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:
Around 2.8212 sec
Explanation:
Given the eqn x=1/2at^2+vot
your vo=0
39=1/2(-9.8)t^2
=7.95=t^2
=2.82sec
I think the puck pushes the stick backwards
Answer:
Solution
λ=v/n
Here, v=344 m s−1
n=22 MHz =22×106 Hz
λ=344/22×106=15.64×10−6m=15.64μm.
I’m gonna have to say “Ocean waves” as the answer
