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kotegsom [21]
3 years ago
8

The back wall of a home aquarium is a mirror that is a distance of 46.0 cm away from the front wall. The walls of the tank are n

egligibly thin. A fish, swimming midway between the front and back walls, is being viewed by a person looking through the front wall. The index of refraction of air is nair = 1.000 and that of water is nwater = 1.333. (a) Calculate the apparent distance between the fish and the front wall. (b) Calculate the apparent distance between the image of the fish and the front wall.
Physics
1 answer:
monitta3 years ago
5 0

Answer:

i know the questin but i got to try and find it

Explanation:

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oee [108]
Weight=mg
g=GM/r^2
g on venus is 0.80w as radius is kept constant
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w α g
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3 years ago
Calcite (CaCO_3) is a crystal with abnormally large birefringence. The index of refraction for light with electric field paralle
katrin2010 [14]

Answer:

(a) 42.28°

(b) 37.08°

Explanation:

From the principle of refraction of light, when light wave travels from one medium to another medium, we have:

\frac{n_{b} }{n_{a} } = sinθ_{a}/sinθ_{b}

In the given problem, we are given the refractive indices of light which are parallel and perpendicular to the axis of the optical lens as 1.4864 and 1.6584 respectively.

For critical angle θ_{a} = θ_{c}, θ_{b} = 90°; n_{b} = 1

(a) n_{a} = 1.4864

\frac{1 }{1.4864 } = sinθ_{c}/sin90°

0.6728 = sinθ_{c}θ[tex]_{c} = sin^(-1) 0.6728 = 42.28°(b) [tex]n_{a} = 1.6584

\frac{1 }{1.6584} = sinθ_{c}/sin90°

0.60299 = sinθ[tex]_{c}

θ[tex]_{c} = sin^(-1) 0.60299 = 37.08°

7 0
4 years ago
two forces act at an angle of 135 degree.the forces are F1 = 50N and F2 = 25N respectively.graphically construct a parallelogram
olya-2409 [2.1K]

Answer:gay

Explanation:

gay

8 0
3 years ago
The quantity of charge q (in coulombs) that has passed through a surface of area 2.00 cm2 varies with time
Makovka662 [10]

Explanation:

We have,

Surface area, A=2\ cm^2=0.0002\ m^2

The current varies wrt time t as :

q(t) = 4t^3 + 5t + 6

(a) At t = 2 seconds, electrical charge is given by :

q(t) = 4t^3 + 5t + 6\\\\q(2) = 4(2)^3 + 5(2) + 6\\\\q=48\ C

(b) Current is given by :

I=\dfrac{dq}{dt}\\\\I=\dfrac{d(4t^3 + 5t + 6)}{dt}\\\\I=12t^2+5

Instantaneous current at t = 1 s is,

I=12(1)^2+5=17\ A

(c) Current is, I=12t^2+5

Current density is given by electric current per unit area.

J=\dfrac{I}{A}\\\\J=\dfrac{(12t^2+5)}{0.0002}\\\\J=5000(12t^2+5)\ A/m^2

Therefore, it is the required explanation.

7 0
3 years ago
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