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inessss [21]
3 years ago
14

A lecture hall having a volume of 106 ft3 contains air at 80°F, 1 atm, and a humidity ratio of 0.01 lb of water vapor per lb of

dry air. Using the appropriate equations, determine(a) the relative humidity.(b) the dew point temperature, in °F.

Engineering
1 answer:
Lyrx [107]3 years ago
3 0

Answer:

(a) %RH = 45%

(b) Dp Temp = 58 °F (14.44 °C)

Explanation:

We can obtain the results easily and faster with a psychometric chart. (See the attachment). In the graph we can see how we obtain results crossing the respective data (marked in red).

First, for the relative humidity, we cross the humidity ratio (starting from the right side) with the dry temperature (horizontal axis). In that way we can read from the relative humidity curves the value of 45%.

After that, we extend the horizontal line from the right side until the left side and cross it with the 100% saturation curve in order to read the dew point temperature. Here we obtain the value 58 °F (14.44 °C)

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​Write 83,120 in expanded form using powers of 10.
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8*10000+3*1000+1*00+2*10+2

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Why do you think the Combination Square is better to use than the tape
Tatiana [17]

Answer:

The correct answer to the following question will be "Because the head acts as a stop".

Explanation:

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8 0
3 years ago
What’s the number of gold atoms in a nanogram? a picogram?
zvonat [6]

Answer :

The number of gold atoms in nanogram is, 3.057\times 10^{12}

The number of gold atoms in picogram is, 3.057\times 10^{9}

Explanation :

As we know that the molar mass of gold is, 196.97 g/mole. That means, 1 mole of gold has 196.97 grams of mass of gold.

As we know that,

1 mole contains 6.022\times 10^{23} number of atoms.

First we have to determine the number of gold atoms in a nanogram.

As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

And, 1 grams of gold contains \frac{1g}{196.97g}\times (6.022\times 10^{23}) number of gold atoms

So, 10^{-9} nanograms of gold contains \frac{1g}{196.97g}\times (10^{-9})\times (6.022\times 10^{23})=3.057\times 10^{12} number of gold atoms

The number of gold atoms in nanogram is, 3.057\times 10^{12}

Now we have to determine the number of gold atoms in a picogram.

As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

And, 1 grams of gold contains \frac{1g}{196.97g}\times (6.022\times 10^{23}) number of gold atoms

So, 10^{-12} picograms of gold contains \frac{1g}{196.97g}\times (10^{-12})\times (6.022\times 10^{23})=3.057\times 10^{9} number of gold atoms

The number of gold atoms in picogram is, 3.057\times 10^{9}

8 0
3 years ago
A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a co
NNADVOKAT [17]

Answer:

The volume up to cylindrical portion is approx  32355 liters.

Explanation:

The tank is shown in the attached figure below

The volume of the whole tank is is sum of the following volumes

1) Hemisphere top

Volume of hemispherical top of radius 'r' is

V_{hem}=\frac{2}{3}\pi r^3

2) Cylindrical Middle section

Volume of cylindrical middle portion of radius 'r' and height 'h'

V_{cyl}=\pi r^2\cdot h

3) Conical bottom

Volume of conical bottom of radius'r' and angle \theta is

V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}

Applying the given values we obtain the volume of the container up to cylinder is

V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}

Hence the capacity in liters is V=32.355\times 1000=32355Liters

3 0
3 years ago
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