A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that numerically e
qual to the instantaneous velocity. The mass is initially released from a point 1 ft above the equilibrium position with a downward velocity of 14 feet/second. Determine the time at which the mass passes through the equilibrium position
1 answer:
Answer:
the time at which it passes through the equilibrum position is:
t = 0.1 second
Explanation:
given
w= 4pounds
k(spring constant) = 2lb/ft
g(gravitational constant) = 10m/s² = 32ft/s²
β(initial point above equilibrum) = 1
velocity = 14ft/s
attached is an image showing the calculations, because some of the parameters aren't convenient to type.
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Explanation:
It is given that,
Length of rod, l = 14 cm = 0.14 m
Total charge,
We need to find the magnitude and direction of the net electric field produced by the charged rod at a point 36.0 cm to the right of its center along the axis of the rod, z = 36 cm = 0.36 m
Electric field at the axis of rod is given by :
Where
is the linear charge density,
So,
E = −7849988.22 N/C
or
Negative sign shows the direction of electric field is inward in all direction. Hence, this is the required solution.