The hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would produce is apparently as a result of <u>Eddington luminosity</u>
<h3>What are stars?</h3>
Stars are a fixed luminous point in the sky which is a large and remote incandescent body
So therefore, the hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would produce is apparently as a result of Eddington luminosity
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Answer:
It is (1/5)th as much.
Explanation:
If we apply the equation
F = G*m*M / r²
where
m = mass of a man
M₀ = mass of the planet Driff
M = mass of the Earth
r₀ = radius of the planet Driff
r = radius of the Earth
G = The gravitational constant
F = The gravitational force on the Earth
F₀ = The gravitational force on the planet Driff
g = the gravitational acceleration on the surface of the earth
g₀ = the gravitational acceleration on the surface of the planet Driff
we have
F₀ = G*m*M₀ / r₀² = G*m*(5*M) / (5*r)²
⇒ F₀ = G*m*M / (5*r²) = (1/5)*F
If
F₀ = (1/5)*F
then
W₀ = (1/5)*W ⇒ m*g₀ = (1/5)*m*g ⇒ g₀ = (1/5)*g
It is (1/5)th as much.
For the given question above, I think there is an associated choice of answer for it. However, the answer for this is London Dispersion Forces. <span>Dipole-dipole forces and hydrogen bonding are much stronger, leading to higher melting and boiling points.</span>
Answer:
1.36 x 10^-3 cm
Explanation:
Area = 50 ft^2 = 46451.5 cm^2
mass = 6 oz = 170.097 g
density = 2.70 g/cm^3
Let t be the thickness of foil in cm.
mass = volume x density
mass = area x thickness x density
170.097 = 46451.5 x t x 2.70
t = 1.36 x 10^-3 cm
Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.
In my view, correct answer should look like this: Although wave power does not produce pollution, some people may not want to invest in it because it is <span>prone to storm damage and limited to particular areas of the ocean.</span>
