Θ is the angular displacement = ωt
ω is the angular velocity = θ/t
α is the angular acceleration = ω/t
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
Answer
Given,
Average speed of Malcolm and Ravi = 260 km/h
Let speed of the Malcolm be X and speed of the Ravi Y.
From the given statement
....(i)
....(ii)
Adding both the equations
3 X = 600
X = 200 km/h
Putting value in equation (i)
Y = 520 - 200
Y = 320 Km/h
Speed of Malcolm = 200 Km/h
Speed of Ravi = 320 Km/h
Hi there!
We can use the rotational equivalent of Newton's Second Law:
Στ = Net Torque (Nm)
I = Moment of inertia (kgm²)
α = Angular acceleration (rad/sec²)
We can plug in the given values to solve.
I would make the ramp flatter. In doing so the ramp would have to be longer.
