In which of the following materials is the velocity of light greatest?
2 answers:
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Answer:
the final angular velocity of the platform with its load is 1.0356 rad/s
Explanation:
Given that;
mass of circular platform m = 97.1 kg
Initial angular velocity of platform ω₀ = 1.63 rad/s
mass of banana = 8.97 kg
at distance r = 4/5 { radius of platform }
mass of monkey = 22.1 kg
at edge = R
R = 1.73 m
now since there is No external Torque
Angular momentum will be conserved, so;
mR²/2 × ω₀ = [ mR²/2 + (
R)² +
R² ]w
m/2 × ω₀ = [ m/2 + (
)² +
]w
we substitute
w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1
w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )
w = 48.55 × [ 1.63 / ( 76.3908 ) ]
w = 48.55 × 0.02133
w = 1.0356 rad/s
Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s
We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.
1) Equilibrium of forces:
where
is the weight of the person
is the weight of the scaffold
Re-arranging, we can write the equation as
(1)
2) Equilibrium of torques:
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using
and replacing T1 with (1), we find
from which we find
And then, substituting T2 into (1), we find
1) Equilibrium of forces:
where
Re-arranging, we can write the equation as
2) Equilibrium of torques:
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using
from which we find
And then, substituting T2 into (1), we find
Answer:
Magnitude of induced emf is 0.00635 V
Explanation:
Radius of circular loop r = 45 mm = 0.045 m
Area of circular loop
Magnetic field is increases from 250 mT to 350 mT
Therefore change in magnetic field
Emf induced is given by
Magnitude of induced emf is equal to 0.00635 V