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3 years ago

Can a golf ball bounce like a rubber ball? Due in 1 hour. 15 points.

Physics

1 answer:

klasskru [66]3 years ago

5 0

Answer:

So, I tested it out and came to a conclusion.

The rubber ball is quite bouncy due to the fact that it is made of rubber which is what makes it bounce back and high, but although the golfball does in fact bounce back it does not bounce as high as the rubber ball due to the fact that it is made of plastic.

Explanation:

Hope this helps you out!

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Aleonysh [2.5K]

Answer:

all the details are in the attached picture, the answers are marked with colour.

5 0

2 years ago

Rocks A and B are located at the same height on top of a hill. The mass of rock A is twice the mass of rock B. How does the pote

Ronch [10]

Potential energy is defined by formula

$U = mgh$

here

m = mass

g = acceleration due to gravity

h = height

Now here two different stones are located at same height

while mass of stone A is twice that of stone B

so here we can say potential energy of A is

$U_a = (2m)gh$

Similarly potential energy of B is

$U_b = mgh$

now if we take the ratio of two energy

$\frac{U_a}{U_b} = 2$

so we can say potential energy of stone A is two times the potential energy of B

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3 years ago

The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de

spin [16.1K]

Answer:

the spring constant k = $5.409*10^4 \ N/m$

the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

Explanation:

From Hooke's Law

$F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m$

Thus; the spring constant k = $5.409*10^4 \ N/m$

The amplitude is decreasing 37% during one period of the motion

$e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}$

$b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s$

Therefore; the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

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3 years ago

How do the forces on a skydiver affect the velocity of the skydiver as they fall?<br><br><br> Hellp

solmaris [256]

Answer:

Explanation:

As a skydiver falls, he accelerates downwards, gaining speed with each second. The increase in speed is accompanied by an increase in air resistance. This force of air resistance counters the force of gravity.

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3 years ago

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IgorLugansk [536]

New force or Old Force

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3 years ago