Answer:
too much exposure to the sun's rays
Answer:
The correct answer to the question is
B. It always decreases
Explanation:
To solve the question, we note that the foce of gravity is given by
where
G= Gravitational constant
m₁ = mass of first object
m₂ = mass of second object
r = the distance between both objects
If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have
= ![\frac{2}{4} \frac{Gm_1m_2}{r^2}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B4%7D%20%5Cfrac%7BGm_1m_2%7D%7Br%5E2%7D)
Therefore the gravitational force is halved. That is it will always decrease
Answer:
The velocity of mass 2m is ![v_B = 0.67 m/s](https://tex.z-dn.net/?f=v_B%20%3D%200.67%20m%2Fs)
Explanation:
From the question w are told that
The mass of the billiard ball A is =m
The initial speed of the billiard ball A =
=1 m/s
The mass of the billiard ball B is = 2 m
The initial speed of the billiard ball B = 0
Let the final speed of the billiard ball A = ![v_A](https://tex.z-dn.net/?f=v_A)
Let The finial speed of the billiard ball B = ![v_B](https://tex.z-dn.net/?f=v_B)
According to the law of conservation of Energy
![\frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20m%20%28v_1%29%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%202m%20%280%29%20%5E%202%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%20%28v_A%29%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%202m%20%28v_B%29%5E2)
Substituting values
![\frac{1}{2} m (1)^2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20m%20%281%29%5E2%20%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%20%28v_A%29%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%202m%20%28v_B%29%5E2)
Multiplying through by ![\frac{1}{2}m](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dm)
![1 =v_A^2 + 2 v_B ^2 ---(1)](https://tex.z-dn.net/?f=1%20%3Dv_A%5E2%20%2B%202%20v_B%20%5E2%20---%281%29)
According to the law of conservation of Momentum
![mv_1 + 2m(0) = mv_A + 2m v_B](https://tex.z-dn.net/?f=mv_1%20%2B%202m%280%29%20%3D%20mv_A%20%2B%202m%20v_B)
Substituting values
![m(1) = mv_A + 2mv_B](https://tex.z-dn.net/?f=m%281%29%20%20%3D%20mv_A%20%2B%202mv_B)
Multiplying through by ![m](https://tex.z-dn.net/?f=m)
![1 = v_A + 2v_B ---(2)](https://tex.z-dn.net/?f=1%20%3D%20v_A%20%2B%202v_B%20---%282%29)
making
subject of the equation 2
![v_A = 1 - 2v_B](https://tex.z-dn.net/?f=v_A%20%3D%201%20-%202v_B)
Substituting this into equation 1
![(1 -2v_B)^2 + 2v_B^2 = 1](https://tex.z-dn.net/?f=%281%20-2v_B%29%5E2%20%2B%202v_B%5E2%20%3D%201)
![1 - 4v_B + 4v_B^2 + 2v_B^2 =1](https://tex.z-dn.net/?f=1%20-%204v_B%20%2B%204v_B%5E2%20%2B%202v_B%5E2%20%3D1)
![6v_B^2 -4v_B +1 =1](https://tex.z-dn.net/?f=6v_B%5E2%20%20-4v_B%20%2B1%20%3D1)
![6v_B^2 -4v_B =0](https://tex.z-dn.net/?f=6v_B%5E2%20-4v_B%20%3D0)
Multiplying through by ![\frac{1}{v_B}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bv_B%7D)
![6v_B -4 = 0](https://tex.z-dn.net/?f=6v_B%20-4%20%3D%200)
![v_B = \frac{4}{6}](https://tex.z-dn.net/?f=v_B%20%3D%20%5Cfrac%7B4%7D%7B6%7D)
![v_B = 0.67 m/s](https://tex.z-dn.net/?f=v_B%20%3D%200.67%20m%2Fs)
Answer:
conservative
Explanation:
Nonconservative force is the force that depends on a path, however conservative does not depend on a path and it is not associated with the potential energy. When the work is done by an unconservative force, mechanical energy is added or removed. Friction is the best example for a non-conservative force. When these non-conservative forces are acting, the mechanical energy changes but these are not preserved.
hope this helped!