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:
a= 0.22 m/s²
Explanation:
Given that
M = 3.5 kg
θ = 30°
m = 1 kg
μ= 0.3
The force due to gravity
F₁= M g sinθ
F₁=3.5 x 10 x sin 30
F₁= 17.5 N
F₂ = m g
F₂ = 1 x 10 = 10 N
The maximum value of the friction force on the incline plane
Fr = μ M g cosθ
Fr = 0.3 x 2.5 x 10 cos30°
Fr= 6.49 N
Lets take acceleration of the system is a m/s²
F₁ - F₂ - Fr = (M+m) a
17.5 - 10 - 6.49 = (3.5+1)a
a= 0.22 m/s²
