Explanation:
radio waves, which include visible light waves.
The force between the two objects is 19.73 nN.
<u>Explanation:
</u>
Any force acting between two objects tends to be directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects. And this kind of attraction force between two objects is termed as gravitational force.
So if we consider
and
as the masses of both objects and let d be the distance of separation of two objects. Then the force between the two objects can be determined as below:
![\text {Gravitational force}=\frac{G \times M_{1} \times M_{2}}{d^{2}}](https://tex.z-dn.net/?f=%5Ctext%20%7BGravitational%20force%7D%3D%5Cfrac%7BG%20%5Ctimes%20M_%7B1%7D%20%5Ctimes%20M_%7B2%7D%7D%7Bd%5E%7B2%7D%7D)
As gravitational constant
,
= 20 kg and
= 100 kg, while d = 2.6 m, then
![\text {Gravitational force}=\frac{6.67 \times 10^{-11} \times 20 \times 100}{(2.6)^{2}}=\frac{6.67 \times 20 \times 10^{-9}}{6.76}](https://tex.z-dn.net/?f=%5Ctext%20%7BGravitational%20force%7D%3D%5Cfrac%7B6.67%20%5Ctimes%2010%5E%7B-11%7D%20%5Ctimes%2020%20%5Ctimes%20100%7D%7B%282.6%29%5E%7B2%7D%7D%3D%5Cfrac%7B6.67%20%5Ctimes%2020%20%5Ctimes%2010%5E%7B-9%7D%7D%7B6.76%7D)
Thus, we get finally,
![\text {Gravitational force}=19.73 \times 10^{-9} \mathrm{N}](https://tex.z-dn.net/?f=%5Ctext%20%7BGravitational%20force%7D%3D19.73%20%5Ctimes%2010%5E%7B-9%7D%20%5Cmathrm%7BN%7D)
As we know, nano denoted by letter 'n' equals to ![10^{-9}](https://tex.z-dn.net/?f=10%5E%7B-9%7D)
So the force acting between two objects is 19.73 nN.
Explanation:
It is given that,
Focal length of the concave mirror, f = -13.5 cm
Image distance, v = -37.5 cm (in front of mirror)
Let u is the object distance. It can be calculated using the mirror's formula as :
![\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bv%7D%2B%5Cdfrac%7B1%7D%7Bu%7D%3D%5Cdfrac%7B1%7D%7Bf%7D)
![\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bu%7D%3D%5Cdfrac%7B1%7D%7Bf%7D-%5Cdfrac%7B1%7D%7Bv%7D)
![\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bu%7D%3D%5Cdfrac%7B1%7D%7B%28-13.5%29%7D-%5Cdfrac%7B1%7D%7B%28-37.5%29%7D)
u = -21.09 cm
The magnification of the mirror is given by :
![m=\dfrac{-v}{u}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B-v%7D%7Bu%7D)
![m=\dfrac{-(-37.5)}{(-21.09)}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B-%28-37.5%29%7D%7B%28-21.09%29%7D)
m = -1.77
So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.