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]
At the same speed because it will slow down as it approaches the peak then speed up as it goes down again
it will be going 15m/s when it gets to the same height if we neglect air resistance and the object doesn't hit something
<u>Answer</u>
48 Volts
<u>Explanation</u>
The question can be solve using the turn rule of a transformer that states;
Np/Ns = Vp/Vs
Where Np ⇒ number of turns in the primary coil.
Ns ⇒number of turns in the seconndary coil
Vp ⇒ primary voltage
Vs ⇒secondary voltage
Np/Ns = Vp/Vs
10/4 = 120/Vp
Vp = (120 × 4)/10
= 480/10
= 48 Volts
Answer:
The power dissipated by the meter is 1188W
Explanation:
Here we have a circuit constituted with a power source and two resistors in series, we can calculate the power dissipated by the meter using the following formula:
We first need to fin the current going through the circuit:
because they are connected in series. So:
