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xeze [42]
2 years ago
8

A gauge is attached to a pressurized nitrogen tank reads a gauge pressure of 28 in of mercury. If atmospheric pressure is 14.4 p

sia, what is the absolute pressure in the tank
Physics
1 answer:
Gekata [30.6K]2 years ago
7 0

Answer:

The absolute pressure is 28.15 psi.

Explanation:

Gauge pressure = 28 inch of Mercury

Absolute pressure, Po = 14.4 psi

The absolute pressure is the sum of the gauge pressure and the absolute pressure.  

gauge pressure = 28 inch = 0.7112 m of Mercury

= 0.7112 x 13.6 x 1000 x 9.8 = 94788.736 Pa

= 13.75 psi

The absolute pressure is

P = 14.4 + 13.75 = 28.15 psi

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Rina8888 [55]

Answer:first law

Explanation:

it states the a body in motion or rest maintain its state until an external force is acted on it

8 0
3 years ago
Which statement is a scientific theory?
Dmitry [639]

Answer:

I think it  is D

Explanation:

7 0
2 years ago
Read 2 more answers
A tube is sealed at both ends and contains a 0.0100-m long portion of liquid. The length of the tube is large compared to 0.0100
Ahat [919]

Answer:

31.321 rad/s

Explanation:

L = Tube length

A = Area of tube

\rho = Density of fluid

v = Fluid velocity

m = Mass = \rho Al

Centripetal force is given by

F=\dfrac{mv^2}{L}\\ F=\dfrac{m(\omega L)^2}{L}\\ F=m\omega^2\\ F= 0.01A\rho\omega^2L

Pressure is given by

P=\dfrac{F}{A}=\rho gL\\\Rightarrow \dfrac{0.01A\rho\omega^2L}{A}=\rho gL\\\Rightarrow 0.01\omega^2=g\\\Rightarrow \omega^2=\dfrac{g}{0.01}\\\Rightarrow \omega=\sqrt{\dfrac{g}{0.01}}\\\Rightarrow \omega=\sqrt{\dfrac{9.81}{0.01}}\\\Rightarrow \omega=31.321\ rad/s

The angular speed of the tube is 31.321 rad/s

5 0
3 years ago
A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the m
konstantin123 [22]

Answer:

the mass of the body is 0.02 kg.

Explanation:

Given;

relative density of the oil, \gamma _0 = 0.875

mass of the object in oil, M_o = 0.013 kg

mass of the object in water, M_w = 0.012 kg

let the mass of the object in air = M_a

weight of the oil, W_0 = M_a - 0.013

weight of the water, W_w = M_a - 0.012

The relative density of the oil is given as;

\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg

Therefore, the mass of the body is 0.02 kg.

6 0
2 years ago
A car owner forgets to turn off the headlights of his car while it is parked in his garage. If the 12.0-V battery in his car is
avanturin [10]

Answer: 22.6 hours

Explanation:

The power is the measure of the rate of energy.

In this problem, the 12.0 V battery is rated at 51.0 Ah, which means it delivers 51.0 A of current in a time of t = 1 h = 3600 s. The power delivered by the battery can be written as

P=IV

where

I is the current

V = 12.0 V is the voltage of the battery

So the energy delivered by the battery can be written as

E=Pt=VIt

Where

It=51.0 A\cdot h = 51.0 A \cdot 3600 s/h=183,600 A\cdot s

So the energy delivered is

E=(12.0)(183,600)=2.2\cdot 10^6 J

At the same time, the headlight consumes 27.0 W of power, so 27 Joules of energy per second; Therefore, it will remain on for a time of:

t=\frac{2.2\cdot 10^6 J}{27.0 W}=81481 s = 22.6 h

3 0
3 years ago
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