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3 years ago

I NEED HELP WITH THIS PLEASE

Physics

1 answer:

Ann [662]3 years ago

6 0

Answer:

Explanation:

One of the factors of increasing the rate of a reaction is increasing concentration. Therefore adding more people increases the number of people on the dance floor , therefore increasing the concentration increases the rate of reaction.

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(n = 1.33). When visible light (400nm – 700 nm) shines normally on the film an observed from above, the bright reflected light l

Yuki888 [10]

Complete Question

A thin film in air is made by putting a thin layer of acetone (n = 1.25) on a layer of water

(n = 1.33). When visible light (400 nm – 700 nm) shines normally on the film an observed from above, the bright reflected light looks kind of “purplish” with red light of wavelength 650 nm mixed some blue giving it that appearance and “greenish” light of wavelength 520 nm is completely destroyed. Determine the thickness of the acetone film. (you only need to worry about the red and green wavelengths, not the blue)

Answer:

The thickness of acetone is $d = 104 nm$

Explanation:

A diagram showing this process is shown on the first uploaded image

From the question we are told that

The refractive index of acetone is $n_a =1.25$

The refractive index of water is $n_w = 1.33$

The wavelength of the reflected light is $\lambda_r = 650nm = 650 *10^{-9}m$

The wavelength of the destroyed light is $\lambda_g = 520nm = 520 *10^{-9}m$

Looking at the given data we can see that the

$n_a < n_w$

This means that the light which the acetone-water layer would reflect will have a phase shift of $\pi$

Again this make us to understand that the light reflected at the acetone layer will also have a phase shift of $\pi$

Since they would be having the same phase shift the two light would interfere

For interference the condition for minima is mathematically represented as

$2 n_a d = (m + \frac{1}{2} ) \lambda_g$

Where d is the thickness of acetone

$d = \frac{\lambda_g}{4 n_a}$

Substituting values

$d = \frac{520 *10^{-9}}{4 * 1.25}$

$d = 104 *10^{-9}m$

$d = 104 nm$

5 0

3 years ago

The tallest volcano on Mars

Nonamiya [84]

Answer: Is three times the height of Mt. Everest

Explanation:

Using the process of elimination we can rule out “D” Because its a volcano, C because it says we have visited mars 44 times not the volcano, B because in the passage is literally says “3 times the size of Mt. Everest”.

8 0

3 years ago

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A train travels 55km, south along a straight track in 34minutes. What is the train Ls average velocity in kilometers per hour

Westkost [7]

Answer:

96.5 km/h

Explanation:

The average velocity of the train is given by:

$v=\frac{d}{t}$

where

d is the displacement

t is the time taken

For this train, we have:

d = 55 km south (displacement is a vector, so we must also consider the direction)

$t= 34 min \cdot \frac{1}{60 min/h}=0.57 h$

Substituting into the equation, we find the average velocity:

$v=\frac{55 km}{0.57 h}=96.5 km/h$

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3 years ago

What is the formula to calculate acceleration

yan [13]

Explanation:

 your formula is here 

hope it's helps u

6 0

3 years ago

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The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl

Ulleksa [173]

Hi there!

We can begin by using the equation for energy density.

$U = \frac{1}{2}\epsilon_0 E^2$

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
$\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}$

Creating a Gaussian surface being the lateral surface area of a cylinder:
$A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}$

Now, we can calculate the energy density using the equation:
$U = \frac{1}{2} \epsilon_0 E^2$

Plug in the expression for the electric field and solve.

$U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}$

Now, we can integrate over the volume with respect to the radius.

Recall:
$V = \pi r^2L \\\\dV = 2\pi rLdr$

Now, we can take the integral of the above expression. Let:
$r_i$ = inner cylinder radius

$r_o$ = outer cylindrical shell inner radius

Total energy-field energy:

$U = \int\limits^{r_o}_{r_i} {U_D} \, dV = \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr$

Plug in the equation for the electric field energy density and solve.

$U = \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\$

Bring constants in front and integrate. Recall the following integration rule:
$\int {\frac{1}{x}} \, dx = ln(x) + C$

Now, we can solve!

$U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})$

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

$\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}$

And here's our equation!

3 0

2 years ago