What is the relationship between temperature, dew point temperature, and air pressure

1 answer:

6 0

Dew point is the temperature at which the water vapor in the air condenses , then evaporates. The barometric or air pressure is independent from the dew point.

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Helpppppppp pleaseeeee!!

BabaBlast [244]

Im pretty sure its nucleus !

3 0

2 years ago

10. Suppose the tern travels 1.70*10^ km south, only to encounter bad weatherInstead of trying to fly around the stormthe tem tu

Ghella [55]

We can define average speed as the quotient between the whole distance traveled and the time it takes to travel that distance, and the velocity as the quotient between the displacement and the time it takes to displace that quantity.

We will find that:

Average speed = 2.6×10^2 km/day

Velocity = 2.52×10^2 km/day

So we know that:

The tent travels:

1.70×10^4 km to the south

6.00×10^2 km to the north

1.44×10^4 km to the south.

The total distance traveled is just the sum of the 3 above distances:

total distance = 1.70×10^4 km + 6.00×10^2 km + 1.44×10^4 km

Notice that the second term has a different exponent than the others, so to have the same exponent we can write:

6.00×10^2 km = 0.06×10^4 km

Now we get:

total distance = 1.70×10^4 km + 0.06×10^4 km + 1.44×10^4 km

total distance = (1.70 + 0.06 + 1.44)×10^4 km

<u>total distance = 3.2×10^4 km</u>

And it travels that distance in 122 days, then the average speed is:

AS = (3.2×10^4 km)/(122 days) = 2.6×10^2 km/day

Now we need to find the displacement, which is the difference between the final position and the initial position, the displacement will just be:

d = 1.70×10^4 km - 0.06×10^4 km + 1.44×10^4 km

(We add the distance traveled to south and subtract the distance that it travels due north)

d = (1.70 - 0.06 + 1.44)×10^4 km = 3.08×10^4 km

Then the velocity will be:

v = (3.08×10^4 km)/(122 days) = 2.52×10^2 km/day

Because we defined south as positive and north as negative, having a positive velocity means that the direction of the <u>velocity is due south.</u>

If you want to learn more, you can read:

brainly.com/question/12322912

3 0

2 years ago

An electric hoist is used to lift a 235.0 kg load to a height of 69.0 m in 38.1 s. (a) what is the power of the hoist in kw?

Ugo [173]

$Given:\\m=235.0kg\\h=69.0m\\t=38.1s\\g=9.81 \frac{m}{s^2} \\\\Find:\\P=?\\\\Solution:\\\\P= \frac{W}{t}\\\\W=\Delta E_p\\\\E_p_0=0\Rightarrow \Delta E_p=E_p\\\\E_p=mgh \\\\P= \frac{mgh}{t} \\\\P= \frac{235kg\cdot9.81 \frac{m}{s^2}\cdot69m}{38.1s} \approx 4175W=4.175kW$