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2 years ago

13

A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum o

f reflection with normally incident green light (wavelength 500 nm in air). Assuming the maximum occurs in the first order, determine the thickness of the oil slick.

1 answer:

Ad libitum [116K]2 years ago

4 0

Answer:

The thickness of the oil slick is $1.95\times10^{-7}\ m$

Explanation:

Given that,

Index of refraction = 1.28

Wave length = 500 nm

Order m = 1

We need to calculate the thickness of oil slick

Using formula of thickness

$2nt= m\lambda$

Where, n = Index of refraction

t = thickness

$\lambda$ = wavelength

Put the value into the formula

$2\times1.28 \times t=1\times\times500\times10^{-9}$

$t = \dfrac{1\times\times500\times10^{-9}}{2\times1.28 }$

$t=1.95\times10^{-7}\ m$

Hence, The thickness of the oil slick is $1.95\times10^{-7}\ m$

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Explanation:

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3 years ago

A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's

blondinia [14]

Answer:

a) $z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

b) $v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

Explanation:

The particle position is given by:

$z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0$

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

$z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

Part b

For this case we can find the average velocity with the following formula:

$v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

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3 years ago

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barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

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(a) Change in truck's kinetic energy, $\Delta E=\dfrac{1}{2}m(v^2-u^2)$

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$\Delta p=2000\ kg\times (16.11-9.44)$

$\Delta p=13340\ kg-m/s$

Hence, this is the required solution.

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3 years ago