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Zolol
3 months ago
9

A tennis ball bounces on the floor three times. If each time it loses 22.0% of its energy due to heating, how high does it rise

after the third bounce, provided we released it 2.3 m from the floor?
Physics
1 answer:
Ostrovityanka [3.2K]3 months ago
6 0

Answer:

H = 109.14 cm

Explanation:

Given,                                                            

Assume that the total energy equals 1 unit.                                

Energy remaining after the first collision = 0.78 x 1 unit

Balance after the first impact = 0.78 units

Remaining energy after the second impact = 0.78 ^2 units

Balance after the second impact = 0.6084 units

Remaining energy after the third impact = 0.78 ^3 units

Balance after the third impact = 0.475 units

The height reached after the third collision is equivalent to the remaining energy.

Let H denote the height achieved after three bounces.

0.475 (m g h) = m g H                  

H = 0.475 x h                                    

H = 0.475 x 2.3 m                          

H = 1.0914 m                      

H = 109.14 cm                      

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4 0
3 months ago
Can pockets of vacuum persist in an ideal gas? Assume that a room is filled with air at 20∘C and that somehow a small spherical
kicyunya [3294]

Answer:

The time required is 20 μs

Explanation:

Here is the data provided:

temperature = 20°C  = 293 K

radius = 1 cm

atomic mass of air = 29 u

To determine

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solution:

We calculate the root mean square velocity of air particles. This can be expressed as:

\frac{1}{2}mv^2 = \frac{3}{2}RT

where m indicates mass, t is temperature, v is speed, and R is the ideal gas constant, which is approximately 8.3145 (kg·m²/s²)/K·mol.

v = \sqrt{\frac{3RT}{M} }............................1

v = \sqrt{\frac{3(8.314)293}{29*10{-3}kg} }

Resulting in v = 501.99 m/s.

<pNow, to cover the distance of 1 cm,<pThe duration needed for air is calculated as:

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so, time taken = 19.92 × 10^{6} seconds = 20μs.

Thus, the required time is 20 μs.

3 0
4 months ago
A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p
Keith_Richards [3271]

Response:

Q = 8,345 * v

Clarification:

We need an expression that shows how much water has been drained from the tub. This is represented by v, which indicates how many gallons have flowed out since the plug was taken out. Each gallon removed equates to 8.345 pounds of water, so the weight of the drained water Q in pounds as a function of v can be expressed as:

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Where v signifies the number of gallons emptied from the tub.

Have a great day! Let me know if there's anything else I can assist with.

8 0
3 months ago
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