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kifflom [539]
3 years ago
8

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r

efraction of 1.38, while the eyedrops have a refractive index of 1.45. After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 nm has been reinforced in the reflected light.Part A What is the minimum thickness of the film of eyedrops on your cornea?t= nm.
Physics
1 answer:
mina [271]3 years ago
6 0

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

2 n t = (m + \dfrac{1}{2})\lambda

At m = 0 for the minimum thickness, so

2\times 1.45 \times t = (0 + 0.5)\times 600

2.9 \times t =300

t =\dfrac{300}{2.9}

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

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Answer:

1. As rocket mass increases, acceleration decreases.

2. The inverse of the mass of the boat.

Explanation:

1. Newton's second law of motion states;

         F = ma

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Therefore, increasing the mass of a rocket increases its weight which would reduce its acceleration provided that the force is constant. Thus, as rocket mass increases, acceleration decreases.

2. The slope of the graph can be expressed as;

From Newton's second law,

F = ma

Slope = (Δa) ÷ (ΔF)

Slope = \frac{a}{F}

⇒ \frac{1}{m} = \frac{a}{F}

Therefore, the slope of the graph is the reciprocal of the mass of the boat.

7 0
3 years ago
A student pushes a 0.2 kg box against a spring causing the spring to compress 0.15 m. When the spring is released, it will launc
german

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x  = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

Mass of box = 0.2 kg

Work, W, done by the spring = \frac{1}{2} kx^2 and the kinetic energy gained by the box is given by KE = \frac{1}{2} mv^2

Since work done by the spring = kinetic energy gained by the box we have

\frac{1}{2} mv^2 =  \frac{1}{2} kx^2  therefore we have v = \sqrt{\frac{kx^2}{m} } = x\sqrt{\frac{k}{m} } = 0.15\sqrt{\frac{300}{0.2} } = 5.81 m/s

Therefore the maximum height is given by

v² = 2·g·h or h = \frac{v^2}{2g} = \frac{5.81^{2} }{2*9.81} = 1.72 m

6 0
4 years ago
Calculate the total resistance in a series circuit made up of resistances of 3Ω, 4Ω, and 5Ω.
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The total resistance in a series circuit is equal to the sum of all resistors (R total = ΣRi).

R total = R1 + R2 + R3 = (3 + 4 + 5) Ω = 12 Ω
5 0
3 years ago
A certain spring has a spring constant k1 = 660 N/m as the spring is stretched from x = 0 to x1 = 35 cm. The spring constant the
pantera1 [17]

(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².

(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.

(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.

<h3>Work done in the spring</h3>

The work done in stretching the spring is calculated as follows;

W = ¹/₂kx²

W(1 to 2) = ¹/₂K₂Δx²

W(1 to 2)  =  ¹/₂(250)(0.65 - 0.35)²

W(1 to 2)  = 11.25 J

W(0  to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃

W(0  to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)

W(0  to 3) = 64.28 J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

6 0
2 years ago
Ph11_UnitPacket2019
frozen [14]

Let's see

Use snells law

\\ \rm\Rrightarrow \dfrac{n_1}{n_2}=\dfrac{sini}{sinr}

\\ \rm\Rrightarrow \mu=\dfrac{sin30}{sin19.9}

\\ \rm\Rrightarrow \mu=0.5/0.34

\\ \rm\Rrightarrow \mu=1.47

It may be glass

3 0
3 years ago
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