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chubhunter [2.5K]
3 years ago
11

A puddle of water has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavel

ength of light incident on the film is 560 nm and the indices of refraction of gasoline and water are 1.40 and 1.33, respectively, what must be the minimum thickness of the film to see a bright reflection? 100 nm 200 nm 300 nm 400 nm 500 nm
Physics
1 answer:
g100num [7]3 years ago
7 0

Answer:

option A

Explanation:

Given,

wavelength of light,\lambda = 560\ nm

refractive index of gasoline, n₁ = 1.40

Refractive index of water, n₂ = 1.33

thickness of the film, t = ?

Condition of constructive interference is given by

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

For minimum thickness of the film m = 0

From the question we can clearly observe that phase change from gasoline to air

so, n = 1.4

2 \times 1.4 \times t = \dfrac{560}{2}

t = 100\ nm

Hence, the correct answer is option A

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There is a refrigerator running in a room, the heat flowing into the refrigerator from the outside is 40 J/s, and the refrigerat
Mamont248 [21]

Answer:

(A) 140 j/sec (b) 1.26 K

Explanation:

We have given the heat heat flowing into the refrigerator = 40 J/sec

Work done = 40 W

(a) So the heat discharged from the refrigerator =heat\ flowing\ in\ refrigerator+work\ done=40+100=140j/sec

(b) Total heat absorbed =140 j/sec =140\times 3600=504000j/hour

Let the temperature be \Delta T

Heat absorbed per hour =504000 [tex]=400\times 10^3\times \Delta T

So  \Delta T=\frac{504000}{400000}=1.26K

8 0
3 years ago
A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

8 0
2 years ago
A silver bar 0.125 meter long is subjected to a temperature change from 200°C to 100°C. What will be the length of the bar after
Delicious77 [7]
\Delta L=  \alpha L_0 (T_f-T_i)

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248

D
5 0
3 years ago
Read 2 more answers
Please help me out i'm so depressed and such a failure
aksik [14]

Answer:

I reckon towards b. Let me know if im right

4 0
3 years ago
The owner of a company that manufactures drinking cups decides it would be impressive to build an inground swimming pool that is
garri49 [273]

Answer:

The depth is 5.15 m.

Explanation:

Lets take the depth of the pool = h m

The atmospheric pressure ,P = 101235 N/m²

The area of the top = A m²

The area of the bottom =  a m²

Given that A= 1.5 a

The force on the top of the pool = P A

The total pressure on the bottom = P + ρ g h

ρ =Density of the  water = 1000 kg/m³

The total pressure at the bottom of the pool =  (P + ρ g h) a

The bottom and the top force is same

(P + ρ g h) a = P A

P a +ρ g h a =  P A

ρ g h a =  P A - P a

h=\dfrac{P ( A-a)}{\rho g a}

h=\dfrac{P ( 1.5 a-a)}{\rho g a}

h=\dfrac{P ( 1.5- 1)}{\rho g}

h=\dfrac{101235 ( 1.5- 1)}{1000\times 9.81}\ m

h=5.15 m

The depth is 5.15 m.

7 0
3 years ago
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