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Katen [24]
2 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]2 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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Energy of our sun. It composed of a lot of hydrogen and by the two fusion of isotopes. Helium is then released.

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What affect, if any, does increasing the speed of the plunger have on the wavelength of the waves
mojhsa [17]

As the speed of the plunger increases, the wavelength of the waves decreases. The greater the frequency, the smaller the wavelength. The smaller the frequency, the greater the wavelength. When we increase the speed of the plunger, the frequency of the waves also increases, and just like with the size of the ball, it’s the speed of the plunger and the frequency of the waves are directly related.

6 0
2 years ago
An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the accelerati
lubasha [3.4K]

Answer:

The correct option is

a. v = g (1-e^{-bt})/b

Explanation:

Time at which the object start fall t = 0

The acceleration a is given by a = g - bV

Where V = Speed of the object

Speed V² = u² + 2·a·h

However with the drag force the object will approach terminal velocity as t becomes progressively larger whereby v will stop increasing

Option a. is the only option that has  limiting value of v which is in the range of g as t increases ∴ option a. is the correct option.

v = g (1-e^{-bt})/b  as t increases (1-e^{-bt}) → 1 s and v→ g/b m/s

6 0
3 years ago
A fence 8 ft high​ (w) runs parallel to a tall building and is 24 ft​ (d) from it. Find the length​ (L) of the shortest ladder t
AveGali [126]

Answer:

25.3 ft

Explanation:

The illustration of the problem is shown the attached image.

The length of the ladder can be calculated using the Pythagoras theorem:

Hypotenuse^2 = opposite^2 + adjacent^2

The hypotenuse is the length of the ladder.

Hypotenuse = \sqrt{opposite^2 + adjacent^2}

L^2 = BC^2 + (24 + AE)^2........1

Triangle ABC is similar to triangle AEF, hence:

\frac{BC}{8} = \frac{AE + 24}{AE}

BC = \frac{8(AE + 24)}{AE}.................2

Substitute 2 into 1

L^2 = (\frac{8(AE + 24)}{AE})^2 + (24 + AE)^2

Let AE = x

L^2 = (\frac{8x + 192}{x})^2 + (24 + x)^2

    = (8 + \frac{192}{x})^2 + (24 + x)^2

Minimize L with respect to x.

2\frac{dL}{dx} = 2(8 + \frac{192}{x})(-\frac{192}{x^2}) + 2(24 + x)

       =

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3 years ago
Lực hút trái đất là dì
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