Answer:
Magnitude of the force on proton = F = 1.1085 × 10^-15 N
Explanation:
Charge on proton = q = 1.60 × 10^-19 C
Velocity of proton = V = 4.0 × 10^4 m/s
Magnetic field = B = 0.20 T
Angle between V and B = θ = 60
We know that,
F = qVBsin θ = (1.60 × 10^-19)( 4.0 × 10^4)( 0.20)sin(60)
F = 1.1085 × 10^-15 N
Answer:
sorry I don't know I am too bad a t this coz I am only at class 7
Strength of the magnetic field: 20 T
Explanation:
For a conductive wire moving perpendicular to a magnetic field, the electromotive force (voltage) induced in the wire due to electromagnetic induction is given by

where
B is the strength of the magnetic field
v is the speed of the wire
L is the length of the wire
For the wire in this problem, we have:
(induced emf)
L = 0.20 m (length of the wire)
v = 3.0 m/s (speed)
Solving for B, we find the strength of the magnetic field:

Learn more about magnetic fields:
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Answer:


Explanation:
Knowing that the truck travels up a hill with a 5.4º incline
The horizontal component of the truck's velocity is:

The vertical component of the truck's velocity is:
