Answer:
a) α = 1.875 
b) t = 8 s
Explanation:
Given:
ω1 = 0 
ω2 = 15 
theta (angular displacement) = 60 rad
*side note: you can replace regular, linear variables in kinematic equations with angular variables (must entirely replace equations with angular variables)*
a) α = ?
(ω2)^2 = (ω1)^2 + 2α(theta)
=
+ 2(α)(60)
225 = 120α
α = 1.875 
b)
α = (ω2-ω1)/t
t = (ω2-ω1)/α = (15-0)/1.875 = 8
t = 8 s
Complete Question
The complete question is shown on the first uploaded image
Answer:
The maximum emf is 
The emf induced at t = 1.00 s is 
The maximum rate of change of magnetic flux is 
Explanation:
From the question we are told that
The number of turns is N = 44 turns
The length of the coil is 
The width of the coil is 
The magnetic field is 
The angular speed is 
Generally the induced emf is mathematically represented as

Where
is the maximum induced emf and this is mathematically represented as

Where
is the magnetic flux
N is the number of turns
A is the area of the coil which is mathematically evaluated as

Substituting values


substituting values into the equation for maximum induced emf


given that the time t = 1.0sec
substituting values into the equation for induced emf 


The maximum induced emf can also be represented mathematically as

Where
is the magnetic flux and
is the maximum rate at which magnetic flux changes the value of the maximum rate of change of magnetic flux is

Answer:4.39 s
Explanation:
Given
initial velocity 
acceleration 
velocity acquired by sled in
time


distance traveled by sled in 



distance traveled in
time with velocity 




----1


substitute the value of
in 1
we get

thus 

Answer:
Explanation:
Given that,
Mass of the thin hoop
M = 2kg
Radius of the hoop
R = 0.6m
Moment of inertial of a hoop is
I = MR²
I = 2 × 0.6²
I = 0.72 kgm²
Period of a physical pendulum of small amplitude is given by
T = 2π √(I / Mgd)
Where,
T is the period in seconds
I is the moment of inertia in kgm²
I = 0.72 kgm²
M is the mass of the hoop
M = 2kg
g is the acceleration due to gravity
g = 9.8m/s²
d is the distance from rotational axis to center of of gravity
Therefore, d = r = 0.6m
Then, applying the formula
T = 2π √ (I / MgR)
T = 2π √ (0.72 / (2 × 9.8× 0.6)
T = 2π √ ( 0.72 / 11.76)
T = 2π √0.06122
T = 2π × 0.2474
T = 1.5547 seconds
T ≈ 1.55 seconds to 2d•p
Then, the period of oscillation is 1.55seconds