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dedylja [7]
3 years ago
13

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​

Physics
1 answer:
serg [7]3 years ago
4 0

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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3 years ago
Missy watched a storm last night with her brother. During the storm, she witnessed a large lightening bolt. What energy transfor
devlian [24]

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3 years ago
Read 2 more answers
A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?
Romashka [77]

Answer:

The displacement of the car after 6s is 43.2 m

Explanation:

Given;

velocity of the car, v = 12 m/s

acceleration of the car, a = -1.6 m/s² (backward acceleration)

time of motion, t = 6 s

The displacement of the car after 6s is given by the following kinematic equation;

d = ut + ¹/₂at²

d = (12 x 6) + ¹/₂(-1.6)(6)²

d = 72 - 28.8

d = 43.2 m

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6 0
2 years ago
An alpha particle collides with an oxygen nucleus, initially at rest. The alpha particle is scattered at an angle of 25.0° above
Greeley [361]

Answer:

v_{i}= 19\times 10^5\ m/s

Explanation:

given,

scattering angle of alpha particle = 25.0°  above its initial direction of motion

oxygen nucleus recoils at = 50.0° below this initial direction.

final speed of the oxygen = 2.08×10⁵ m/s

mass of alpha particle = 4.0 u

mass o oxygen nucleus = 16 u

momentum conservation along x- axis

m_{a}v_{i} = m_a v_a cos\theta + m_o v_o cos\theta

4v_{i} = 4\times v_a cos25^0 + 16\times 2.08 \times 10^5 cos50^0

v_{i}= \dfrac{3.625\times v_a+ 21.39\times 10^5}{4}....(1)

Along y-direction

0 = m_av_a sin \theta - m_ov_o sin\theta

0 = 4\times v_a sin 25 - 16\times  2.08 \times 10^5 sin50^0

v_a = \dfrac{25.49 \times 10^5}{1.69}

v_a = 15.082\times 10^5\ m/s

putting value in equation (1)

v_{i}= \dfrac{3.625\times 15.082\times 10^5+ 21.39\times 10^5}{4}

v_{i}= 19\times 10^5\ m/s

5 0
3 years ago
Aluminum has a coefficient of linear expansion of 2.4x10^-5 C^-1 and a coefficient of volume expansion of 7.2x10^-5 C^-1. An alu
Alla [95]

Answer:

New length = 10.2 m

Explanation:

Let ΔL  = change in length

     ΔT = change in the temperature

     L(0) = intial length

     α = coeffient of linear expansion

Then

ΔL = αL(0)ΔT

    =(2.4×10^-5)(10)(100-20)

    =0.0192 m

New length = L(0) + ΔL

                   = 10 + 0.0192

                   = 10.2 m

6 0
3 years ago
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