Answer:
The maximum emf that can be generated around the perimeter of a cell in this field is 
Explanation:
To solve this problem it is necessary to apply the concepts on maximum electromotive force.
For definition we know that
Where,
N= Number of turns of the coil
B = Magnetic field
Angular velocity
A = Cross-sectional Area
Angular velocity according kinematics equations is:
Replacing at the equation our values given we have that
Therefore the maximum emf that can be generated around the perimeter of a cell in this field is
Answer:
Explanation:
We apply Newton's second law at the crate :
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
m=90kg : crate mass
F= 282 N
μk =0.351 :coefficient of kinetic friction
g = 9.8 m/s² : acceleration due to gravity
Crate weight (W)
W= m*g
W= 90kg*9.8 m/s²
W= 882 N
Friction force : Ff
Ff= μk*N Formula (2)
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W = 0
N = W
N = 882 N
We replace the data in the formula (2)
Ff= μk*N = 0.351* 882 N
Ff= 309.58 N
We apply the formula (1) in x direction:
∑Fx = m*ax , ax=0
282 N - 309.58 N = 90*a
a= (282 N - 309.58 N ) / (90)
a= - 0.306 m/s²
Kinematics of the crate
Because the crate moves with uniformly accelerated movement we apply the following formula :
vf²=v₀²+2*a*d Formula (3)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
v₀ = 0.850 m/s
d = 0.75 m
a= - 0.306 m/s²
We replace the data in the formula (3)
vf²=(0.850)²+(2)( - 0.306 )(0.75 )
Answer:
α=0.625rad/s^2
v=340m/s
w=10rad/s
θ=320rad
Explanation:
Constant angular acceleration = ∆w/∆t
angular acceleration = 20/32
α=0.625rad/s^2
Linear velocity v=wr
v = 20×17= 340m/s
Average angular velocity
w0+w1/2
w= 0+20/2
w= 20/2
w=10rad/s
What angle did it rotate with
θ=wt
θ= 10×32
=320rad
