Answer:
Answer:
r=\dfrac{1}{B}\sqrt{\dfrac{2Vm}{e}}r=
B
1
e
2Vm
Explanation:
Let m and e are the mass and charge of an electron. It is accelerated from rest through a potential difference V and are then deflected by a magnetic field that is perpendicular to their velocity. Let v is the velocity of the electron. It can be calculated as :
\dfrac{1}{2}mv^2=eV
2
1
mv
2
=eV
v=\sqrt{\dfrac{2eV}{m}}v=
m
2eV
When the electron enters the magnetic field, the centripetal force is balanced by the magnetic force as :
\dfrac{mv^2}{r}=evB
r
mv
2
=evB
r=\dfrac{mv}{eB}r=
eB
mv
or
r=\dfrac{1}{B}\sqrt{\dfrac{2Vm}{e}}r=
B
1
e
2Vm
So, the radius of the resulting electron trajectory is \dfrac{1}{B}\sqrt{\dfrac{2Vm}{e}}
B
1
e
2Vm
. Hence, this is the required solution.
Given :
Mass of box , m = 250 kg.
Force applied , F = 285 N.
The value of the incline angle is 30°.
the coefficient of dynamic friction is
.
To Find :
The speed with which the box is moving with, assuming it takes 4 seconds to reach the top of the incline.
Solution :
Net force applied in box is :
![F=285 - mgsin\ \theta - \mu mg cos \ \theta\\ \\F=285-mg( sin \ \theta - \mu cos\ \theta)\\\\F=285 - 20\times 10( \dfrac{1}{2}+0.72\times \dfrac{\sqrt{3}}{2})\\\\F=60.29\ N](https://tex.z-dn.net/?f=F%3D285%20-%20mgsin%5C%20%5Ctheta%20-%20%5Cmu%20mg%20cos%20%5C%20%5Ctheta%5C%5C%20%5C%5CF%3D285-mg%28%20sin%20%5C%20%5Ctheta%20-%20%5Cmu%20cos%5C%20%5Ctheta%29%5C%5C%5C%5CF%3D285%20-%2020%5Ctimes%2010%28%20%5Cdfrac%7B1%7D%7B2%7D%2B0.72%5Ctimes%20%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%29%5C%5C%5C%5CF%3D60.29%5C%20N)
Acceleration ,
.
By equation of motion :
![v=u+at\\\\v=0+3.01\times 4\\\\v=12.04\ m/s](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5C%5Cv%3D0%2B3.01%5Ctimes%204%5C%5C%5C%5Cv%3D12.04%5C%20m%2Fs)
Therefore, the speed of box is 12.04 m/s.
Hence, this is the required solution.
I'd Say B :) Your Welcome LOL!!!!!
Because real electrons are in an electron cloud, not layers