<span>The energy of the Sun is produced by nuclear fusion going on in its core.</span>
On Earth, solar radiation at the frequencies of visible light largely passes through the atmosphere to warm the planetary surface. The surface itself emits energy at the lower frequencies of infrared thermal radiation. Infrared radiation is absorbed by greenhouse gases in the atmosphere. These gases also radiate energy, some of which is directed to the surface and lower atmosphere. The mechanism is named after the effect of solar radiation passing through glass and warming a greenhouse, but the way it retains heat is fundamentally different as a greenhouse works by reducing airflow, isolating the warm air inside the structure so that heat is not lost by convection
Answer:
1.024 × 10⁸ m
Explanation:
The velocity v₀ of the orbit 8RE is v₀ = 8REω where ω = angular speed.
So, ω = v₀/8RE
For the orbit with radius R for it to maintain a circular orbit and velocity 2v₀, we have
2v₀ = Rω
substituting ω = v₀/8RE into the equation, we have
2v₀ = v₀R/8RE
dividing both sides by v₀, we have
2v₀/v₀ = R/8RE
2 = R/8RE
So, R = 2 × 8RE
R = 16RE
substituting RE = 6.4 × 10⁶ m
R = 16RE
= 16 × 6.4 × 10⁶ m
= 102.4 × 10⁶ m
= 1.024 × 10⁸ m
You will have to google these constants for yourself. I am warned not to give you the references.
Givens
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The given constant acceleration of the sun = 273.7 m/s^2
The Radius of the sun = 695,700 km
Radius of the star = 20 km
acceleration on the star's surface = a1
Formula
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m_test * a_sun = G * m_test * mass_sun /( r_sun)^2
=========== . . .=========================
m_test * a_sun. . . G* m_test* mass_star / (r_star)^2
There is a lot oc cancellation. The result is.
a_sun / a_star = (r_star)^2 / r^2 Now all you have to do is substitute
Calculations
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273.7 / a _ star = (20 000)^2 / (695700 )^2
a_star = 273.7 * (695700)^2 / (20000)^2
a_star = 331176 m/s^2 which is pretty big
Answer:
180 N
Explanation:
Use Newton's second law, F = ma.
m = 4.5 kg; a = 40 m/s/s
F = (4.5)(40) = 180 N