A collection of 1.25×1019 electrons has the charge -3c, -1c, -2c
1 answer:
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Explanation:
It is given that,
Initial speed of sprinter, u = 0
Final speed of sprinter, v = 10 m/s
Time taken, t = 1.28 s
a. We need to find the acceleration of sprinter. It can be calculated using first equation of motion as :
b. Final speed of the sprinter, v = 36 km/h
Time, t = 0.000355 h
Acceleration,
Hence, this is the required solution.
Answer:
Explanation:
Given the following :
Speed (V) = speed of 2.30×10^7 m/s
Acceleration (a) = 1.70×10^13 m/s^2
Using the right hand rule provided by Lorentz law:
B = F / qvSinΘ
Where B = magnitude of the magnetic field
v = speed of the particle
Θ = 90° (perpendicular to the field)
q = charge of the particle
SinΘ = sin90° = 1
Note F = ma
Therefore,
B = ma / qvSinΘ
Mass of proton = 1.67 × 10^-27
Charge = 1.6 × 10^-19 C
B = [(1.67 × 10^-27) × (1.70 × 10^13)] / (1.6 × 10^-19) × (2.30 × 10^7) × 1
B = 2.839 × 10^-14 / 3.68 × 10^-12
B = 0.7715 × 10^-2
B = 7.72 × 10^-3 T
2) Magnetic field will be in the negative y direction according to the right hand thumb rule.
Since Velocity is in the positive z- direction, acceleration in the positive x - direction, then magnetic field must be in the negative y-direction.
