The voltage across an inductor ' L ' is
V = L · dI/dt .
I(t) = I(max) sin(ωt)
dI/dt = I(max) ω cos(ωt)
V = L · ω · I(max) cos(ωt)
L = 1.34 x 10⁻² H
ω = 2π · 60 = 377 /sec
I(max) = 4.80 A
V = L · ω · I(max) cos(ωt)
V = (1.34 x 10⁻² H) · (377 / sec) · (4.8 A) · cos(377 t)
<em>V = 24.25 cos(377 t)</em>
V is an AC voltage with peak value of 24.25 volts and frequency = 60 Hz.
Answer:
(A) Distance will be equal to 1.75 km
(B) Displacement will be equal to 1.114 km
Explanation:
We have given circumference of the circular track = 3.5 km
Circumference is given by 
r = 0.557 km
(a) It is given that car travels from southernmost point to the northernmost point.
For this car have to travel the distance equal to semi perimeter of the circular track
So distance will be equal to 
(b) If car go along the diameter of the circular track then it will also go from southernmost point to the northernmost point. and it will be equal to diameter of the track
So displacement will be equal to d = 2×0.557 = 1.114 m
Answer:
Explanation:
Time taken to accelerate to 28 m /s
= 28 / 2 = 14 s
a ) Total length of time in motion
= 14 + 41 + 5
= 60 s .
b )
Distance covered while accelerating
s = ut + 1/2 at²
= 0 + .5 x 2 x 14²
= 196 m .
Distance covered while moving in uniform motion
= 28 x 41
= 1148 m
distance covered while decelerating
v = u - at
0 = 28 - a x 5
a = 5.6 m / s²
v² = u² - 2 a s
0 = 28² - 2 x 5.6 x s
s = 28² / 2 x 5.6
= 70 m .
Total distance covered
= 196 + 1148 + 70
= 1414 m
total time taken = 60 s
average velocity
= 1414 / 60
= 23.56 m /s .
Answer:
<h2>66.67 km/hr</h2>
Explanation:
The average velocity of the car can be found by using the formula

d is the distance
t is the time taken
From the question we have

We have the final answer as
<h3>66.67 km/hr</h3>
Hope this helps you
No because your opinion and beliefs answers many questions