The RDS-220 <span>hydrogen bomb, soviet </span>
Let us consider two bodies having masses m and m' respectively.
Let they are separated by a distance of r from each other.
As per the Newtons law of gravitation ,the gravitational force between two bodies is given as -
where G is the gravitational force constant.
From the above we see that F ∝ mm' and 
Let the orbital radius of planet A is
= r and mass of planet is
.
Let the mass of central star is m .
Hence the gravitational force for planet A is 
For planet B the orbital radius
and mass
Hence the gravitational force 
![f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }](https://tex.z-dn.net/?f=f_%7B2%7D%20%3DG%5Cfrac%7Bm%2A3m_%7B1%7D%20%7D%7B%5B2r_%7B1%7D%5D%20%5E%7B2%7D%20%7D)

Hence the ratio is 
[ ans]
B. truthfully it's the only one that makes sense
Answer:
Given, Apparent weight(W₂)=4.2N
Weight of liquid displaced (u)=2.5N
Let weight of body in air = W₁
Solution,
U=W₁-W₂
W₁=4.2=2.5=6.7N
∴Weight of body in air is 6.7N
Answer:
C, A and E
Explanation:
In theory you always assume that the prostive is going to the negtive and since C, A and E are in series they all will turn off will turn off