Answer:
explained
Explanation:
When the intensity of light is increased on a piece of metal only the number of electron ejected will increase because all other things independent of intensity of light.
Light below certain frequency will not cause any electron emission no matter how intense.
The intensity produces more electron but does not change the maximum kinetic energy of electrons.
Work function is independent of the intensity of light, because it is an intrinsic property of a material.
<span>So the question is what gases form the atmosphere of Uranus. So Uranus and Neptune are classified as "ice giants". They have a similar atmosphere to "gas giants" Saturn and jupiter and Neptunes atmosphere is primarily composed out of hidrogen and helyum. So the correct answer is a.</span>
You're talking about a grain of sand or a stone or a rock that's drifting in space, and then the Earth happens to get in the way, so the stone falls down to Earth, and it makes a bright streak of light while it's falling through the atmosphere and burning up from the friction.
-- While it's drifting in space, it's a <em>meteoroid</em>.
-- While it's falling through the atmosphere burning up and making a bright streak of light, it's a <em>meteor</em>.
-- If it doesn't completely burn up and there's some of it left to fall on the ground, then the leftover piece on the ground is a <em>meteorite</em>.
Answer:
B. 175 N
Explanation:
Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.
Mathematically, net force is given by the formula;
Where;
Fnet is the net force
Fapp is the applied force
Fg is the force due to gravitation
In this scenario, we observed that both forces are acting in the same direction.
Therefore:
Net force = 100 N + 75 N
Net force = 175 Newton
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)