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OLga [1]
3 years ago
14

Is radiant energy potential energy ?

Physics
2 answers:
sveta [45]3 years ago
8 0

Answer:No

Explanation:

Radiant energy is kinetic energy.

Brums [2.3K]3 years ago
5 0

Answer:

Energy can be potential or kinetic. Radiant energy is a form of Kinetic energy. Movement of atoms, molecules, objects, wave, and substance is connected to a Kinetic energy.

I hope this help!:)

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entrained air is drier than the air within a cloud. considering the phase change of water, how could this lead to further coolin
Wittaler [7]

Dry air adjacent to the cloud is entrained air is drier than the air within a cloud. The evaporation occurs in the cloud which cools the air.  The cooling of air increases its density and creates a downdraft.

<h3>How clouds are formed?</h3>

A cloud can be described as a mass of ice crystals or water drops suspended in the atmosphere. Clouds can be formed when the water condenses in the atmosphere. The sky possesses some quantity of water vapours and it is invisible to us.

Clouds can be formed when an area of air gets cooler until the water vapour there condenses to liquid form. At this point, the air gets saturated with water vapours.

A cloud can never be perfectly adiabatic. Therefore, after mixing the environmental air with the clouds, its boundaries will not stay well defined and this process is called entrainment.

Learn more about cloud formation, here:

brainly.com/question/1242352

#SPJ1

5 0
1 year ago
Where are you most likely to find electrons?
Gelneren [198K]

Answer:

nucleus maybe .........

8 0
3 years ago
The vibrations along a transverse wave move in a direction _________.
GrogVix [38]

Answer: perpendicular to it oscillations.

Explanation: A transverse wave is a wave whose oscillations is perpendicular to the direction of the wave.

By perpendicular, we mean that the wave is oscillating on the vertical axis (y) of a Cartesian plane and the vibration is along the horizontal axis (x) of the plane.

Examples of transverse waves includes wave in a string, water wave and light.

Let us take a wave in a string for example, you tie one end of a string to a fixed point and the other end is free with you holding it.

If you move the rope vertically ( that's up and down) you will notice a kind of wave traveling away from you ( horizontally) to the fixed point.

Since the oscillations is perpendicular to the direction of wave, it is a transverse wave

5 0
3 years ago
Football player 1 has a mass of 80 kg and a velocity of 2 m/s east while player 2 has a mass of 70 kg and a velocity of 3 m/s we
sukhopar [10]

The total momentum of the system is equal to 50 Kgm/s.

<u>Given the following data:</u>

  • Mass 1 = 80 kg
  • Velocity 1 = 2 m/s east.
  • Mass 2 = 70 kg
  • Velocity 2 = 3 m/s west.

To determine the total momentum of the system:

Mathematically, momentum is given by the formula;

Momentum = mass \times velocity

<u>For Football player 1:</u>

Momentum = 80 \times 2

Momentum 1 = 160 Kgm/s.

<u>For Football player 2:</u>

Momentum = 70 \times 3

Momentum 1 = 210 Kgm/s.

Now, we can calculate the total momentum of the system:

Total\;momentum = momentum \;2 - momentum \;1\\\\Total\;momentum = 210-160

Total momentum = 50 Kgm/s.

<u>Note:</u> We subtracted because the football players were moving in opposite directions.

Read more: brainly.com/question/15517471

6 0
3 years ago
A 320-g ball and a 400-g ball are attached to the two ends of a string that goes over a pulley with a radius of 8.7 cm. Because
Gnom [1K]

To solve this problem, it is necessary to apply the concepts related to force described in Newton's second law, so that

F = ma

Where,

m = mass

a = Acceleration (Gravitational acceleration when there is action over the object of the earth)

Torque, as we know, is the force applied at a certain distance, that is,

\tau = F*d

Where

F= Force

d = Distance

Our values are given as,

m_1 = 0.32Kg

m_2 = 0.4Kg

d = 8.7*10^{-2}m

Since the system is in equilibrium the difference of the torques is the result of the total Torque applied, that is to say

\tau = T_2-T_1

\tau = F_2*d-F_1*d

\tau = m_2g*d-m_1*g*d

\tau = (m_2-m_1)g*d

\tau = (0.4-0.32)(9.8)(8.7*10^{-2})

\tau = 0.068N\cdot m

Therefore the magnitude of the frictional torque at the axle of the pulley if the system remains at rest when the balls are released is \tau = 0.068N\cdot m

8 0
3 years ago
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