Answer:
crust, and upper mantle
Explanation:
i had a question like this
earth's lithosphere includes these things
The volume of the sphere is given by V = 4/3 * pie * r^3. We seek dr/dt that the rate of change of the radius with respect to time.
V = 4/3 * pie * r^3
Since we know the rate at which the volume changes wrt time. We can plug it in to find dr/dt.
dV/dt = 36pie. But dV/dt = 4/3 * pie * 3r^2 * dr/dt
So we have that dr/dt = (dV/dt) / 4/3 * pie * 3r^2
So dr/dt = (36pie)/ 4/3 * pie * 3(4)^2
dr/dt = 113.112/ 201.0864 = 0.5624
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https://www.siyavula.com/read/science/grade-10/sound/10-sound-03
The answer is w=p.t. Substitution:hz 110.20.
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
