Answer: first blank- 26.17° C, second- 1.17 °C, third- 30.86°C, and last fourth- 5.86°C
Explanation:
Physics, got it right on edg. 2020:)
Answer:
t = 103.45 n m
Explanation:
given,
refractive index of cornea = 1.38
refractive index of eye drop = 1.45
wavelength of refractive index = 600 nm
refractive index of eye drop is greater than refractive index of cornea and the air.
Formula used in this case
for constructive interference

At m = 0 for the minimum thickness, so
t = 103.45 n m
the minimum thickness of the film of eyedrops t = 103.45 n m
(a) We can find the current flowing between the walls by using Ohm's law:

where

is the potential difference and

is the resistance. Substituting these values, we get

(b) The total charge flowing between the walls is the product between the current and the time interval:

The problem says

, so the total charge is

The current consists of Na+ ions, each of them having a charge of

. To find the number of ions flowing, we can simply divide the total charge by the charge of a single ion:
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
Use newtons second law F=ma, plug in the given values which gives us the answer of 22 kg for the mass