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Fed [463]
4 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]4 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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8 0
3 years ago
Heather and Jerry are standing on a bridge 49 mm above a river. Heather throws a rock straight down with a speed of 17 m/sm/s .
lara [203]

Answer:

3.467 s

Explanation:

given,

distance , d = 49 mm = 0.049 m

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t₁ = -3.47 s , 0.0028 s

rejecting negative values

t₁ = 0.0028 s

now, time taken by the ball of Jerry

using equation of motion

s = ut +\dfrac{1}{2}at^2

taking downward as negative

-0.049 = 17 t -\dfrac{1}{2}\times 9.8\times t^2

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now,

t_2 = \dfrac{-(-17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}

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rejecting negative values

t₂ = 3.47 s

now, time elapsed is = t₂ - t₁ = 3.47 - 0.0028 = 3.467 s

5 0
3 years ago
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