I believe this is what you have to do:
The force between a mass M and a point mass m is represented by
So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ = 
G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ = F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.
Answer:
i think it would be B, a large factory
Explanation:
Answer:
hello your question is not properly arranged attached below is the arranged table and solution
answer : attached table below
Explanation:
Given data:
02 molecules size = 10^-10m
smoke particles size = 0.3 mm
cloud droplets size = 20 mm
Rain droplets size = 3 mm
Attached below is a table showing the kind of scattering that is expected to occur at various wave lengths
Note : For Rayleigh scattering the wave particle is smaller than the wave length while for Non-selective scattering the wave particle is greater than the wavelength.
and For Mie scattering the wavelength is the same as the wavelength.
When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted light, and all the light is reflected. The value of this angle is given by
where n2 and n1 are the refractive indices of the second and first medium, respectively.
In the first part of the problem, light moves from glass to air (
) and the critical angle is
. This means that we can find the refractive index of glass by re-arranging the previous formula:
Now the glass is put into water, whose refractive index is
. If light moves from glass to water, the new critical angle will be
