Answer:
Part a)

Part b)

Part c)

Explanation:
Part a)
As we know that it starts from rest and moves on runway by total distance 165 m
so we will have



now we have

Part b)
Now for take off time we will have



Part c)



Answer:
i = 0.00077A
Explanation:
Given:
loop radius, r = 3.0 cm = 0.03 m
Area, A = π x r² = π x 0.03² = 0.0028 m²
Magnetic Field, B = 0.75 T
Loop resistance, R = 18 Ω
time, t = 0.15 seconds
Now,
the induced emf is given as:
EMF = - BA/t .......1
Likewise,
EMF = iR.......2
Equate 1 and 2
iR = - BA/t
i = - BA/tR
i = 0.75×0.0028/0.15×18
i = 0.0021/2.7
i = 0.00077A
The definition is a nuclear reactor that generates more fissile material than it consumes.
Answer:
W_net = μ 5.58, μ = 0.1 W_net = 0.558 J
Explanation:
The work is defined by the related
W = F. d = F d cos θ
where bold indicates vectors.
In the case, the work of the friction force on a circular surface is requested.
The expression for the friction force is
fr = μ N
the friction force opposes the movement, therefore the angle is 180º and the cos 180 = -1
W = - fr d
the path traveled half the length of the circle
L = 2 π R
d = L / 2
d = π R
we substitute
W = - μ N d
Total work is initial to
W_neto = - μ π R (N_b - N_a)
let's calculate
W_net = - μ π 0.550 (0.670 - 3.90)
W_net = μ 5.58
for the complete calculation it is necessary to know the friction coefficient, if we assume that μ = 0.1
W_net = 0.1 5.58
W_net = 0.558 J