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Katen [24]
3 years ago
14

A small but bright light is at the bottom of a pool 2.2 m  deep. How wide is the circle of light that exits the surface of the

water when that light shines in the middle of the night? Remember that nwater=1.33.
Physics
1 answer:
mixas84 [53]3 years ago
5 0

Answer: 1.65m

Explanation:

Refractive index in terms of the depth of liquid is the ratio of the real depth to the apparent depth of the liquid i.e Refractive index =Real depth/apparent depth

Refractive index of water given = 1.33

Real depth is the measure of how deep is the liquid while apparent depth is the depth at the surface of the liquid.

Real depth = 2.2m

Apparent depth =?

Applying the formula above

Apparent depth =Real depth/refractive index

= 2.2/1.33

= 1.65m

Therefore, the circle of light that exits the surface of the water when that light shines in the middle of the night is 1.65m wide

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and since here it is a sine curve so we know that

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Which statement is correct? Theories are accepted as true when a single experiment yields similar results to another one. When a
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The third statement is correct.

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The distance recorded for riding a motorcycle on its rear wheel without stopping is more than 320 km! Suppose the rider in this
antiseptic1488 [7]

Answer:

<h3>14.97m/s</h3>

Explanation:

Given

Initial velocity of the car u = 8m/s

Distance travelled by the rider S = 40m

Acceleration a = 2m/s²

Required

rider's velocity after the acceleration v

Using the equation of motion

v² = u²+2as

v² = 8²+2(2)(40)

v² = 64+160

v² = 224

v = √224

v = 14.97m/s

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5 0
3 years ago
How is velocity different from speed? Based on distance and/or direction
Mnenie [13.5K]

Answer:

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6 0
2 years ago
when a metal ball is heated through 30°c,it volume becomes 1.0018cm^3 if the linear expansivity of the material of the ball is 2
Vlad1618 [11]

Answer:

The original volume of the metal sphere is approximately 1 cm³

Explanation:

The given parameters are;

The temperature change of the metal ball, ΔT = 30°C = 30 K

The new volume of the metal ball, V₂  = 1.0018 cm³

The linear expansivity of the material ball, α = 2.0 × 10⁻⁵ K⁻¹

We have;

d₂ = d₁·(1 + α·ΔT)

Where;

d₁ = The original diameter of the metal ball

d₂ = The new diameter of the ball

From the volume of the ball, V₂, we have;

V₂ = 1.0018 cm³ = (4/3)×π×r₂³

Where;

r₂ = The new radius = d₂/2

∴ V₂ = 1.0018 cm³ = (4/3)×π×(d₂/2)³

∴ d₂ = ∛(2³ × 1.0018 cm³/((4/3) × π)) ≈ 1.241445 cm

d₁ = d₂/(1 + α·ΔT)

∴ d₁ ≈ 1.241445 cm/(1 + 2.0 × 10⁻⁵·K × 30 K) ≈ 1.24070058 cm

The original volume of the metal ball, V₁ = (4/3)×π×(d₁/2)³

∴ V₁ = (4/3)×π×(1.24070058/2)³ ≈ 0.99999902845 cm³ ≈ 1 cm³

The original volume of the metal sphere, V₁ ≈ 1 cm³.

8 0
3 years ago
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