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Alecsey [184]
2 years ago
6

How does the sun's energy most directly influence precipitation in an area?

Physics
1 answer:
topjm [15]2 years ago
5 0
The sun's energy influences climate in various ways. For example the latitudes at the equator receive more energy from the sun and therefore have warmer temperatures, On the other hand the sun's energy influences precipitation in a climate by driving the water cycle which determines precipitation.The sun is what makes the water cycle take place. That is the sun provides energy or heat to the earth; the heat causes liquid and frozen water to evaporate into water vapor gas, which rises high in the sky to form clouds ( precipitation), that in turn give us rain
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2 years ago
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A 50.0 kg object rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and
MariettaO [177]

Answer:

f=140\ N

Explanation:

Given:

  • mass of the object on a horizontal surface, m=50\ kg
  • coefficient of static friction, \mu_s=0.3
  • coefficient of kinetic friction, \mu_k=0.2
  • horizontal force on the object, F=140\ N

<u>Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:</u>

F_s=\mu_s.N

where:

N= normal force of reaction acting on the body= weight of the body

F_s=0.3\times (50\times 9.8)

F_s=147\ N

As we know that the frictional force acting on the body is always in the opposite direction:

So, the frictional force will not be at its maximum and will be equal in magnitude to the applied external force and hence the body will not move.

so, the frictional force will be:

f=140\ N

8 0
3 years ago
What is the unique geological feature found on Mercury surface?
Cloud [144]

Answer:

The surface of Mercury has landforms that indicate its crust may have contracted. They are long, sinuous cliffs called lobate scarps. These scarps appear to be the surface expression of thrust faults, where the crust is broken along an inclined plane and pushed upward.

Explanation:

I hope this helps a little bit.

3 0
3 years ago
A) what is the light source for a plant growing in the shade?
Ulleksa [173]

a.) Plants that thrive in the shade are often able to hold on to sunlight for extensive periods of time; they're in a sense like the camels of the plaNt WoRld.

b.) Though artificial lights are not nearly as beneficial as the sun, one could invest in one of them plant growing light thingies, but sun-loving plants might be sad if u do this instead of letting them soak in ePic rays from the sun.

6 0
2 years ago
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
3 years ago
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