First choice: the inability of current technology to capture
large amounts of the
Sun's energy
Well, it's true that large amounts of it get away ... our 'efficiency' at capturing it is still rather low. But the amount of free energy we're able to capture is still huge and significant, so this isn't really a major problem.
Second choice: the inability of current technology to store
captured solar
energy
No. We're pretty good at building batteries to store small amounts, or raising water to store large amounts. Storage could be better and cheaper than it is, but we can store huge amounts of captured solar energy right now, so this isn't a major problem either.
Third choice: inconsistencies in the availability of the resource
I think this is it. If we come to depend on solar energy, then we're
expectedly out of luck at night, and we may unexpectedly be out
of luck during long periods of overcast skies.
Fourth choice: lack of
demand for solar energy
If there is a lack of demand, it's purely a result of willful manipulation
of the market by those whose interests are hurt by solar energy.
Answer:
A- Astronomical body
C- Galaxy
D- Comet
B- Moon
Hope this helps you! Have a great day!
To find the ratio of planetary speeds Va/Vb we need the orbital velocity formula:
V=√({G*M}/R), where G is the gravitational constant, M is the mass of the distant star and R is the distance of the planet from the star it is orbiting.
So Va/Vb=[√( {G*M}/Ra) ] / [√( {G*M}/Rb) ], in our case Ra = 7.8*Rb
Va/Vb=[ √( {G*M}/{7.8*Rb} ) ] / [√( {G*M}/Rb )], we put everything under one square root by the rule: (√a) / (√b) = √(a/b)
Va/Vb=√ [ { (G*M)/(7.8*Rb) } / { (G*M)/(Rb) } ], when we cancel out G, M and Rb we get:
Va/Vb=√(1/7.8)/(1/1)=√(1/7.8)=0.358 so the ratio of Va/Vb = 0.358.
Complete Question
The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .
What eyepiece focal length will give the microscope an overall angular magnification of 300?
Answer:
The eyepiece focal length is 
Explanation:
From the question we are told that
The focal length is 
This negative sign shows the the microscope is diverging light
The angular magnification is 
The distance between the objective and the eyepieces lenses is 
Generally the magnification is mathematically represented as
![m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]](https://tex.z-dn.net/?f=m%20%20%3D%20%20%5B%5Cfrac%7BZ%20-%20f_e%20%7D%7Bf_e%7D%5D%20%5B%5Cfrac%7B0.25%7D%7Bf_0%7D%20%5D)
Where 
is the eyepiece focal length of the microscope
Now making the subject of the formula
![f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }](https://tex.z-dn.net/?f=f_e%20%20%3D%20%5Cfrac%7BZ%7D%7B1%20-%20%5B%5Cfrac%7BM%20%20%2A%20%20f_o%20%7D%7B0.25%7D%5D%20%7D)
substituting values
![f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }](https://tex.z-dn.net/?f=f_e%20%20%3D%20%5Cfrac%7B%200.19%20%7D%7B1%20-%20%5B%5Cfrac%7B300%20%20%2A%20%20-0.0055%20%7D%7B0.25%7D%5D%20%7D)
