solution,
In case of simple harmonic motion, the time period of oscillation is given by :
m is the mass of object
m = 0.25 kg
k = 24.09 N/m
or
k = 24.11 N/m
So, the spring constant of the spring is 24.1 N/m.
Answer:
Explanation:
We are given that
Linear charge density of wire=
Radius of hollow cylinder=R
Net linear charge density of cylinder=
We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R
By Gauss theorem
Where surface area of cylinder=