Solution :
We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :
If we put an electron on point P, then force on point e is :
![F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}](https://tex.z-dn.net/?f=F%3D%20%5Cfrac%7B-eKQx%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B3%2F2%7D%7D%3D%20%5Cfrac%7B-eKQx%7D%7Br%5E3%5B1%2B%5Cfrac%7Bx%5E2%7D%7Br%5E2%7D%5D%5E%7B3%2F2%7D%7D)
If r >> x , then 
Then, 
Compare, a = -ω²x
We get,
Answer:
TRUE
Explanation:
Low mass stars last lots longer.
Answer:
A) 12.57 m
B) 5 RPM
C) 3.142 m/s
Explanation:
A) Distance covered in 1 Revolution:
The formula that gives the relationship between the arc length or distance covered during circular motion to the angle subtended or the revolutions, is given as follows:
s = rθ
where,
s = distance covered = ?
r = radius of circle = 2 m
θ = Angle = 2π radians (For 1 complete Revolution)
Therefore,
s = (2 m)(2π radians)
<u>s = 12.57 m</u>
B) Angular Speed:
The formula for angular speed is given as:
ω = θ/t
where,
ω = angular speed = ?
θ = angular distance covered = 15 revolutions
t = time taken = 3 min
Therefore,
ω = 15 rev/3 min
<u>ω = 5 RPM</u>
C) Linear Speed:
The formula that gives the the linear speed of an object moving in a circular path is given as:
v = rω
where,
v = linear speed = ?
r = radius = 2 m
ω = Angular Speed in rad/s = (15 rev/min)(2π rad/1 rev)(1 min/60 s) = 1.571 rad/s
Therefore,
v = (2 m)(1.571 rad/s)
<u>v = 3.142 m/s</u>
Answer:
1995 and 2000 , 4 trillions
Explanation:
