In a displacement versus time graph, the slope of the line at any point on the graph indicates the <em>magnitude of velocity</em>.
(It can't indicate velocity completely, because the graph shows nothing about the direction of the motion.)
Answer:
It becomes a giant or supergiant.
Explanation:
Once all the hydrogen supply is gone, fusion of hydrogen into helium stops. The core starts to contract and liberates energy, which heats the superior layer until it becomes hot enough to start the fusion of hydrogen into helium.
Answer:
d) 2Fr
Explanation:
We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².
We now evaluate the integral from r = +r to -r
W = ∫Fdr
= ∫(-e²/4πε₀r²)dr
= -∫e²dr/4πε₀r²
= -e²/4πε₀∫dr/r²
= -e²/4πε₀ × -[1/r] from r = +r to -r
W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.
Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.
So W = -2e²/4πε₀r = 2Fr.
So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr
<u>We are given:</u>
Mass of Neptune = 1.03 * 10²⁶ kg
Distance from the center of Neptune (r) = 2.27 * 10⁷
now, computing the value of the acceleration due to gravity (g)
<u>Finding g:</u>
We know the formula:
g = G(mass of planet) / (r)²
g = [6.67 * 10⁻¹¹ * 1.03*10²⁶] / (2.27*10⁷) [since G is 6.67*10⁻¹¹]
g = (6.87 * 10¹⁵) / (5.15 * 10¹⁴)
which can be rewritten as:
g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15
g = (6.87 * 10¹⁵⁻¹⁴) / 5.15
g = (6.87/5.15) * 10
g = 1.34 * 10
g = 13.4 m/s² <em>(approx)</em>
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