Base your answer to the question on the information below.A go-cart travels around a flat, horizontal, circular track with a rad
1 answer:
Explanation:
It is given that,
Radius of the circular track, r = 25 m
Mass of the go cart with the rider, m = 200 kg
The magnitude of the maximum centripetal force exerted by the track on the go-cart is 1200 N.
The centripetal force exerted by the track o the go cart is given by :
v is the maximum speed at which the go-cart could travel without sliding off this track.
The maximum speed is directly proportional to the force and radius of track. On increasing force and radius and decreasing the mass would increase maximum speed. It is given by :
v = 12.24 m/s
Hence, this is the required solution.
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Answer:
magnitude of the magnetic field 0.692 T
Explanation:
given data
rectangular dimensions = 2.80 cm by 3.20 cm
angle of 30.0°
produce a flux Ф = 3.10 × Wb
solution
we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm
and here angle between magnitude field and area will be ∅ = 90 - 30
∅ = 60°
and flux is express as
flux Ф = .................1
and Ф = BA cos∅ ............2
so B =
and we know
A = ab
so
B = ..............3
put here value
B =
solve we get
B = 0.692 T
<h2>Thus the force of friction is 235 N</h2>
Explanation:
When the bear was at the height of 14 m . Its potential energy = m g h
here m is the mass of bear , g is acceleration due to gravity and h is the height .
Thus P.E = 27 x 10 x 14 = 3780 J
The K.E of the bear just before hitting = m v²
= x 27 x ( 6.1 )² = 490 J
The force of friction f = P.E - K.E = 3290 J
Because the work done = Force x Distance
Thus frictional force = = 235 N