So you can have a understanding of what you are doing and figure out your overall goal
The speed of the rock at 20 m is 34.3 m/s
Explanation:
We can solve this problem by using the law of conservation of energy: the mechanical energy of the rock, sum of its potential energy + its kinetic energy) must be conserved in absence of air resistance. So we can write:
where
:
is the initial potential energy
is the initial kinetic energy
is the final potential energy
is the final kinetic energy
The equation can also be rewritten as follows:
where:
m = 100 kg is the mass of the rock
is the acceleration of gravity
is the initial height
u = 0 is the initial speed (the rock starts at rest)
is the final height of the rock
v is the final speed when h = 20 m
And solving for v, we find:
Learn more about kinetic energy and potential energy here:
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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
Answer:
I think the awnser is B (but don't qoute me on that) if its right then yay but if its wrong im sorry
Explanation:
In a displacement/time graph, the slope of the line is equal to the velocity
