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2 years ago

List out some use of atmospheric pressure

2 answers:

7 0

The atmospheric air pressure acts in isosteric manner. That is air pressure is exerted in every direction on the object.

How does it help….?

1 - We suck out air in the straw so that air pressure within the straw drops and becomes less than the atmospheric air pressure. The liquid rises in the straw reaches our mouth.

2 - Domestic Vacuum cleaners suck in air and create low pressure area, around the point of suction. The surrounding air is still at the atmospheric air pressure. The atmospheric air pressure pushes in more air to the low pressure area. But the suction continues. This results in an air current formation. The air current carries dust particles along with. The dust particles are trapped in the trash bags and the air is let out.

3 - Siphons work because of air pressure.

4 - The injection syringes utilise atmospheric air pressure. In the first step the piston is drawn in. This creates low pressure within the syringe. The atmospheric air rushes in. The needle is inserted through the cap and the air is pushed in to the bottle. It creates high pressure area in the bottle. The piston is gradually withdrawn it creates low pressure area in the syringe. Medicine in the bottle is at higher pressure so it flows in through the needle into the syringe.

5 - Automobile servicing - uses air pressure to create strong water current to wash away stubborn sticky dust.

6 - Air Brakes - Air pressure is used to activate brakes in trains.

inna [77]2 years ago

4 0

Answer:

<h2>1)Straw : When a man sucks up the fluid from straw the pressure inside is moderately low and accordingly atmospheric pressure outside powers up the fluid into straw. 2)Vacuum cleaner : When a vacuum cleaner is exchanged on, the pressure inside tumble off because of air inside.</h2>

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A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period 2.68 s, at what ampli

fredd [130]

Answer:

Part a)

$A = 1.78 m$

Part b)

$f = 2 rev/s$

Explanation:

Part A)

As we know that time period of the motion is given as

$T = 2.68 s$

so we have

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{2.68}$

$\omega = 2.34 rad/s$

now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

$mg = m\omega^2 A$

so we have

$A = \frac{g}{\omega^2}$

$A = \frac{9.81}{2.34^2}$

$A = 1.78 m$

Part b)

Now if the amplitude of the SHM is 6.23 cm

and now at this amplitude if object will lose the contact then in that case again we have

$mg = m\omega^2 A$

$g = \omega^2 (0.0623)$

$\omega = 12.5 rad/s$

so now we have

$2\pi f = 12.5$

$f = 2 rev/s$

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2 years ago

If it takes 50.0 seconds to lift 10.0 Newtons of books to a height of 7.0

Ilya [14]

Work of the force = 10 N

Time required for the work = 50 sec

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We are given with the value of work and time in the question.

Substitute the values in the formula of power and then you'll get the power required.

We know that,

w = Work

p = Power

t = Time

By the formula,

Given that,
Work (w) = 7 m = 70 Joules
Time (t) = 50 sec
Substituting their values,

p = 70/50

p = 1.4 watts

Therefore, the power required is 1.4 watts.

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scoundrel [369]

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List out some use of atmospheric pressure​

List out some use of atmospheric pressure