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)
D - Most likely. Those who read the magazine can choose whether or not to return the survey.
It produces only virtual images is the answer
Answer:
12 cm and 0.4
Explanation:
f = - 20 cm, u = - 30 cm
Let v be the position of image and m be the magnification.
Use lens equation
1 / f = 1 / v - 1 / u
- 1 / 20 = 1 / v + 1 / 30
1 / v = - 5 / 60
v = - 12 cm
m = v / u = - 12 / (-30) = 0.4
-- Take a sample of the first fluid.
-- Measure its mass.
-- Measure its volume.
-- Divide its mass by its volume.
This gives you the density of the first fluid.
-- Take a sample of the second fluid.
-- Measure its mass.
-- Measure its volume.
-- Divide its mass by its volume.
This gives you the density of the second fluid.
You want their average ?
OK
-- Add (Density of the first fluid) + (Density of the second fluid).
-- Divide the sum by 2 .
Now you have the average of the two densities.
Note:
That's NOT necessarily the density of a mixture when you
pour some of fluid-1 and fluid-2 into a jar. The density of the
fluid in the jar is going to depend on how much of each fluid is
in there.
I started to calculate how much of each one has to be there in order
for the density of the mixture to be equal to the average of their two
densities. But then I sat up straight, asked myself "Why ? !" .
Then I stopped, and went into the kitchen and ate some meatloaf.