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Ahat [919]
2 years ago
13

What is the proportionality between pressure and temperature, and the proportionality between atmospheric pressure and tempratur

eIs it a direct relationship or inverse relationship?
Physics
1 answer:
Alborosie2 years ago
6 0

According to Ideal gasTo solve this problem, the fastest relationship allows us to observe the proportionality between the two variables would be the one expressed in the ideal gas equation, which is

PV= NRT

Here

P = Pressure

V = Volume

N = Number of moles

R = Gas constant

T = Temperature

We can see that the pressure is proportional to the temperature, then

P \propto T

This relationship can be extrapolated to all the scenarios in which these two variables are related. As the pressure increases the temperature increases. The same goes for the pressure in the atmosphere, for which an increase in this will generate an increase in temperature. This variable can be observed in areas of different altitude. At higher altitude lower atmospheric pressure and lower temperature.

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As a marble rolls across the floor, it gradually slows to a stop due to friction. Which statement best describes the change in m
Olegator [25]

Answer:

I belive it would be "C"

Explanation:

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3 0
3 years ago
A circular radar antenna on a Coast Guard ship has a diameter of 2.10 m and radiates at a frequency of 16.0 GHz. Two small boats
Anna35 [415]

Answer:

d = 76.5 m

Explanation:

To find the distance at which the boats will be detected as two objects, we need to use the following equation:

\theta = \frac{1.22 \lambda}{D} = \frac{d}{L}

<u>Where:</u>

θ: is the angle of resolution of a circular aperture

λ: is the wavelength

D: is the diameter of the antenna = 2.10 m

d: is the separation of the two boats = ?

L: is the distance of the two boats from the ship = 7.00 km = 7000 m

To find λ we can use the following equation:

\lambda = \frac{c}{f}

<u>Where:</u>

c: is the speed of light = 3.00x10⁸ m/s

f: is the frequency = 16.0 GHz = 16.0x10⁹ Hz

\lambda = \frac{c}{f} = \frac{3.00 \cdot 10^{8} m/s}{16.0 \cdot 10^{9} s^{-1}} = 0.0188 m            

Hence, the distance is:

d = \frac{1.22 \lambda L}{D} = \frac{1.22*0.0188 m*7000 m}{2.10 m} = 76.5 m

Therefore, the boats could be at 76.5 m close together to be detected as two objects.

 

I hope it helps you!

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2 years ago
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Can someone please help me it would mean alot​
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