Answer:
(a) ![v=22.177\times 10^{12}\ m.s^{-1}](https://tex.z-dn.net/?f=v%3D22.177%5Ctimes%2010%5E%7B12%7D%5C%20m.s%5E%7B-1%7D)
(b) ![B=420.855\ T](https://tex.z-dn.net/?f=B%3D420.855%5C%20T)
(c) ![f=11.7647\times 10^{14}\ Hz](https://tex.z-dn.net/?f=f%3D11.7647%5Ctimes%2010%5E%7B14%7D%5C%20Hz)
(d) ![T=8.5\times 10^{-14}s](https://tex.z-dn.net/?f=T%3D8.5%5Ctimes%2010%5E%7B-14%7Ds)
Explanation:
Given:
Kinetic Energy of an electron, ![KE=1.4\ keV=1400\ eV=1400\times 1.6\times 10^{-19}\ J](https://tex.z-dn.net/?f=KE%3D1.4%5C%20keV%3D1400%5C%20eV%3D1400%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%5C%20J)
radius of the orbit, ![r=0.3\ m](https://tex.z-dn.net/?f=r%3D0.3%5C%20m)
we have:
mass of an electron, ![m=9.109\times 10^{-31}\ kg](https://tex.z-dn.net/?f=m%3D9.109%5Ctimes%2010%5E%7B-31%7D%5C%20kg)
charge on an electron, ![q=1.6\times 10^{-19}\ C](https://tex.z-dn.net/?f=q%3D1.6%5Ctimes%2010%5E%7B-19%7D%5C%20C)
(a)
we know:
![KE=\frac{1}{2} m.v^2](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7B1%7D%7B2%7D%20m.v%5E2)
![1400\times 1.6\times 10^{-19}=0.5\times (9.109\times 10^{-31})\times v^2](https://tex.z-dn.net/?f=1400%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%3D0.5%5Ctimes%20%289.109%5Ctimes%2010%5E%7B-31%7D%29%5Ctimes%20v%5E2)
![v=22.177\times 10^{12}\ m.s^{-1}](https://tex.z-dn.net/?f=v%3D22.177%5Ctimes%2010%5E%7B12%7D%5C%20m.s%5E%7B-1%7D)
(b)
<em>We also have the relation after the comparison of forces(centripetal and magnetic) on a moving charge in a magnetic field as:</em>
...........................(1)
where:
B = magnetic field normal to the plane of circulating charge
putting respective values in eq. (1)
![(9.109\times 10^{-31})\times (22.177\times 10^{12})=(1.6\times 10^{-19})\times 0.3\times B](https://tex.z-dn.net/?f=%289.109%5Ctimes%2010%5E%7B-31%7D%29%5Ctimes%20%2822.177%5Ctimes%2010%5E%7B12%7D%29%3D%281.6%5Ctimes%2010%5E%7B-19%7D%29%5Ctimes%200.3%5Ctimes%20B)
![B=420.855\ T](https://tex.z-dn.net/?f=B%3D420.855%5C%20T)
(d)
angular speed:
![\omega=\frac{v}{r}](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7Bv%7D%7Br%7D)
![\omega=\frac{22.177\times 10^{12}}{0.3}](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B22.177%5Ctimes%2010%5E%7B12%7D%7D%7B0.3%7D)
![\omega=73.92\times 10^{12}\ rad.s^{-1}](https://tex.z-dn.net/?f=%5Comega%3D73.92%5Ctimes%2010%5E%7B12%7D%5C%20rad.s%5E%7B-1%7D)
∴Time taken for 1 radian:
![t=\frac{1}{\omega}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B1%7D%7B%5Comega%7D)
![t=\frac{1}{73.92\times 10^{12}}\ s.rad^{-1}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B1%7D%7B73.92%5Ctimes%2010%5E%7B12%7D%7D%5C%20s.rad%5E%7B-1%7D)
Now time take for 1 circulation i.e. 2π radians(Time period):
![T=2\pi\times t](https://tex.z-dn.net/?f=T%3D2%5Cpi%5Ctimes%20t)
![T=2\pi\times \frac{1}{73.92\times 10^{12}}](https://tex.z-dn.net/?f=T%3D2%5Cpi%5Ctimes%20%5Cfrac%7B1%7D%7B73.92%5Ctimes%2010%5E%7B12%7D%7D)
![T=8.5\times 10^{-14}s](https://tex.z-dn.net/?f=T%3D8.5%5Ctimes%2010%5E%7B-14%7Ds)
(c)
we know frequency :
![f=\frac{1}{T}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7BT%7D)
![f=\frac{1}{8.5\times 10^{-14}}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7B8.5%5Ctimes%2010%5E%7B-14%7D%7D)
![f=11.7647\times 10^{14}\ Hz](https://tex.z-dn.net/?f=f%3D11.7647%5Ctimes%2010%5E%7B14%7D%5C%20Hz)
The water molecules would slow down, and as they slow down, the heat created from their movement would cease.
The weight of an object is (mass) x (gravity).
The weight of Mr. McDonald's object is (112) x (9.8) = <u>1,097.6 newtons</u>.
(about 247 pounds)
That's the force pulling the object down, because it is near the Earth, and
the Earth and the object are attracting each other with forces of gravity.
In order to move the object away from the center of the Earth ("lift" it), a force greater than 1,097.6 newtons must be applied to it <em>in the other direction</em> ... <u>upwards</u>.
<em><u>Any</u></em> force greater than its weight will lift it. The more the upward force exceeds the minimum of 1,097.6 newtons, the faster Mr. McDonald's object will <u>accelerate</u> upwards.
The formula for the force of Kinetic friction is:
![F_{k} =](https://tex.z-dn.net/?f=F_%7Bk%7D%20%3D%20)
μ*N
Please consider μ as the coefficient of the kinetic friction = 0.2
N = Normal to the desk = m*g = 3.5 * 9.8 = 34.3 N
![F_{k} =](https://tex.z-dn.net/?f=F_%7Bk%7D%20%3D%20)
Force of Kinetic Friction = ?
Therefore,
The force of Kinetic Friction =
![F_{k} =](https://tex.z-dn.net/?f=F_%7Bk%7D%20%3D%20)
0.2 * 34.3
![F_{k} =](https://tex.z-dn.net/?f=F_%7Bk%7D%20%3D%20)
6.86N
The size of force of Kinetic Friction = 6.86N.