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2 years ago

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A thin film of polystyrene is used as an antireflective coating for fabulite (known as the substrate). the index of refraction o

f the polystyrene is 1.49, and the index of refraction of the fabulite is 2.409. what is the minimum thickness of film required? assume that the wavelength of the light in air is 500 nanometers.

1 answer:

kvasek [131]2 years ago

6 0

To solve this problem, we assume that the wavelength of the light in air is 500 nanometers.

For this case we only need the refractive index of the polystyrene. For an antireflective coating, we need a quarter of wave thickness at the wavelength in the air. Which means that the antireflective coating needs to be as thick as 1/4 of the wavelength, divided by the coating’s refractive index. This is expressed mathematically in the form:

x = λ / (4 * n)

where,

x = thickness

λ = wavelength of light

n = index of refraction of polystyrene

Substituting:

x = 500 nm / (4 * 1.49)
x = 500 nm / 5.96
x = 83.90 nm

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The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve

Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

$\varepsilon _t=\dfrac{D-d}{d}$

$\varepsilon _t=\dfrac{32-30}{30}$

$\varepsilon _t=0.0667$

We know that

$n=-\dfrac{\epsilon _t}{\varepsilon _{long}}$

$\varepsilon _{long}=-\dfrac{\epsilon _t}{n}$

$\varepsilon _{long}=-\dfrac{0.0667}{0.45}$

$\varepsilon _{long}=-0.148$

So the axial pressure

$P=E\times \varepsilon _{long}$

$P=5\times 0.148$

P=740 KPa

The movement in the sleeve

$\Delta =\varepsilon _{long}\times L$

$\Delta =0.148\times 50$

Δ=7.4 mm

6 0

2 years ago

In the sport of curling, large smooth stones are slid across an ice court to land on a target. Sometimes the stones need to move

lara31 [8.8K]

Answer:

To increase kinetic friction, the amount of fine water droplets sprayed before the game is limited.

To reduce kinetic friction. increase the amount of fine water droplets during pregame preparation and sweeping in front of the curling stones.

Explanation:

In curling sports, since the ice sheets are flat, the friction on the stone would be too high and the large smooth stone would not travel half as far. Thus controlling the amount of fine water droplets sprayed before the game is limited pregame is necessary to increase friction.

On the other hand, reducing ice kinetic friction involves two ways. The first way is adding bumps to the ice which is known as pebbling. Fine water droplets are sprayed onto the flat ice surface. These droplets freeze into small "pebbles", which the curling stones "ride" on as they slide down the ice. This increases contact pressure which lowers the friction of the stone with the ice. As a result, the stones travel farther, and curl less.

The second way to reduce the kinetic friction is sweeping in front of the large smooth stone. The sweeping action quickly heats and melts the pebbles on the ice leaving a film of water. This film reduces the friction between the stone and ice.

8 0

2 years ago

A professor's office door is 0.99 m wide, 2.2 m high, 4.2 cm thick; has a mass of 27 kg, and pivots on frictionless hinges. A "d

ANEK [815]

Answer:

$I=8.8209\ kg.m^2$

$\alpha=0.6348\ rad.s^{-2}$

Explanation:

Given:

width of door, $w=0.99\ m$

height of the door, $h=2.2\ m$

thickness of the door, $t=4.2\ cm$

mass of the door, $m=27\ kg$

torque on the door, $\tau=5.6\ N.m$

<em>∵Since the thickness of the door is very less as compared to its other dimensions, therefore we treat it as a rectangular sheet.</em>

For a rectangular sheet we have the mass moment of inertia inertia as:

$I=\frac{1}{3} m.w^2$

$I=\frac{1}{3}\times 27\times 0.99^2$

$I=8.8209\ kg.m^2$

We have a relation between mass moment of inertia, torque and angular acceleration as:

$\alpha=\frac{\tau}{I}$

$\alpha=\frac{5.6}{8.8209}$

$\alpha=0.6348\ rad.s^{-2}$

6 0

2 years ago

Pamela drove her car 999999 kilometers and used 999 liters of fuel. she wants to know how many kilometers (k)(k)left parenthesis

Vanyuwa [196]

When the relationship between two variables are said to be proportional, it means that one variable is a constant multiple of the other variable. They are related by a constant of proportionality, usually denoted as k.

In this problem, the dependent variable is the distance in kilometers. Your mileage is limited with the amount of fuel you have. Thus, the independent variable is the liters of fuel. When these two are proportional, it could be expressed as

distance = k * liters of fuel, such that
distance/liters of fuel = k

By variation,

distance,1/liters of fuel,1 = distance,2/liters of fuel,2, where 1 denotes situation 1 and 2 denotes situation 2. Therefore,

999999 km /<span>999 liters = x km /</span><span>121212 liters, where x is the unknown distance. We can now therefore find the value of x.

x = (999999*121212)/999
x = 121333212 kilometers</span>

3 0

2 years ago

Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the

astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

$B = \frac{\mu_oI}{2\pi R}$

Where $\mu_o$ is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

$B_1 = \frac{\mu_oI_1}{2\pi R}$

$B_2 = \frac{\mu_oI_2}{2\pi R}$

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T = $\frac{\mu_oI_1}{2\pi R}$ - $\frac{\mu_oI_2}{2\pi R}$

or

4.0×10⁻⁶T = $\frac{\mu_o}{2\pi R}\times (I_1-I_2 )$

also

$\frac{I_1}{I_2} = \frac{3}{1}$

⇒ $I_1 = 3\times I_2$

substituting the values in the above equation we get

4.0×10⁻⁶T = $\frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)$

⇒ $I_2 = 1A$

also

$I_1 = 3\times I_2$

⇒ $I_1 = 3\times 1A$

⇒ $I_1 = 3A$

Hence, the larger of the two currents is 3A

3 0

2 years ago