<span>light colored and smooth surface would most likely be the best reflector of electromagnetic energy.Light, shiny surfaces are the best reflectors of radiation and they will allow the waves to reflect and bounce off rather than absorb. we can consider mirror as the example ,it will only reflect the light energy falling on them and it will not absorb. The darker coloured and rough surfaced substances will definitely absorb some amount of light falling on it. so light coloured smooth or shiny surfaced material would be the best reflector for electromagnetic energy.</span>
The correct answer is Latent. Latent heat provides energy for thunderstorms and hurricanes. Hurricanes are a low pressure that is formed in warm oceans and the source of this energy is water vapor. Water vapor serves to be the food to the hurricanes because they release the latent heat of condensation.
Answer:
The direction of the momentum of the large ball after the collision with respect to east is 146.58°.
Explanation:
Given that,
Mass of large ball = 3.0 kg
Mass of steel ball = 1.0 kg
Velocity = 3.0 kg
After collision,
Velocity = 2.0 m/s
Using conservation of momentum




The direction of the momentum



The direction of the momentum with respect to east

Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.
Taking the distance of Tarzan from the ground before and after he makes the swing:
Ho (initial height) = L(1 - cos45) = 20 (1 - 0.707) = 5.86 meters
Hf (final height) = L(1 - cos30) = 20 (1 - 0.866<span>) = 2.68 meters
</span>
Difference in height = 5.86 - 2.68 = 3.18 meters
PE = KE
mgh = (1/2)mv^2
Solving for v:
v = sqrt (2*g*h)
v = sqrt (2*9.8*3.18)
v = 7.89 m/s
With Tarzan going that fast, it is likely that he will knock Jane off.
This question can be solved from the Kepler's law of planetary motion.
As per this law the square of time period of a planet is proportional to the cube of semi major axis.
Mathematically it can be written as 
⇒
Here K is the proportionality constant.
If
and
are the orbital periods of the planets and
and
are the distance of the planets from the sun, then Kepler's law can be written as-

⇒ 
Here we are asked to calculate the the distance of Saturn from sun.It can solved by comparing it with earth.
Let the distance from sun and orbital period of Saturn is denoted as
and
respectively.
Let the distance from sun and orbital period of earth is denoted as
and
respectively.
we are given that
we know that
1 AU and
1 year.
1 AU is the mean distance of earth from the sun which is equal to 150 million kilometre.
Hence distance of Saturn from sun is calculated as -
From Kepler's law as mentioned above-

=![[1 ]^{3} *\frac{[29.46]^{2} }{[1]^{2} } AU](https://tex.z-dn.net/?f=%5B1%20%5D%5E%7B3%7D%20%2A%5Cfrac%7B%5B29.46%5D%5E%7B2%7D%20%7D%7B%5B1%5D%5E%7B2%7D%20%7D%20AU)

⇒![R_{1} =\sqrt[3]{867.8916}](https://tex.z-dn.net/?f=R_%7B1%7D%20%3D%5Csqrt%5B3%5D%7B867.8916%7D)
=9.5386 AU [ans]