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Rina8888 [55]
3 years ago
14

Great amounts of heat are found at hotspots. What is the source of this heat?

Physics
1 answer:
Mice21 [21]3 years ago
6 0
The Earth's mantle is the source of heat at hot spots.

Hope this helped. Have a great day!
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3 years ago
:What will be the value of the refractive index of the medium? Critical angle of that medium is 30 degree
earnstyle [38]

Answer:

Let the second medium be air (n₁=1)

The refractive index n₂ of the medium where first medium is air is found (a)

(a) n₂ = 2

Explanation:

Critical angle can be defined as the angle of incidence that provides the angle of refraction of 90°.

Refractive index of a medium can be defined as a number that describes that how fast a light will travel through that medium.

Critical angle and Refractive index are related by:

\theta_{critical}= sin^{-1}(\frac{n_1}{n_2})

sin \theta_{critical}=\frac{n_1}{n_2}

To find refractive index of medium with respect to air, substitute n₁=1 (Refractive index of air is 1)

Also θ(critical)=30°

Find n₂ :

sin30= \frac{1}{n_2}\\0.5=\frac{1}{n_2}\\n_2=\frac{1}{0.5}\\n_2=2

8 0
3 years ago
A 10 N force is used to push a 30 kg box across the floor, moving it a distance of 20 m. There is no friction. What is the magni
tekilochka [14]
Work=Force • Distance
W= 10N • 20m
W= 200 J

5 0
3 years ago
Suppose you observe two stars and you know they have the same luminosity. If one star is twice as far away as the other, the mor
rosijanka [135]

Answer:

The farther star will appear 4 times fainter than the star that is near to the observer.

Explanation:

Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time

Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)

This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

e_{1}=\frac{E}{4\pi r^{2}}

For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by

e_{2}=\frac{E}{4\pi (2r)^{2}}=\frac{E}{8\pi r^{2}}=\frac{e_{1}}{4}

Hence we sense it as 4 times fainter than the nearer star.

5 0
3 years ago
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