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3 years ago

What is true about radio waves

1 answer:

6 0

They are a type of electromagnetic radiation with wavelengths in the electromagnetic spectrum longer than infrared light

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A pressure vessel that has a volume of 10m3 is used to store high-pressure air for operating a supersonic wind tunnel. If the ai

Anna71 [15]

Answer:

The mass of air stored in the vessel is 235.34 kilograms.

Explanation:

Let supossed that air inside pressure vessel is an ideal gas, The density of the air ( $\rho$ ), measured in kilograms per cubic meter, is defined by following equation:

$\rho = \frac{P\cdot M}{R_{u}\cdot T}$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$M$ - Molar mass, measured in kilomoles per kilogram.

$R_{u}$ - Ideal gas constant, measured in kilopascal-cubic meters per kilomole-Kelvin.

$T$ - Temperature, measured in Kelvin.

If we know that $P = 2026.5\,kPa$ , $M = 28.965\,\frac{kg}{kmol}$ , $R_{u} = 8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K}$ and $T = 300\,K$ , then the density of air is:

$\rho = \frac{(2026.5\,kPa)\cdot \left(28.965\,\frac{kg}{kmol} \right)}{\left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} \right)\cdot (300\,K)}$

$\rho = 23.534\,\frac{kg}{m^{3}}$

The mass of air stored in the vessel is derived from definition of density. That is:

$m = \rho \cdot V$ (2)

Where $m$ is the mass, measured in kilograms.

If we know that $\rho = 23.534\,\frac{kg}{m^{3}}$ and $V = 10\,m^{3}$ , then the mass of air stored in the vessel is:

$m = \left(23.534\,\frac{kg}{m^{3}} \right)\cdot (10\,m^{3})$

$m = 235.34\,kg$

The mass of air stored in the vessel is 235.34 kilograms.

7 0

3 years ago

Please Help!

wariber [46]

The block is ACCELERATING to the left, but there's not enough information to tell which direction it's moving.

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3 years ago

Help please, last ride guys

Ierofanga [76]

Answer:

Solution ( for fourth attachment ) : 38°C

Tip : Remember the units °C when submitting answer

Explanation:

As you mentioned, we only need the solution for the fourth attachment.

The idea here is that the heat lost by the metal will be equal to the heat gained by the water. We know that the specific heat gained or lost will always be represented by the following formula,

q = m $*$ c

Therefore if we substitute the know values and equate the two equations knowing that " q " is common among them --- ( 1 )

0.33 $*$ 448

Remember that the change in temperature of iron (ΔT) would be represented by final temperature - initial temperature, or final temperature - 693. Similarly the change in temperature of water will be final temperature - 39. Now we can pose the final temperature as a, and solve for a through substitution --- ( 2 )

0.33 $*$ 448

From here on take a look at the attachment. It represents how to receive get a through simple algebra. Here a, the final temperature, is about 38°C. In exact terms it will be $38.03617\dots$ °C.