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ANTONII [103]
3 years ago
8

A 360-nm thick oil film floats on the surface of a pool of water. the indices of refraction of the oil and the water are 1.5 and

1.33, respectively. when the surface of the oil is illuminated from above at normal incidence with white light, what are the two wavelengths of light between 400 nm to 800 nm wavelength that are most strongly reflected?a) 410 nm and 700 nmb) 430 nm and 720 nmc) 450 nm and 740 nmd) 470 nm and 760 nme) 490 nm and 780 nm
Physics
1 answer:
jenyasd209 [6]3 years ago
7 0
The correct answer is 432, and 720.
The thickness of a film is t= 360nm
the refractive index of oil n₀t = (m +1/2) λ
For m =0
λ = 4n₀t
= 4(1.50)(360)
= 2160nm
for m = 1
λ = 4n₀t
= 4(1.50)(360)/3
= 720nm
m = 2
λ = 4n₀t/5 = 4(1.50)(360)/5
= 432nm
The wavelength which are most strongly reflected are
432nm, 720nm.
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Answer:

The maximum kinetic energy is 100 j.    

Explanation:

<h3>The kinetic energy = (potential energy) + (kinetic energy) and the potential energy of 0 J implying its kinetic energy is 100 J, which is its maximum. </h3>

4 0
3 years ago
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5 0
4 years ago
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 m /
Sidana [21]

Answer:

The maximum height is 2881.2 m.

Explanation:

Given that,

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Time = 7.00 s

We need to calculate the distance

Using equation of motion

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Put the value into the formula

s=0+\dfrac{1}{2}\times29.4\times7^2

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We need to calculate the velocity

Using formula of velocity

v=a\times t

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v=205.8\ m/s

We need to calculate the height

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Put the value into the formula

H=\dfrac{(205.8)^2}{2\times9.8}

H=2160.9\ m

We need to calculate the maximum height

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H'=H+s

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4 0
3 years ago
The four-wheel-drive all-terrain vehicle has a mass of 385 kg with center of mass G2. The driver has a mass of 75 kg with center
jenyasd209 [6]

The coefficient of friction is missing and it has a value of μ = 0.4

Answer:

a = 3.924 m/s²

Explanation:

I've attached the kinematic free body diagram.

Taking the sum of all upward and downward forces,

EFy = 0;

N1 + N2 - m_p•g - m_v•g = 0

N1 + N2 = m_p•g + m_v•g

Where;

N1 and N2 are the normal reactions at the wheels

m_p is the mass of the driver

m_v is the mass of the vehicle

g is the acceleration due to gravity.

Plugging in the relevant values in the question,we obtain;

N1 + N2 = (385 + 75) x 9.81

N1 + N2 = 4512.6N - - - (eq1)

Now, taking sum of all horizontal forces;

EFx = (m_p + m_v) x a

So,

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0.4(N1 + N2) = (385 + 75)a

0.4(N1 + N2) = 460a

N1 + N2 = 1150a

From eq(1),N1 + N2 = 4512.6N

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1150a = 4512.6N

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a = 3.924 m/s²

Therefore, the acceleration, a = 3.924 m/s²

7 0
3 years ago
If pressure is increased from 200 kPa to 300 kPa, and the original volume of gas was 1.5 L, what is the new volume? Assume the t
Oxana [17]

Answer:

The answer to your question is:      V2 = 1 l

Explanation:

Data

P1 = 200 kPa

P2 = 300 kPa

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Formula

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                          V2 = (200 x 1.5) / 300

                          V2 = 1 l

6 0
4 years ago
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