How long does a car (1000 kg) have a speed of 30 m/s from a rest if the engine power is 10kw
1 answer:
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Answer:
Explanation:
radius of the solenoid, r = 0.05 m
length of the solenoid, l = 0.39 m
Magnetic field of the solenoid, B = 2 x 10^-5 T
Number of turns, N = 200
The magnetic field of the solenoid is given by
where, i be the current and n be the number of turns per unit length
n = N / l = 200 / 0.39 = 512.8
i = 0.031 A
Answer:
Explanation:
q = Charge of proton =
r = Radius of circle =
v = Velocity of proton =
Magnetic moment is given by
The magnetic moment associated with this motion is
The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.