A fish inside the water 12cm below the surface looking up through the water sees the outside world contained in a circular horiz
on. If the refractive index of water is 4/3 the radius of circle is
1 answer:
Answer:
13.6 cm
Explanation:
From Snell's law:
n₁ sin θ₁ = n₂ sin θ₂
In the air, n₁ = 1, and light from the horizon forms a 90° angle with the vertical, so sin θ₁ = sin 90° = 1.
Given n₂ = 4/3:
1 = 4/3 sin θ
sin θ = 3/4
If x is the radius of the circle, then sin θ is:
sin θ = x / √(x² + 12²)
sin θ = x / √(x² + 144)
Substituting:
3/4 = x / √(x² + 144)
9/16 = x² / (x² + 144)
9/16 x² + 81 = x²
81 = 7/16 x²
x ≈ 13.6
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After adding them, you will get the reciprocal again and then compute for the value. The problem says that there are 4 resistors in the circuit that have a resistance of 75.
Add up the numerator and copy the denominator:
Then get the reciprocal to get the total resistance:
The answer to your question then is A. 18.8.
After adding them, you will get the reciprocal again and then compute for the value. The problem says that there are 4 resistors in the circuit that have a resistance of 75.
Add up the numerator and copy the denominator:
Then get the reciprocal to get the total resistance:
The answer to your question then is A. 18.8.
Answer:
(a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is
Explanation:
Given that,
Charge q₁ = -4.00 μC
Inner radius = 3.13 m
Outer radius = 4.13 cm
Net charge q₂ = -6.43 μC
We need to calculate the charge on the outer surface
Using formula of charge
The charge on the inner surface is q.
We need to calculate the electric field outside the shell
Using formula of electric field
Put the value into the formula
Hence, (a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is