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insens350 [35]
3 years ago
12

A tennis ball is released from a height of 4.0 m above the floor. After its third bounce off the floor, it reaches a height of 1

83 cm. What percentage of its mechanical energy does it lose with each bounce?
23%

27%

20%

15%
Physics
1 answer:
diamong [38]3 years ago
5 0

Answer:

The percentage of its mechanical energy does the ball lose with each bounce is 23 %

Explanation:

Given data,

The tennis ball is released from the height, h = 4 m

After the third bounce it reaches height, h' = 183 cm

                                                                       = 1.83 m

The total mechanical energy of the ball is equal to its maximum P.E

                                      E = mgh

                                          = 4 mg

At height h', the P.E becomes

                                      E' = mgh'

                                           = 1.83 mg

The percentage of change in energy the ball retains to its original energy,

                                 \Delta E\%=\frac{1.83mg}{4mg}\times100\%

                                  ΔE % = 45 %

The ball retains only the 45% of its original energy after 3 bounces.

Therefore, the energy retains in each bounce is

                                   ∛ (0.45) = 0.77

The ball retains only the 77% of its original energy.

The energy lost to the floor is,

                                E = 100 - 77

                                   = 23 %

Hence, the percentage of its mechanical energy does the ball lose with each bounce is 23 %      

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In the decreasing order of magnitude, which of the following is correct?
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A soap bubble appears red (λ = 633nm) at the point on its front surface nearest to the viewer. Assuming n = 1.35, what is the sm
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Answer:

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Explanation:

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A communications channel has a bandwidth of 4,000 hz and a signal-to-noise ratio (snr of 30 db. what is the maximum possible dat
levacccp [35]
Through Shannon's Theorem, we can calculate the capacity of the communications channel using the value of its bandwidth and signal-to-noise ratio. The capacity, C, can be expressed as 

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Since the given SN ratio is in decibels, we must first express it as a ratio with no units as

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4 0
3 years ago
A rock is dropped from a vertical cliff. The rock takes 3.00 s to reach the ground below the cliff. A second rock is thrown vert
Phantasy [73]

Answer:

Velocity v= 12.25 \frac{m}{s}

Explanation:

The first rock dropped give the distance Y in meters

Y_{f}=Y_{o}+v_{o}*t +\frac{1}{2}*a*t^{2}\\   Y_{f}=\frac{1}{2}* 9.8 \frac{m}{s^{2} }* 3^{2}   \\Y_{f}=44.1 m

Now the motion of the second rock the time change so to know the velocity

Y_{f}= Y_{o} +v_{o}*t +\frac{1}{2}*a *t^{2} \\v_{o}*t= -Y_{f} +\frac{1}{2} *a*t^{2} \\v_{o} =\frac{-44.1 +0.5 * 9.8* s^{2} }{2} \\v_{o}=  12.25 \frac{m}{s}

7 0
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