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vlabodo [156]
3 years ago
9

What do you mean by peltier coefficient? ​

Physics
2 answers:
vredina [299]3 years ago
5 0

Answer:

The Peltier coefficient is a measure of the amount of heat carried by electrons or holes.

Explanation:

devlian [24]3 years ago
4 0

Answer:

The Peltier coefficient is a measure of the amount of heat carried by electrons or holes

Explanation:

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Which type of surface would most likely be the best reflector of electromagnetic energy?
Len [333]
<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>
5 0
3 years ago
_____ heat provides energy for thunderstorms and hurricanes. Polar Geothermal Nuclear Latent
NikAS [45]
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.
4 0
3 years ago
A small 1.0 kg steel ball rolls west at 3.0 m/s collides with a large 3.0 kg ball at rest. After the collision, the small ball m
77julia77 [94]

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

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}

-3i+2j=3.0\times v_{2}

v_{2}=-i+0.66j

The direction of the momentum

tan\theta=\dfrac{0.66}{-1}

\theta=tan^{-1}\dfrac{0.66}{-1}

\theta=-33.42^{\circ}

The direction of the momentum with respect to east

\theta=180-33.42=146.58^{\circ}

Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

7 0
3 years ago
Tarzan, in one tree, sights Jane in another tree
taurus [48]
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.
8 0
3 years ago
Read 2 more answers
Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr
vivado [14]

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   T^{2} \alpha R^{3}

                                                          ⇒T^{2} = KR^{3}

Here K is the proportionality constant.

If T_{1} andT_{2} are the orbital periods of the planets and

R_{1} and R_{2} are the distance of the planets from the sun, then Kepler's law can be written as-

          \frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{2} }

      ⇒ R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }

  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 R_{1} and T_{1} respectively.

Let the distance  from sun and orbital period of earth is denoted as R_{2} and T_{2} respectively.

we are given thatT_{1} =29.46 years

we know that R_{2} = 1 AU and T_{2} = 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-

                                    R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }

                                             =[1 ]^{3} *\frac{[29.46]^{2} }{[1]^{2} } AU

                                    =867.8916 AU^{3}

                                        ⇒R_{1} =\sqrt[3]{867.8916}

                                           =9.5386 AU [ans]

5 0
3 years ago
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