Help me please please
1 answer:
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The speed of light in a material is given by:
where
is the speed of light in vacuum
n is the refractive index of the material
The lens in this problem has a refractive index of n=1.50, therefore the speed of light in the lens is
And the correct answer is C).
where
n is the refractive index of the material
The lens in this problem has a refractive index of n=1.50, therefore the speed of light in the lens is
And the correct answer is C).
Answer:
205 V
V = 2.05 V
Explanation:
L = Inductance in Henries, (H) = 0.500 H
resistor is of 93 Ω so R = 93 Ω
The voltage across the inductor is
w = 500 rad/s
IwL = 11.0 V
Current:
I = 11.0 V / wL
= 11.0 V / 500 rad/s (0.500 H)
= 11.0 / 250
I = 0.044 A
Now
V = IR
= (0.044 A) (93 Ω)
V = 4.092 V
Deriving formula for voltage across the resistor
The derivative of sin is cos
V = V
cos (wt)
Putting V = 4.092 V and w = 500 rad/s
V = V
cos (wt)
= (4.092 V) (cos(500 rad/s )t)
So the voltage across the resistor at 2.09 x 10-3 s is which means
t = 2.09 x 10⁻³
V = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))
= (4.092 V) (cos (500 rads/s)(0.00209))
= (4.092 V) (cos(1.045))
= (4.092 V)(0.501902)
= 2.053783
V = 2.05 V