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____ [38]
4 years ago
9

When we take infrared photos of Jupiter we see that it is glowing brightly with heat. In fact, Jupiter emits more energy (in the

form of infrared light) than it receives from the Sun in the form of sunlight. What is the source of most of Jupiter's heat?
Physics
1 answer:
dangina [55]4 years ago
4 0

Answer:

Jupiter has an internal heat source.

Explanation:

It is believed that most of this heat is the residual (left over) heat from the time when formation of early Solar nebula took place. Jupiter absorbs energy from the Sun in the form of light and the coverts it into heat and later releases this heat in the form of thermal radiation.

It is recorded that Jupiter emits almost twice amount of energy from the amount which it receives from the sun. It is this internal source of energy and sunlight which provide energy to Jupiter's atmosphere.

However, this phenomenon of internal heat source is also seen in other celestial bodies as well such as Saturn, Neptune, Stars, etc.

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Give three examples about a force that deforms the shape of a body.​
AURORKA [14]

Answer:

  • Squeezing of a plastic bottle changes the shape of the bottle.
  • Deformation of clay by pressing it between the two hands
  • stretching of rubber band

hope it becomes helpful to you ☺️☺️

good luck

8 0
3 years ago
Two objects are moving at equal speed along a level, frictionless surface. The second object has twice the mass of the first obj
denis23 [38]

Answer:

E)The two objects rise to the same height : h₁=h₂ =( v²) / (2g)

Explanation:

With  coefficient of kinetic friction,  μk =0:

We apply the principle of energy conservation :

E₀ = Ef Formula (1)

K₀+U₀ = Kf + Uf Formula (2)

K = (1/2) *m*v² Formula (3)

U = m*g*h Formula (4)

Where:

E₀:  

Initial total energy (J)

Ef:   Final total energy  (J)

K₀:   Initial kinetic energy (J)

U₀:  Initial potential energy (J)

Kf:  Final kinetic energy (J)

Uf:

Final kinetic energy (J)

v : speed (m/s)

m: mass (kg)

h : hight (m)

Known data

v₁=v₂=v

m₁=m

m₂ =2m

μk =0 : coefficient of kinetic friction

Problem development

We apply formulas (1), (2), (3) and (4)  for the  first object ,(1):

E₀₁ = Ef₁

K₀₁+U₀₁ = Kf₁ + Uf₁  U₀₁=0 ,  Kf₁ =0 , m₁=m

(1/2) *m*v²+0 =0+m*g*h₁    :We eliminate m,then,

(1/2) *v²=g*h₁

h₁ =( v²) /(2g)

We apply formulas (1), (2), (3) and (4)  for the  second object ,(2):

E₀₂ = Ef₂

K₀₂+U₀₂ = Kf₂ + Uf₂  U₀₂=0 ,  Kf₂ =0 , m₂=2m

(1/2) *2m*v²+0 =0+2m*g*h₂    :We eliminate m,then,

(1/2) *2*v² =2*g*h₂  : We divide by 2 on both sides of the equation,then,

(1/2) *v²=g*h₂

h₂ =( v²) / (2g)

8 0
3 years ago
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Nata [24]
When the child is moving fastest
6 0
4 years ago
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A light beam in glass (n = 1.5) reaches an air-glass interface, at an angle of 60 degrees from the surface. What is the angle of
tester [92]

Answer:

θ₂ = 35.26°

Explanation:

given,

refractive index of air, n₁ = 1

refractive index of glass, n₂ = 1.5

angle of incidence, θ₁ = 60°

angle of refracted light, θ₂ = ?

using Snell's Law

n₁ sin θ₁ = n₂ sin θ₂

1 x sin 60° = 1.5 sin θ₂

sin θ₂ = 0.577

θ₂ = sin⁻¹(0.577)

θ₂ = 35.26°

Hence, the refracted light is equal to  θ₂ = 35.26°

7 0
3 years ago
1) what is a calorie?<br>​
Bogdan [553]
I read it’s a unit of energy
3 0
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