A car (mass of 830 kg) is sitting on a car lift in a shop (neglect the mass of the lift itself). While the car is being lifted u
1 answer:
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Answer:
(a) 0.345 T
(b) 0.389 T
Solution:
As per the question:
Hall emf,
Magnetic Field, B = 0.10 T
Hall emf,
Now,
Drift velocity,
Now, the expression for the electric field is given by:
(1)
And
Thus eqn (1) becomes
where
d = distance
(2)
(a) When
(b) When
Answer:
a)Velocity of car =v=16 m/s
b)Force against the track at point B=1.15*N
Explanation:
Given mass of roller coaster=m=350 kg
Position of A=Ha=25 m
Position of B=Hb=12 m
Net potential energy=mg(ha-hb)
Net potential energy=(350)(9.80)(25-12)
Net potential energy=44590 J
Using energy conservation
net kinetic energy=net potential energy
(1/2)mv^2=mg(ha-hb)
m=350
velocity=v=16 m/s
b)There two force acting,centripetal force upward and gravity downward.
Thus net force acting will be
Net force=(mv^2/r)-mg
Net force=14933.33-3430
Net force=1.15* N