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3 years ago

How does solar altitude affect the length of the path of solar radiation through earth’s atmosphere?

1 answer:

7 0

The angle of the Sun above the horizon, which is the solar altitude, influences the intensity ofsolar radiation received at Earth’s surface. At the place on Earth where the Sun is directlyoverhead, the local solar altitude has its maximum value of 90 degrees and solar rays are mostconcentrated. Whenever the Sun is positioned lower in the sky, solar radiation spreads over alarger area of Earth’s horizontal surface and thus is less intense. Solar radiation reaches theplanet essentially as parallel beams of uniform intensity. The nearly spherical Earth presents acurved surface to incoming solar radiation so that the noon solar altitude always varies withlatitude. The intensity of solar radiation actually striking Earth’s atmosphere is greatest at thelatitude where the noon Sun is in the zenith and decreases with distance north and south of thatlatitude. Decreasing solar altitude lengthens the path of the Sun’s rays through the atmosphere.As the path lengthens, the greater interaction of solar radiation with clouds, gases and aerosols<span>reduces its intensity</span>

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What elements are good conductors of electricity?

kramer

In solids: All metals are good conductors of electricity as they contain free moving electrons. Non-metals doesn't conduct , but we consider Graphite the only non-metal that can conduct electricity for the presence of free moving electrons.

In Liquids ; Ionic compunds contains free moving ions , so they conduct electricity as well .

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3 years ago

Work is the transfer of _______ that occurs when a force makes an object move.

Vlad1618 [11]

I think it is energy

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3 years ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

$v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s$

Then, to get $y_2$ , we do:

$2a_2(y_2-y_{02})=-v_{02}^2$

$y_2-y_{02}=-\frac{v_{02}^2}{2a_2}$

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}$

And we substitute the values:

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m$

3 0

3 years ago

Help with a length problem

Andru [333]

It would be B

Explanation:
Because if you're not measuring in inches you want to go the next one down other than inches which would be millimeters!(: hope this helps.

6 0

2 years ago

Read 2 more answers

A plane drops a hamper of medical supplies from a height of 5000 m during a practice run over the ocean. The plane’s horizontal

Marizza181 [45]

Answer:

346.70015 m/s

Explanation:

In the x axis speed is

$v_x=149\ m/s$