A particle with a charge of 5.13 μC has a velocity of 8.64 x106 m/s in a direction perpendicular to a magnetic fieldof 1.99 x 10
1 answer:
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Refer to the diagram shown.
There are twelve 5-minute divisions.
Each 5-minute division is equal to 360°/12 = 30°.
By convention, angles are measured counterclockwise from the positive x-axis.
The angular position of the minute hand at 2:55 is
θ = 90° + 30° = 120°
Because 360° = 2π radians, therefore
θ = (120/360)*2π = (2π)/3 radians = 2.0944 radians
Answer: (2π)/3 radians ofr 2.0944 radians.
There are twelve 5-minute divisions.
Each 5-minute division is equal to 360°/12 = 30°.
By convention, angles are measured counterclockwise from the positive x-axis.
The angular position of the minute hand at 2:55 is
θ = 90° + 30° = 120°
Because 360° = 2π radians, therefore
θ = (120/360)*2π = (2π)/3 radians = 2.0944 radians
Answer: (2π)/3 radians ofr 2.0944 radians.
Answer:
v = 5.75 x 10⁶ m/s
Explanation:
The radius (r) of the circular orbit taken by a charged particle is related to its speed perpendicular to a magnetic field of strength B, and is given by
r = --------------(i)
Where,
q = charge of the particle
m = mass of the particle
Making v subject of the formula in equation (i) above gives
v = -------------------(ii)
Given;
r = 20cm = 0.2m
B = 0.3T
v = unknown
q = charge of proton = 1.6 x 10⁻¹⁹ C
m = mass of the proton = 1.67 x 10⁻²⁷kg
Substitute the values of m, q, B and r into equation (ii) above to get;
v =
Solving for v gives:
v = 5.75 x 10⁶ m/s
Therefore, the velocity of the proton is 5.75 x 10⁶ m/s