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Sindrei [870]
3 years ago
5

A very thin oil film (n = 1.25) floats on water (n = 1.33). What is the thinnest film that produces a strong reflection for gree

n light with a wavelength of 500 nm?
Physics
1 answer:
mixer [17]3 years ago
6 0

Answer:

200 nm is the thinnest film that produces a strong reflection for green light with a wavelength of 500 nm

Explanation:

If two reflected waves interfere constructively ,strong reflection is produced. Two reflected waves will experience a phase change

For constructive interference

2\times n\times t=m\lambda

for thinnest film m=1

refractive index should be taken for film n=1.25

thickness of the thinnest film is

t=\frac{m\lambda}{2n} \\t=\frac{1\times 500}{2\times 1.25} \\t=200 nm

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The diagram shows forces acting on a boat.
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Two automobiles traveling at right angles to each other collide and stick together. Car A has a mass of 1200 kg and had a speed
sergij07 [2.7K]

Answer:

v_{B0}=15.73 m/s

Explanation:

We can use the conservation of momentum. The initial momentum is equal to the final momentum:

x-coordinate

p_{0x}=p_{fx}

m_{A}v_{A0}=(m_{A}+m_{B})v_{cx}  

m_{A}v_{A0}=(m_{A}+m_{B})v_{c}cos(40) (1)

y-coordinate

p_{0y}=p_{fy}

m_{B}v_{B0}=(m_{A}+m_{B})v_{cy}  

m_{B}v_{B0}=(m_{A}+m_{B})v_{c}sin(40) (2)

We can divide equations (2) and (1):

\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=\frac{sin(40)}{cos(40)}

\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=tan(40)

v_{B0}=\frac{m_{A}v_{A0}}{m_{B}}*tan(40)

v_{B0}=\frac{1200*25}{1600}*tan(40)

v_{B0}=15.73 m/s

I hope it helps you!

           

4 0
4 years ago
Read 2 more answers
Please answer correctly to be marked as brainly.
lidiya [134]

A- PRICE

B-QUANTITY

C-SUPPLY

D-DEMAND

E-EQUILIBRIUM POINT

Explanation:

It is the Supply Demand curve in Economics. It gives relationship between price and quantity

7 0
3 years ago
A typical nuclear fission power plant produces about 1.00 GW of electrical power. Assume the plant has an overall efficiency of
mamaluj [8]

Answer:

mass consumed by 235U each day = 2 kg

Explanation:

electrical power produced = 1 GW = 1 × 10⁹ × (6.24151 × 10¹⁸ ) eV

                                            = 6.24151× 10²¹ MeV/s

thermal energy =  0.420 * 250 = 105 MeV

\dfrac{1 GW}{150 MeV}= \dfrac{6.24151\times 10^{21}}{105}

                                      = 5.94 × 10¹⁹ fission/second

                                       =5.94 × 10¹⁹× 24 × 60 ×60)

                                      =  5.13 × 10²⁴ fission/day

mu = 235.04393 ×  1.660× 10 ⁻²⁷ = 390.1729× 10⁻²⁷ Kg

M = mu ×5.13 × 10²⁴

   = 390.1729× 10⁻²⁷ ×5.13 × 10²⁴

M   =  2 kg(approx.)

mass consumed by 235U each day = 2 kg

3 0
3 years ago
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