Answer:
16 cm
Explanation:
For protons:
Energy, E = 300 keV
radius of orbit, r1 = 16 cm
the relation for the energy and velocity is given by

So,
.... (1)
Now,

Substitute the value of v from equation (1), we get

Let the radius of the alpha particle is r2.
For proton
So,
... (2)
Where, m1 is the mass of proton, q1 is the charge of proton
For alpha particle
So,
... (3)
Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle
Divide equation (2) by equation (3), we get

q1 = q
q2 = 2q
m1 = m
m2 = 4m
By substituting the values

So, r2 = r1 = 16 cm
Thus, the radius of the alpha particle is 16 cm.
Answer:
Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all - option A
Explanation:
Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all; the reason being that: no magnetic field is being produced by a charge that is static. Only a moving charge can produce a magnetic effect. And the magnet can not have any torque due to its own magnetic lines of force.
The answers is 30 miles per hour, the driver is speeding the car up, section-H, 12 minutes, section-D, and 65 miles per hour.
Reword the questions and you’ll get it right