Answer:
9.47 rad/s^2
Explanation:
Diameter = 15 cm, radius, r = diameter / 2 = 7.5 cm = 0.075 m, u = 0, v = 7.1 m/s,
s = 35.4 m
let a be the linear acceleration.
Use III equation of motion.
v^2 = u^2 + 2 a s
7.1 x 7.1 = 0 + 2 x a x 35.4
a = 0.71 m/s^2
Now the relation between linear acceleration and angular acceleration is
a = r x α
where, α is angular acceleration
α = 0.71 / 0.075 = 9.47 rad/s^2
I think the answer is B true
Answer:
Explanation:
A proton of charge
q=+1.609×10^-19C
Orbit a radius of 12cm
r=0.12m
Magnetic Field of 0.31T
Angle between velocity and field is 90°
a. Because the magnetic force F supplies the centripetal force Fc.
The magnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by
F = qvB sin θ
And the centripetal force is given as
Fc=mv²/r
Where m is mass of proton
m=1.673×10^-27kg
Then, F=Fc
qvB sin θ=mv²/r
qBSin90=mv/r
rqB=mv
Then, v=rqB/m
v=0.12×1.609×10^-19×0.31/1.673×10^-23
v=3577692.78m/s
v=3.58×10^6m/s
b. Since,
F=qVBSin90
F=1.609×10^-19×3.58×10^6×0.31
F=1.785×10^-13 N.