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lisov135 [29]
3 years ago
12

Describe what happens to radiant energy once it enters the atmosphere? How does Radiant energy cause weather?

Physics
1 answer:
Triss [41]3 years ago
6 0
Once energy from the Sun gets to Earth, several things can happen to it:

Energy can be scattered or absorbed by aerosols in the atmosphere. Aerosols are dust, soot, sulfates and nitric oxides. When aerosols absorb energy, the atmosphere becomes warmer. When aerosols scatter energy, the atmosphere is cooled.
Short wavelengths are absorbed by ozone in the stratosphere.
Clouds may act to either reflect energy out to space or absorb energy, trapping it in the atmosphere.
The land and water at Earth's surface may act to either reflect energy or absorb it. Light colored surfaces are more likely to reflect sunlight, while dark surfaces typically absorb the energy, warming the planet.
Albedo is the percentage of the Sun's energy that is reflected back by a surface. Light colored surfaces like ice have a high albedo, while dark colored surfaces tend to have a lower albedo. The buildings and pavement in cities have such a low albedo that cities have been called "heat islands" because they absorb so much energy that they warm up.
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A train traveling at 27.5 accelerates to 42.4 m s over 75.0 s What is the displacement of the train in this time period A train
jarptica [38.1K]

Answer:

The displacement of the train in this time period is 2,616.86 m.

Explanation:

A Uniformly Varied Rectilinear Motion is Rectilinear because the mobile moves in a straight line, Uniformly because of there is a magnitude that remains constant (in this case the acceleration) and Varied because the speed varies, the final speed being different from the initial one.

In other words, a motion is uniformly varied rectilinear when the trajectory of the mobile is a straight line and its speed varies the same amount in each unit of time (the speed is constant and the acceleration is variable).

An independent equation of useful time in this type of movement is:

vf^{2} =vi^{2} +2*a*d <em>Expression A</em>

where:

  • vf = final velocity
  • vi = initial velocity
  • a = acceleration
  • d = distance

The equation of velocity as a function of time in this type of movement is:

vf=vi + a*t

So the velocity can be calculated as: a=\frac{vf-vi}{t}

In this case:

  • vf=42.4 m/s
  • vi=27.5 m/s
  • t=75 s

Replacing in the definition of acceleration:  a=\frac{42.4 m/s-27.5 m/s}{75 s}

a=0.199 m/s²

Now, replacing in expression A:

(42.4 m/s)^{2} =(27.5 m/s)^{2} +2*(0.199 m/s^{2}) *d

Solving:

d= \frac{(42.4 m/s)^{2} - (27.5 m/s)^{2} }{2*(0.199 m/s^{2}) }

d= 2,616.86 m

<u><em>The displacement of the train in this time period is 2,616.86 m.</em></u>

4 0
3 years ago
In the Fresnel circular aperture setup, the distances from the aperture to the light source and the reception screen are 1.5 m a
ivanzaharov [21]

Explanation:

The width of the central maximum is given by

W = 2 λ L / a

where W is the width of the central maximum

λ is the wavelength of the light used.

L is the distance between the aperture and screen

a is the width of the slit or aperture

So we can see that if any one quantity is varied by keeping others constant in the above formula , there would be a change in width of central maximum.

5 0
3 years ago
When developing an experimental design which action could a scientist take to improve the quality of the results
Anni [7]

Answer:

However, scientific findings in biomedical research are not always reproducible. ... that are promoting best practices and helping researchers perform the highest-quality science. ... in life science research and actions that can be taken to improve it. ... when designing experiments, take responsibility for depicting their results ...

Explanation:

4 0
3 years ago
A 3.00-kg object has a velocity 16.00 i ^ 2 2.00 j ^2 m/s.
trapecia [35]

Answer:

390 J

Explanation:

m = 3 kg

u = 16 i + 2 j

(a) Magnitude of velocity = \sqrt{16^{2}+2^{2}} = 16.1245 m/s

KEi = 1/2 m v^2 = 0.5 x 3 x 16.1245 = 390 J

(b) v = 18 i + 14 j

Magnitude of velocity =  \sqrt{18^{2}+14^{2}} = 22.804 m/s

KEf = 1/2 m v^2 = 0.5 x 3 x 22.804 = 780 J

According to the work energy theorem

Work done = change in KE = KEf - KEi = 780 - 390 = 390 J

8 0
4 years ago
A block is given a very brief push up a 20.0 degree frictionless incline to give it an initial speed of 12.0 m/s.(a) How far alo
Orlov [11]

Explanation:

(a)   Net force acting on the block is as follows.

           F_{net} = -mg Sin (\theta)

or,           ma = -mg Sin (\theta)[/tex]

                 a = -g Sin (\theta)

                    = -9.8 \times Sin (20^{o})

                    = -3.35 m/s^{2}

According to the kinematic equation of motion,

             v^{2} - v^{2}_{o} = 2as

Distance traveled by the block before stopping is as follows.

     s = \frac{v^{2} - v^{2}_{o}}{2a}

        = \frac{(0)^{2} - (12.0)^{2}_{o}}{2 \times -3.35}

        = 21.5 m

According to the kinematic equation of motion,

               v = v_{o} + at

      0 = 12.0 m/s + \frac{1}{2} \times -3.35 m/s^{2} \times t

   t_{1} = 7.16 sec

Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.

(b)    When the block is moving down the inline then net force acting on the block is as follows.

                 F_{net} = -mg Sin (\theta)

                ma = mg Sin (\theta)

                    a = g Sin (\theta)

                       = 9.8 m/s^{2} \times Sin (20^{o})

                       = 3.35 m/s^{2}

Kinematics equation of the motion is as follows.

                   s = v_{o}t + \frac{1}{2}at^{2}

      21.5 m = 0 + \frac{1}{2} \times 3.35 m/s^{2} \times t^{2}

     t_{2} = \sqrt{\frac{2 \times 21.5 m}{3.35 m/s^{2}}}

             = 3.58 sec

Hence, total time taken by the block to return to its starting position is as follows.

               t = t_{1} + t_{2}

                 = 7.16 sec + 3.58 sec

                 = 10.7 sec

Thus, we can conclude that 10.7 sec time it take to return to its starting position.

3 0
4 years ago
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