A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60
2 answers:
Answer:
0.012m
Explanation:
To find the radius of the path you can use the formula for the radius od the trajectory of a charge that moves in a constant magnetic field:
m: mass of the proton 9.1*10^{-31}kg
q: charge of the proton 1.6*10^{-19}C
B: magnitude of the magnetic field 0.60T
v: velocity of the proton
In order to use the formula you need to calculate the velocity of the proton. This can be made by using the potential difference and charge, that equals the kinetic energy of the proton:
Then, by replacing in the formula for the radius you obtain:
hence, the radius is 0.012m
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Answer:
a = 7.35 ft / s²
Explanation:
For this exercise we must use the kinematics relations
x = v₀ t + ½ a t²
as the runner leaves the starting line his initial velocity is zero
x = ½ a t²
a =
let's reduce the distance to foot
x = 60 yd (3ft / 1yd) = 180 ft
let's calculate
a = 2 180 / 7²
a = 7.35 ft / s²