If the index of refraction of a medium is 1.4, determine the speed of light in that medium.
1 answer:
The speed of light in that medium is .
<u>Explanation:</u>
It is known that the light's speed is constant when it travels in vacuum and the value is . When the light enters another medium other than vacuum, its speed get decreased as the light gets refracted by an angle.
The amount of refraction can be determined by the index of refraction or refractive index of the medium. The refraction index is measured as the ratios of speed of light in vacuum to that in the medium. It is represented as η =
So, here η is the index of refraction of a medium which is given as 1.4, c is the light's speed in vacuum () and v is the light's speed in that medium which we need to find.
Thus the speed of light in that medium is
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By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.
Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .
now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.
charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.
Now , use vector for solve it.
vector = vector Fe + vector Fn,
⇒ || =
⇒ Fe = Fs = KqQ/(1m)² = KqQ
⇒ = √{Fe² + Fs²} = √{(kqQ)²+(KqQ)²}
⇒ = √2KqQ
Hence, net force act on q {my charge } is √2KqQ and the direction of force is S - W (southwest )direction.
To learn more about positive charges here
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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