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dusya [7]
3 years ago
7

Water does not run out of dropper unless the rubber-bulb is pressed. Why?

Physics
1 answer:
nekit [7.7K]3 years ago
8 0

Answer:

The pressure inside the dropper is same as the atmospheric pressure when the rubber bulb is not pressed. ... But when we press the rubber bulb the pressure inside the dropper increases and hence the water flows out.  Atmospheric pressure acting from outside the dropper balances the pressure exerted by water and water does not come out of a dropper. On pressing the dropper inside pressure of water becomes more than outside atmospheric pressure and water run out. When we press the bulb of a dropper with its nozzle kept in water, air in the dropper is seen to escape in the form of bubbles. Once we release the pressure on the bulb, water gets filled in the dropper.

                                             <u><em>Thank You</em></u>

                                        <u> </u>Please mark me brainliest

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A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle.
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Answer:

 v_{1fy} = - 0.4549 m / s

Explanation:

This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved

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      1.83 = v_{1f} cos θ + 1.15 cos 23.3

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      1.83 = v_{1f} cos θ + 1.0562

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we divide these two equations

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we substitute in one of the two and find the final velocity of the incident ball

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        v_{1f} = -0.7738 / cos 30.45

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the component and this speed is

       v_{1fy} = v1f sin θ

       v_{1fy} = 0.8976 sin (30.45)

       v_{1fy} = - 0.4549 m / s

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