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3 years ago

If there's air in a ball and it bounces how come a room doesn't bounce

Physics

1 answer:

finlep [7]3 years ago

4 0

Answer:

Funny answer but works!

Explanation:

Not enough air. Go outside, crack open the window and blow air into the room like you would a balloon then close the window quickly. If it doesn't start bouncing, try giving the house a push. Good luck.

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PLZ HELP!!WILL MARK THE BRAINLIEST!!The diagram shows a chromosome pair for an offspring.Which best describes the inheritance of

Dmitrij [34]

Answer:

The offspring is homozygous for eye color

Explanation: I had this question on my test and this was the answer, i hope this helps :)

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2 years ago

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Compare the freezing point of water in the aquanaut’s apartment to its value at the surface. Is it higher, lower, or the same?

Lelu [443]

Answer:

Freezing Point - Lower

Boiling Point - Higher

Solid- liquid transition line in the phase diagram has a negative slope, but the liquid-gas transition line has a positive slope. Since there is more air pressure at 100m it will take less to freeze the water but more to boil it since it requires a larger temperature under larger pressures

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2 years ago

What is the mass of an object if it took 270 J of work to move it 15 meters? Please help!

maxonik [38]

That depends upon the direction the object moves

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2 years ago

A 210 Ohm resistor uses 9.28 W of

IgorLugansk [536]

Resistance=R=210Ohm
Power=9.28W=P

Current=I

$\boxed{\sf P=I^2R}$

$\\ \sf\longmapsto I^2=\dfrac{P}{R}$

$\\ \sf\longmapsto I^2=\dfrac{9.28}{210}$

$\\ \sf\longmapsto I^2\approx0.04$

$\\ \sf\longmapsto I\approx\sqrt{0.04}$

$\\ \sf\longmapsto I\approx\sqrt{\dfrac{4}{100}}$

$\\ \sf\longmapsto I\approx\dfrac{\sqrt{4}}{\sqrt{100}}$

$\\ \sf\longmapsto I\approx\dfrac{2}{10}$

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2 years ago

The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, what should be the focal length an

yuradex [85]

Answer:

The focal length of the appropriate corrective lens is 35.71 cm.

The power of the appropriate corrective lens is 0.028 D.

Explanation:

The expression for the lens formula is as follows;

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

Here, f is the focal length, u is the object distance and v is the image distance.

It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.

Put v= -71.4 cm and u= 24.0 cm in the above expression.

$\frac{1}{f}=\frac{1}{24}+\frac{1}{-71.4}$