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.
3 as a single number is considered a sf
We use the formula,
.
Here, v is velocity and its value given 26 mi/h ( in m/s,
) and d is distance and its value is given 80 m.
Substituting these values in above formula we get,

Thus, the time delay between green lights on successive blocks to keep the traffic moving continuously is 6.88 s
Answer:
The transverse component of acceleration is 26.32
where as radial the component of acceleration is 8.77 
Explanation:
As per the given data
u=π/4 rad
ω=u'=2 rad/s
α=u''=4 rad/s

So the transverse component of acceleration are given as

Here


So

The transverse component of acceleration is 26.32 
The radial component is given as

Here

So

The radial component of acceleration is 8.77 