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3 years ago

How is air temperature related to high and low pressure air

1 answer:

3 0

Answer: So, I looked at it to see what was the correct one, and the correct answer is Cool air near surface forms high-pressure areas, warm air forms low pressure areas. I hope this helps :D :)

Explanation:

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Atoms can join together to form ______ substance

ELEN [110]

Answer:

Molecules

Explanation:

7 0

3 years ago

There is a refrigerator running in a room, the heat flowing into the refrigerator from the outside is 40 J/s, and the refrigerat

Mamont248 [21]

Answer:

(A) 140 j/sec (b) 1.26 K

Explanation:

We have given the heat heat flowing into the refrigerator = 40 J/sec

Work done = 40 W

(a) So the heat discharged from the refrigerator $=heat\ flowing\ in\ refrigerator+work\ done=40+100=140j/sec$

(b) Total heat absorbed =140 j/sec $=140\times 3600=504000j/hour$

Let the temperature be $\Delta T$

Heat absorbed per hour =504000 $[tex]=400\times 10^3\times \Delta T$

So $\Delta T=\frac{504000}{400000}=1.26K$

8 0

3 years ago

During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0

oksian1 [2.3K]

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

$d=\frac{1,280,000,000km}{37,000}=34,594.59km$

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

$v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}$

hence, the speed of the Hubble is approximately 384km/min

5 0

3 years ago

Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. Wha

ikadub [295]

We have that the spring constant is mathematically given as

$k=2.37*10^{11}N/m$

Generally, the equation for angular velocity is mathematically given by

$\omega=\sqrt{k}{m}$

Where

k=spring constant

And

$\omega =\frac{2\pi}{T}$

Therefore

$\frac{2\pi}{T}=\sqrt{k}{n}$

Hence giving spring constant k

$k=m((\frac{2 \pi}{T})^2$

Generally

Mass of earth $m=5.97*10^{24}$

Period for on complete resolution of Earth around the Sun

$T=365 days$

$T=365*24*3600$

Therefore

$k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2$

$k=2.37*10^{11}N/m$

In conclusion

The effective spring constant of this simple harmonic motion is

$k=2.37*10^{11}N/m$

For more information on this visit

brainly.com/question/14159361

8 0

3 years ago

How do magnetic trains work and what advantages do they have compared to regular trains

ad-work [718]

Answer:

magnetic trains works at the principle of repel on of the advantage is that they are fast and dont really need diesel

7 0

3 years ago