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
Not at all. Density on its own is not sufficient
Answer:
1. energy lost in the lever due to friction
3. visual estimation of height of the beanbag
5. position of the fulcrum for the lever affecting transfer of energy
Explanation:
Edge 2021
Answer:
d = 76.5 m
Explanation:
To find the distance at which the boats will be detected as two objects, we need to use the following equation:

<u>Where:</u>
θ: is the angle of resolution of a circular aperture
λ: is the wavelength
D: is the diameter of the antenna = 2.10 m
d: is the separation of the two boats = ?
L: is the distance of the two boats from the ship = 7.00 km = 7000 m
To find λ we can use the following equation:
<u>Where:</u>
c: is the speed of light = 3.00x10⁸ m/s
f: is the frequency = 16.0 GHz = 16.0x10⁹ Hz
Hence, the distance is:

Therefore, the boats could be at 76.5 m close together to be detected as two objects.
I hope it helps you!