Consider a thin, spherical shell of radius 14.5 cm with a total charge of 29.5 µC distributed uniformly on its surface. (a) Find
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a. 0.5 T
- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position
- The period T is the time the system takes to complete one oscillation
During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.
So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion
and solving for t we find
b. 1.25T
Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that
- the mass takes a time of 1 T to cover a distance of 4A
we can set the following proportion:
And by solving for t, we find
Given :
A box weighing 12,000 N is parked on a 36° slope.
To Find :
What will be the component of the weight parallel to the plane that balances friction.
Solution :
The component of that will be parallel to the plane to balance friction is :
Therefore, component of force to balance friction is F sin 36° .
Hence, this is the required solution.