Respuesta :
Explanation:
It is given that,
Radius of circular particle accelerator, r = 1 m
The distance covered by the particle is equal to the circumference of the circular path, d = 2πr
d = 2π × 1 m
(a) The speed of satellite is given by total distance divided by total time taken as :
[tex]speed=\dfrac{distance}{time}[/tex]
Let t is the period of the particle.
[tex]t=\dfrac{d}{s}[/tex]
d = distance covered
s = speed of particle
It is given that the charged particle is moving nearly with the speed of light
[tex]t=\dfrac{d}{c}[/tex]
[tex]t=\dfrac{2\pi\times 1\ m}{3\times 10^8\ m/s}[/tex]
[tex]t=2.09\times 10^{-8}\ s[/tex]
(b) On the circular path, the centripetal acceleration is given by :
[tex]a=\dfrac{c^2}{r}[/tex]
[tex]a=\dfrac{(3\times 10^8\ m/s)^2}{1\ m}[/tex]
[tex]a=9\times 10^{16}\ m/s^2[/tex]
Hence, this is the required solution.
