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3 years ago

How an application of atmospheric device work?example siphon

Physics

1 answer:

Vika [28.1K]3 years ago

6 0

Answer:

A practical siphon, operating at typical atmospheric pressures and tube heights, works because gravity pulling down on the taller column of liquid leaves reduced pressure at the top of the siphon (formally, hydrostatic pressure when the liquid is not moving).

I hope it's helpful!

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A bowling ball with a circumference of 27 in. weighs 14.8 lb and has a radius of gyration of 3.43 in. If the ball is released wi

Marrrta [24]

Answer:

Distance covered is 37.63 ft

Solution:

As per the question:

Circumference of the bowling ball, C = 27 in

Weight of the bowling ball, w = 14.8 lb

Radius of gyration, k = 3.43 in

Velocity of release of the ball, v' = 23 ft/s = 276 in/s

Coefficient of friction, $\mu = 0.19$

Now,

We know that the circumference of the circle is given by:

C = $2\pi R$

27 = $2\pi R$

$R = \frac{27}{2\pi } = 4.29\ in$

Also, in case of rolling, we know that:

Angular velocity, $\omega = \frac{v}{R}$

$v = \omega R$

Now, applying the conservation of angular momentum along the floor:

Initial angular momentum = Final angular momentum

$m\omega' = m\omega + I\omega$

Moment of Inertia, I = $\frac{2}{5}mR^{2} = mk^{2}$

$mv'R = mvR + mk^{2}\times \frac{v}{R}$

$v'R = vR + k^{2}\times \frac{v}{R}$

$276\times 4.29 = 4.29v + 3.43^{2}\times \frac{276}{4.29}$

$1184.04 - 756.903 = 4.29v$

v = 99.56 in/s

Now,

Friction force, f = $\mu mg$

Also, acceleration of the ball can be computed as:

$\mu mg = - ma$

$a = - \mu g = - 0.19\times 386.09 = - 73.357\ in/s^{2}$

Now, the distance, d covered by the ball before rolling without slipping:

$v^{2} = v'^{2} + 2ad$

$99.56^{2} = 276^{2} + 2\times (- 73.357)d$

$99.56^{2} - 276^{2} = 2\times (- 73.357)d$

d = 451.65 in = 37.53 ft

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3 years ago

4. The winding ridge of a screw

ZanzabumX [31]

Answer:

Thread.

Explanation:

The most common form consists of a cylindrical shaft with helical grooves or ridges called threads around the outside.

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3 years ago

Due to the tilt of the Earth's axis,

Lynna [10]

6 months with no sunrise and the other 6 without a sunset, at some places on Earth, are the result of the orientation of Earth's axis.

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3 years ago

Which of the following is an example of potential energy?

Papessa [141]

Answer:

Explanation:

potential energy is stored energy so the ball has potential energy to bounce or roll which would then have been converted to kinetic but the rest are in motion meaning energy is no longer stored but used in motion and therefore cannot be potential energy

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3 years ago

A plane flies from alphaville to betaville and then back to alphaville. when there is no wind, the round trip takes 4 hours and

kykrilka [37]

Refer to the diagram shown below.

d = distance (miles) from Alphaville to Betaville.
v = speed (mph) of the plane with no wind.

With no wind:
The time taken to travel a distance of 2d is 4 hrs, 48 min = 4.8 hrs.
Therefore
2d/v = 4.8
v = 2d/4.8 = 0.4167d mph (1)

With the wind:
The velocity from Alphaville to Betaville is (v + 100) mph.
The time of travel is
t₁ = d/(v+100) h
The velocity from Betaville to Alphaville is (v - 100) mph.
The time of travel is
t₂ = d/)v-100) h
Because the return trip takes 5 hours, therefore
t₁ + t₂ = 5
$\frac{d}{v+100} + \frac{d}{v-100} =5 \\ \frac{2vd}{v^{2}-100^{2}} =5 \\ 2vd = 5(v^{2}-10^{4})$ (2)

From (1), obtain
2(0.4167)d² = 5[(0.4167d)² - 10⁴]
0.8334d² = 0.8682d² - 5 x 10⁴
0.0348d² = 5 x 10⁴
d = 1198.7 mi

Answer: 1199 miles (nearest integer)

8 0

3 years ago

How an application of atmospheric device work?example siphon​

How an application of atmospheric device work?example siphon