A calcium-40 ion has a positive charge that is double the charge of a proton, and a mass of 6.64 ✕ 10−26 kg. At a particular ins
tant, it is moving with a speed of 5.00 ✕ 106 m/s through a magnetic field. At this instant, its velocity makes an angle of 61° with the direction of the magnetic field at the ion's location. The magnitude of the field is 0.230 T.
(a)What is the magnitude of the magnetic force (in N) on the ion?
N
B)What is the magnitude of the ion's acceleration (in m/s2) at this instant?
m/s2
(a)What is the magnitude of the magnetic force (in N) on the ion?
N
B)What is the magnitude of the ion's acceleration (in m/s2) at this instant?
m/s2
1 answer:
You might be interested in
Answer:
1000 N
Explanation:
First, we need to find the deceleration of the running back, which is given by:
where
v = 0 is his final velocity
u = 5 m/s is his initial velocity
t = 0.5 s is the time taken
Substituting, we have
And now we can calculate the force exerted on the running back, by using Newton's second law:
so, the magnitude of the force is 1000 N.