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Answer:
Explanation:
Given
See attachment for connection
Required
Determine the time constant in (b)
First, we calculate the total capacitance (C1) in (a):
The upper two connections are connected serially:
So, we have:
Take LCM
Cross Multiply
Make the subject
The bottom two are also connected serially.
In other words, the upper and the bottom have the same capacitance.
So, the total (C) is:
The total capacitance in (b) is calculated as:
First, we calculate the parallel capacitance (Cp) is:
So, the total capacitance (C2) is:
Take LCM
Inverse both sides
Both (a) and (b) have the same resistance.
So:
We have:
Time constant is directional proportional to capacitance:
So:
Convert to equation
Make k the subject
Make T2 the subject
Substitute values for T1, C1 and C2
Hence, the time constance of (b) is 0.592 s
Answer:
(a) v = 5.42m/s
(b) vo = 4.64m/s
(c) a = 2874.28m/s^2
(d) Δy = 5.11*10^-3m
Explanation:
(a) The velocity of the ball before it hits the floor is given by:
(1)
g: gravitational acceleration = 9.8m/s^2
h: height where the ball falls down = 1.50m
The speed of the ball is 5.42m/s
(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.
You use the following formula:
(2)
vo: velocity of the ball where it starts its motion upward
You solve for vo and replace the values of the parameters:
The velocity of the ball is 4.64m/s
(c) The acceleration is given by:
The acceleration of the ball is 2874.28/s^2
(d) The compression of the ball is:
THe compression of the ball when it strikes the floor is 5.11*10^-3m