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MariettaO [177]
3 years ago
5

A tank of gasoline (n = 1.40) is open to the air (n = 1.00). A thin film of liquid floats on the gasoline and has a refractive i

ndex that is between 1.00 and 1.40. Light that has a wavelength of 626 nm (in vacuum) shines perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive interference. The thickness of the film is 290 nm and is the minimum nonzero thickness for which constructive interference can occur. What is the refractive index of the film?
nfilm = 1Your answer is incorrect.I did t=(m)(wavelengthfilm)/(2) solving for the wavelegth of filmThenI did: wavelength film = wavelength vaccum / n and my n comes out as 1.18 which is the wrong answer can anyone help??
Physics
1 answer:
klio [65]3 years ago
5 0

Answer:

1.08

Explanation:

This is the case of interference in thin films in which interference bands are formed due to constructive interference of two reflected light waves , one from upper layer and the other from lower layer . If t be the thickness and μ be the refractive index then

path difference created will be 2μ t.

For light coming from rarer to denser medium , a phase change of π occurs additionally after reflection from denser medium, here, two times, once from upper layer and then from the lower layer ,  so for constructive interference

path diff = nλ , for minimum t , n =1

path diff = λ

2μ t. =  λ

μ = λ / 2t

= 626 / 2 x 290

= 1.08

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A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
Lyrx [107]

Answer:

a) the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta (\theta) be the angle of twist

polar moment of inertia I_p} = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

7 0
3 years ago
Can a material have negative permittivity?
sleet_krkn [62]
No it can't it's material
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3 years ago
An electric bulb is marked 40volts ,230w another bulb is marked 40w,110v
Andrej [43]

Answer:

a. The ratio of their resistance is 2783:64

b. The ratio of their energy is 4:23

c. The charge on the first bulb is 5.75 C

The charge on the second bulb is 0.\overline {36} C

Explanation:

The voltage on one of the electric bulbs, V₁ = 40  volts

The power rating of the bulb, P₁ = 230 w

The voltage on the other electric bulbs, V₂ = 110 volts

The power rating of the bulb, P₂ = 40 w

a. The power is given by the formula, P = I·V = V²/R

Therefore, R = V²/P

For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96

The resistance of the second bulb, R₂ = 110²/40

The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64

∴ The ratio of their resistance, R₂:R₁ = 2783:64

b. The energy of a bulb, E = t × P

Where;

t = The time in which the bulb is powered on

∴ The energy of the first bulb, E₁ = 230 w × t

The energy of the second bulb, E₂ = 40 w × t

The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23

∴ The ratio of their energy, E₂:E₁ = 4:23

c. The charge on a bulb, 'Q', is given by the formula, Q = I × t

Where;

I = The current flowing through the bulb

From P = I·V, we get;

I = P/V

For the first bulb, the current, I = 230 w/40 V = 5.75 amperes

The charge on the first bulb per second (t = 1) is therefore;

Q₁ = 5.75 A × 1 s = 5.75 C

The charge on the first bulb, Q₁ = 5.75 C

Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = 0.\overline {36} C

The charge on the second bulb, Q₂ = 0.\overline {36} C.

d. The question has left out parts

4 0
3 years ago
Strong winds causing a vehicle to lose all control is called
mixer [17]
The answer to this question is "Buffeting". This is an unusual strong wind condition that resulted in the loss of vehicle control. This condition occurs in roads and bridges, across and along mountains that affected vehicles control. The drivers must be alert and put their full attention to overcome this condition.
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3 years ago
A candy-filled piñata is hung from a tree for Elia's birthday. During an unsuccessful attempt to break the 4.4-kg piñata, Tonja
Klio2033 [76]

Answer: v = 0.6 m/s

Explanation: <u>Momentum</u> <u>Conservation</u> <u>Principle</u> states that for a collision between two objects in an isolated system, the total momentum of the objects before the collision is equal to the total momentum of the objects after the collision.

Momentum is calculated as Q = m.v

For the piñata problem:

Q_{i}=Q_{f}

m_{p}v_{p}_{i}+m_{s}v_{s}_{i}=m_{p}v_{p}_{f}+m_{s}v_{s}_{f}

Before the collision, the piñata is not moving, so v_{p}_{i}=0.

After the collision, the stick stops, so v_{s}_{f}=0.

Rearraging, we have:

m_{s}v_{s}_{i}=m_{p}v_{p}_{f}

v_{p}_{f}=\frac{m_{s}v_{s}_{i}}{m_{p}}

Substituting:

v_{p}_{f}=\frac{(0.54)(4.8)}{(4.4)}

v_{p}_{f}= 0.6

Immediately after being cracked by the stick, the piñata has a swing speed of 0.6 m/s.

3 0
3 years ago
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