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Fittoniya [83]
3 years ago
6

On a given day, a barometer at the base of the Washington Monument reads 29.97 in. of mercury. What would the barometer reading

be when you carry it up to the observation deck 500 ft above the base of the monument?
Engineering
1 answer:
Umnica [9.8K]3 years ago
8 0

Answer:

The barometer reading will be 29.43 in

Explanation:

Using the formula of pressure variation

p2 - p1 = -yair * H

= 7.65 * 10^{-2} \frac{lb}{ft^{3} } * 500 ft\\

= 38.5 \frac{lb}{ft^{2} }

According to the relationship between the pressure and the height of the mercury column

p = yHg * h --> where yHg and h is the barometer reading

yHg (\frac{29.97}{12} ft) - yHg * h1 = 38.5 \frac{lb}{ft^{2} }

h1 = (\frac{29.97}{12} ft) - \frac{38.5 \frac{lb}{ft^{2} } }{847 \frac{lb}{ft^{3} } }

     [(\frac{29.97}{12} ft) - 0.0455 ft] - 12 \frac{in}{ft} \\\\h1 = 29.43 in  

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A vacuum pump is used to drain a basement of 20 °C water (with a density of 998 kg/m3 ). The vapor pressure of water at this tem
lord [1]

Answer:

The maximum theoretical height that the pump can be placed above liquid level is \Delta h=9.975\,m

Explanation:

To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature.  As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:

\frac{\Delta P}{\rho}+g\, \Delta h +\frac{1}{2}  \Delta v^2 =0

(\rho stands here for density, h for height)

Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:

\frac{\Delta P}{\rho}+g\, \Delta h  =0

\Delta P= -g\, \rho\, \Delta h

This means that pressure drop is proportional to the suction lift's height.

We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.

That means:

\Delta P = 2.34\,kPa- 100 \,kPa = -97.66 \, kPa\\

We insert that into our last equation and get:

\frac{ \Delta P}{ -g\, \rho\,}= \Delta h\\\Delta h=\frac{97.66 \, kPa}{998 kg/m^3 \, \, 9.81 m/s^2} \\\Delta h=9.975\,m

And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.

8 0
3 years ago
Which one of the following faults cause the coffee in a brewer to keep boiling after the brewing cycle is finished?
MrRa [10]

Answer:

  C.  Welded contacts on the thermostat

Explanation:

Any fault that keeps the heating element heating when it should not is a fault that will cause the symptom described. The details <em>depend on the design of the brewer</em> (not given).

"A short at the terminals" depends on what terminals are being referenced. The device on-off switch terminals are normally connected together when the brewer is turned on, so a short there may not be observable.

"Welded contacts on the thermostat" will have the observed effect if the thermostat is the primary means of ending the brewing cycle. If the thermostat of interest is an overheat protective device not normally involved in ending the brewing cycle, then that fault may not cause the observed symptom.

__

If the heating element is open-circuit, no heating will occur. A gasket leak may cause a puddle, but may have nothing to do with the end of the brewing cycle. (Loss of water can be expected to end boiling, rather than prolong it.)

8 0
3 years ago
A river has an average rate of water flow of 59.6 M3/s. This river has three tributaries, tributary A, B and C, which account fo
Fiesta28 [93]

Answer:

50421.6 m³

Explanation:

The river has an average rate of water flow of 59.6 m³/s.

Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:

Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s

The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken

time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds

The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³

3 0
3 years ago
A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring
ANTONII [103]

Answer:

The volume of the ring shaped solid that remains is 21 unit^3.

Explanation:

The total volume of the sphere is given as:

Volume of Sphere = (4/3)πr^3

where, r = radius of sphere

Volume of Sphere = (4/3)(π)(5)^3

Volume of Sphere = 523.6 unit^3

Now, we find the volume of sphere removed by the drill:

Volume removed = (Cross-sectional Area of drill)(Diameter of Sphere)

Volume removed = (πr²)(D)

where, r = radius of drill = 4

D = diameter of sphere = 2*5 = 10

Therefore,

Volume removed = (π)(4)²(10)

Volume removed = 502.6 unit^3

Therefore, the volume of ring shaped solid that remains will be the difference between the total volume of sphere, and the volume removed.

Volume of Ring = Volume of Sphere - Volume removed

Volume of Ring = 523.6 - 502.6

<u>Volume of Ring = 21 unit^3</u>

5 0
3 years ago
Document the XSS stored exploit script: Use the View Source feature of the web page and create a screenshot of the few lines cod
Natali [406]

Answer:

Hold on let me ask my brother

Explanation:

5 0
3 years ago
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