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tekilochka [14]
3 years ago
12

What is clinograph and how to read it​

Physics
2 answers:
swat323 years ago
7 0

Answer:

Explanation:

A climograph is a graphical representation of a location's basic climate. Climographs display data for two variables:

(a) monthly average temperature  

(b) monthly average precipitation. These are useful tools to quickly describe a location's climate.

zhenek [66]3 years ago
4 0

A climograph is a graphical representation of a location's basic climate. Clinograph is an instrument used to draw parallel lines to the inclined lines. It contains one adjustable wing or strip which can be adjusted to required angle. So, it can be termed as adjustable set square

Hope this helps!:)
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How to do the equation p1v1 = p2v2 as an equation layed out like:
KonstantinChe [14]

P1 and P2 are the pressures, and V1 and V2 are the volumes. So you take the first pressure and volume you are given and place them into the equation P1V1 so the first part of the equation would be 101000*0.5 = P2V2. You then rearrange the equation to find what you want, in this instance you would do 50500/0.25 = P2... therefore P2 = 2020000Pa or 2.02*10^6Pa
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What is the atmospheric pressure and temperature at sea level in a standard<br> atmosphere?
Lady bird [3.3K]

Answer:

The tropospheric tabulation continues to 11,000 meters (36,089 ft), where the temperature has fallen to −56.5 °C (−69.7 °F), the pressure to 22,632 pascals (3.2825 psi), and the density to 0.3639 kilograms per cubic meter (0.02272 lb/cu ft). Between 11 km and 20 km, the temperature remains constant

Explanation:

Hope this helped, Have a wonderful day!!

4 0
3 years ago
Two loudspeakers S1 and S2, 2.20 m apart, emit the same single-frequency tone in phase at the speakers. A listener L is located
seropon [69]

Answer:

the lowest possible frequency of the emitted tone is 404.79 Hz

Explanation:

   Given the data in the question;

S₁ ←  5.50 m → L

↑

2.20 m

↓

S₂

We know that, the condition for destructive interference is;

Δr = ( 2m + \frac{1}{2} ) × λ

where m = 0, 1, 2, 3 .......

Path difference between the two sound waves from the two speakers is;

Δr = √( 5.50² + 2.20² ) - 5.50

Δr = 5.92368 - 5.50

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v = f × λ

f = ( 2m + \frac{1}{2})v / Δr

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Now, for the lowest possible frequency, let m be 0

so

f = ( 0 + \frac{1}{2})v / Δr

f = \frac{1}{2}(v) / Δr

we know that speed of sound in air v = 343 m/s

so we substitute

f = \frac{1}{2}(343) / 0.42368

f = 171.5 / 0.42368

f = 404.79 Hz

Therefore, the lowest possible frequency of the emitted tone is 404.79 Hz

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