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:
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Answer:
The velocity of the ship relative to the earth V = 9.05
Explanation:
The local ocean current is = 1.52 m/s
Direction = 40°
Velocity component in X - direction = 1.52
°
= 1.164
Velocity component in Y - direction = 8 + 1.52
°
= 8.97
The velocity of the ship relative to the earth
Put the values of and
we get,
⇒
⇒ V = 9.05
This is the velocity of the ship relative to the earth.