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ale4655 [162]
3 years ago
8

List out some use of atmospheric pressure​

Physics
2 answers:
Ne4ueva [31]3 years ago
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 The piston is gradually withdrawn it creates low pressure area in the syringe.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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a golfer tees off and hits a golf ball at a speed of 31 m/s and an angle of 35 degrees. how far did the ball travel before hitti
poizon [28]
Ans: R = Ball Travelled = 92.15 meters.

Explanation:
First we need to derive that formula for the "range" in order to know how far the ball traveled before hitting the ground.

Along x-axis, equation would be:
x = x_o + v_o_xt +  \frac{at^2}{2}

Since there is no acceleration along x-direction; therefore,
x = x_o + v_o_xt

Since v_o_x = v_ocos \alpha and x_o=0; therefore above equation becomes,

x = v_ocos \alpha t --- (A)

Now we need to find "t", and the time is not given. In order to do so, we shall use the y-direction motion equation. Before hitting the ground y ≈ 0 and a = -g; therefore,

=> y = y_o + v_o_yt -  \frac{gt^2}{2}
=> t =  \frac{2v_o_y}{g}

Since v_o_y = sin \alpha; therefore above equation becomes,
t = \frac{2v_osin \alpha }{g}

Put the value of t in equation (A):

(A) => x = v_ocos \alpha \frac{2v_osin \alpha }{g}

Where x = Range = R, and 2sin \alpha cos \alpha = sin(2 \alpha ); therefore above equation becomes:

=> R = (v_o)^2 *\frac{sin(2 \alpha )}{g}

Now, as:
v_o = 31 m/s

and \alpha = 35°
and g = 9.8 m/(s^2)

Hence,
R = (31)^2 *\frac{sin(2 *35 )}{9.8}

Ans: R = 92.15 meters.

-i
7 0
3 years ago
A typical laboratory centrifuge rotates at 3700 rpm . Test tubes have to be placed into a centrifuge very carefully because of t
Alexus [3.1K]

Answer:

Explanation:

acceleration of test tube

= ω² R

= (2πn)² R

= 4π²n²R

n = no of rotation per second

= 3700 / 60

= 61.67

R = .10 m

acceleration

= 4π²n²R

= 4 x 3.14² x 61.67² x .10

= 14999 N Approx

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3 years ago
Which of the following is most likely to happen when energy is transferred to
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Explanation:

the object will begin to move

4 0
3 years ago
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One isotope of a metallic element has the mass number 65 and 35 neutrons in the nucleus. the cation derived from the isotope has
Lina20 [59]
The mass number of an isotope is the sum of the numbers of protons and the numbers of the neutron. Given that the mass number is 65 and the number of neutrons is 35, the number of protons is 30. The atom is then Zinc (Zn). The charge is equal to +2, as it lacks 2 more electrons. 
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3 years ago
The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr
Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by  the following equation:

dsin\theta_{n}=n\lambda   (1)

Where:

d is the width of the slit

\lambda  is the wavelength of the light

n is an integer different from zero.

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda   (2)

Now we have to find the value of \theta_{1}:

sin\theta_{1}=\frac{\lambda}{d}  

\theta_{1}=arcsin(\frac{\lambda}{d})   (3)

We know:

\lambda=104nm=104(10)^{-9}m

In addition we are told the diffraction grating has 5000 slits per mm, this means:

d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}

Substituting the known values in (3):

\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})

\theta_{1}=arcsin(0.52)

<u>Finally:</u>

\theta_{1}=31.33\º >>>This is the first-order diffraction angle

4 0
2 years ago
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