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Fed [463]
3 years ago
14

An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spi

lled on the arm connected to the line during a laboratory renovation, it is impossible to see the level of the manometer fluid in this arm. During a period when the gas supply is connected to the line but there is no gas flow, a Bourdon gauge connected to the line downstream from the manometer gives a reading of 7.5 psig. The level of mercury in the open arm is 900mm above the lowest part of the manometer.
(a) When the gas is not flowing, the pressure is the same everywhere in the pipe. How high above the bottom of the manometer would the mercury be in the arm connect to the pipe?
(b) When gas is flowing, the mercury level in the visible arm drops by 25mm. What is the gas pressure (psig) at thismoment?
Physics
1 answer:
almond37 [142]3 years ago
3 0

Answer:

A. Using

Pgauge= Pmanometer

And we know that

Pgauge= deta(hpg)

So deta h = Pgauge/density x g

So

= 7.5(6894.76/1psig)/ 13.6*10^3*9.8m/s²)

= 387.9mm

So to find height of pipe connected to the pipe we say

= h -deta h

= 900-387.97mm

=512.02mm

B. We use manometry principle

Pgas+density xg(25*10^3)-PX density{h-(H-0.25)=0

So

Finally Pgas= 6.54psig

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Read 2 more answers
1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/
mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

8 0
3 years ago
Find the volume of a sphere of radius 10 mm.
vredina [299]

Answer:

Explanation: This is done using the equation:

\frac{4}{3} π R^{3}

Because the Radius is a know value. We have the following.

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Which is:

4188.7902 mm

5 0
3 years ago
The blades in a blender rotate at a rate of 6100 rpm. When the motor is turned off during operation, the blades slow to rest in
MissTica

Answer:

<em>155.80rad/s</em>

Explanation:

Using the equation of motion to find the angular acceleration:

\omega_f = \omega_i + \alpha t

\omega_f is the final angular velocity in rad/s

\omega_i  is the initial angular velocity in rad/s

\alpha is the angular acceleration

t is the time taken

Given the following

\omega_f = 6100rpm

Time = 4.1secs

Convert the angular velocity to rad/s

1rpm = 0.10472rad/s

6100rpm = x

x = 6100 * 0.10472

x  = 638.792rad/s

\omega_f = 638.792rad/s\\

Get the angular acceleration:

Recall that:

\omega_f = \omega_i + \alpha t

638.792 = 0 + ∝(4.1)

4.1∝ = 638.792

∝ = 638.792/4.1

∝ = 155.80rad/s

<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>

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