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Fittoniya [83]
4 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]4 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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6 0
3 years ago
An add tape of 101 ft is incorrectly recorded as 100 ft for a 200-ft distance. What is
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Answer:

the correct distance is 202 ft

Explanation:

The computation of the correct distance is shown below:

But before that correction to be applied should be determined

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Now the correct distance is

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4 0
3 years ago
A program contains the following function definition: int cube(int number) { return number * number * number; } Write a stateme
Nonamiya [84]

Answer:

The statement can be written as

int result = cube(4);

Explanation:

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To call a function, just simply write the function name followed with parenthesis (e.g. <em>cube()</em>). Within the parenthesis, we can include zero or one or more than one values as argument(s) (e.g. <em>cube(4)</em>).

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8 0
4 years ago
An electrical heater is a form of sensible heating process, and heats 0.1m/s of air from 15°C and 80% RH to 50°C? The barometric
lawyer [7]

Answer:

The heater load =35 KJ/kg

Explanation:

Given that

At initial condition

Temperature= 15°C

RH=80%

At final condition

Temperature= 50°C

We know that in sensible heating process humidity ratio remain constant.

Now from chart

At temperature= 15°C and RH=80%

h_1=38 \frac{KJ}{kg},v=0.8 \frac{m^3}{kg}

At  temperature= 50°C

h_2=73 \frac{KJ}{kg}

So\ the\ heater\ load =h_2-h_1

The heater load = 73 - 38 KJ/kg

The heater load =35 KJ/kg

3 0
3 years ago
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