To solve the problem it is necessary to apply conservation of the moment and conservation of energy.
By conservation of the moment we know that
![MV=mv](https://tex.z-dn.net/?f=MV%3Dmv)
Where
M=Heavier mass
V = Velocity of heavier mass
m = lighter mass
v = velocity of lighter mass
That equation in function of the velocity of heavier mass is
![V = \frac{mv}{M}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bmv%7D%7BM%7D)
Also we have that ![m/M = 1/7 times](https://tex.z-dn.net/?f=m%2FM%20%3D%201%2F7%20times)
On the other hand we have from law of conservation of energy that
![W_f = KE](https://tex.z-dn.net/?f=W_f%20%3D%20KE)
Where,
W_f = Work made by friction
KE = Kinetic Force
Applying this equation in heavier object.
![F_f*S = \frac{1}{2}MV^2](https://tex.z-dn.net/?f=F_f%2AS%20%3D%20%5Cfrac%7B1%7D%7B2%7DMV%5E2)
![\mu M*g*S = \frac{1}{2}MV^2](https://tex.z-dn.net/?f=%5Cmu%20M%2Ag%2AS%20%3D%20%5Cfrac%7B1%7D%7B2%7DMV%5E2)
![\mu g*S = \frac{1}{2}( \frac{mv}{M})^2](https://tex.z-dn.net/?f=%5Cmu%20g%2AS%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28%20%5Cfrac%7Bmv%7D%7BM%7D%29%5E2)
![\mu = \frac{1}{2} (\frac{1}{7}v)^2](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28%5Cfrac%7B1%7D%7B7%7Dv%29%5E2)
![\mu = \frac{1}{98}v^2](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B1%7D%7B98%7Dv%5E2)
![\mu = \frac{1}{g(98)(5.1)}v^2](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B1%7D%7Bg%2898%29%285.1%29%7Dv%5E2)
Here we can apply the law of conservation of energy for light mass, then
![\mu mgs = \frac{1}{2} mv^2](https://tex.z-dn.net/?f=%5Cmu%20mgs%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
Replacing the value of ![\mu](https://tex.z-dn.net/?f=%5Cmu)
![\frac{1}{g(98)(5.1)}v^2 mgs = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bg%2898%29%285.1%29%7Dv%5E2%20%20mgs%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
Deleting constants,
![s= \frac{(98*5.1)}{2}](https://tex.z-dn.net/?f=s%3D%20%5Cfrac%7B%2898%2A5.1%29%7D%7B2%7D)
![s = 249.9m](https://tex.z-dn.net/?f=s%20%3D%20249.9m)
Answer:
The momentum of bath cars is 40000 Ns which make the difficulty to stop each car in aspect of fprce is the same.
Explanation:
Momentum (P) =mass(m) × velocity (v)
For car A,
P = m × v = 1000 × 40 = 40000 Ns
For car B,
P = m × v = 4000 × 10 = 40000 Ns
Force (F) = Momentum change(ΔΡ)/ time taken(t)
F = ΔΡ/t
When stopping the car the momentum changes from 40000 Ns to 0
So momentum change in both cars is the same. So to stop the two cars in a given time (t) you need the same force, which means you will feel same difficulty.
Answer:
The frequency of the green light is ![6x10^{14}Hz](https://tex.z-dn.net/?f=6x10%5E%7B14%7DHz)
Explanation:
The visible region is part of the electromagnetic spectrum, any radiation of that electromagnetic spectrum has a speed of
in the vacuum.
Green light is part of the visible region. Therefore, the frequency can be determined by the following equation:
(1)
Where c is the speed of light,
is the wavelength and
is the frequency.
Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave (that is a magnetic field perpendicular to an electric field).
Then,
can be isolated from equation 1
(2)
Notice that it is necessary to express the wavelength in units of meters.
⇒ ![5x10^{-7}m](https://tex.z-dn.net/?f=5x10%5E%7B-7%7Dm)
Hence, the frequency of the green light is ![6x10^{14}Hz](https://tex.z-dn.net/?f=6x10%5E%7B14%7DHz)