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3 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 The piston is gradually withdrawn it creates low pressure area in the syringe.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]3 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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Help please, what does gravitational force do according to these answer choices?

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Answer:

the answer is C

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A motor does a total of 480 J of work if 5 s to lift a 12 clock to the top of a ramp. The average power developed by the motor i

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2 years ago

Read 2 more answers

You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration

jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

$y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}$

Where:

$y=0m$ is the final height of the ball

$y_{o}=40 m$ is the initial height of the ball

$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$\theta=30\°$

Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

$0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}$

Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Where:

$a=4.9$

$b=-5$

$c=-40$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

Solving the equation and choosing the positive result we have:

$t=3.41 s$ This is the time the ball is in air

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4 years ago

A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit

Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

$y = \frac{1}{2} at^2+v_0t+y_0$

Where,

a = acceleration

t = time

$v_0$ = Initial velocity

$y_0$ = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

$v = \frac{x}{t}$

x = Displacement

t = time

With the data we have we can find the time as well

$v = \frac{x}{t}$

$t = \frac{x}{v}$

$t = \frac{18.6}{34}$

$t = 0.547s$

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

$y = \frac{1}{2} at^2+v_0t+y_0$

$y = \frac{1}{2} gt^2+0+0$