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Zinaida [17]
2 years ago
13

If a soap bubble is 115 nm thick, what wavelength is most strongly reflected at the center of the outer surface when illuminated

normally by white light? Assume that n = 1.33. Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
sp2606 [1]2 years ago
7 0

Answer:

611.8 nm

Explanation:

n_{oil} = Index of refraction of soap bubble  = 1.33

t = thickness of the soap bubble = 115 nm = 115 x 10⁻⁹ m

\lambda = wavelength of light = ?

m = order = 0

For reflection , the necessary condition is

2 n_{oil} t = (m + 0.5) \lambda

2 (1.33)(115\times 10^{-9})= (0 + 0.5) \lambda

\lambda = 6.118\times 10^{-7}

\lambda = 611.8 nm

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two masses are kept 2 metre apart there is gravitational force of 2 Newton what is the gravitational force when they are kept at
sukhopar [10]

Answer: 0.5N

Explanation:

Gravitational force is calculated using the formula :

F = Gm1m2/r^2

Where G is the gravitational constant (6.67 × 10^-11)

At a distance 'r' of 2metres apart:

Mass of objects are m1 and m2

Gravitational force 'F1' = 2N

Inputting values into the formula :

2 = Gm1m2 / 2^2 - - - - - (1)

At a distance 'r' of 4meters apart:

Mass of objects are m1 and m2

Gravitational force 'F2' = y

Inputting values

F2 = Gm1m2 / 4^2 - - - - - (2)

Dividing equations 1 and 2

2 = Gm1m2 / 2^2 ÷ F2 = Gm1m2 / 4^2

2 / F2 = (Gm1m2 / 4) / (Gm1m2 / 16)

2 / F2 = (Gm1m2 / 4) × (16 / Gm1m2)

2/F2 = 16 / 4

Cross multiply

2 × 4 = 16 × F2

8 = 16F2

F2 = 8/16

F2 = 0.5N

7 0
3 years ago
Consider a single-slit diffraction pattern for λ=589nm , projected on a screen that is 1.00 m from a slit of width 0.25 mm. How
const2013 [10]

Answer: i) 2.356 × 10^-3 m = 2.356mm, ii) 4.712 × 10^-3 m = 4.712mm

Explanation: The formulae that relates the position of a fringe from the center to the wavelength, distance between slits and distance between slits and screen is given below as

y = R×(mλ/d)

Where y = distance between nth fringes and the center fringe.

m = order of fringe

λ = wavelength of light = 589nm = 589×10^-9m

R = distance between slits and screen = 1.0m

d = distance between slits = 0.25mm = 0.00025m

For distance between the first dark fringe and the center fringe.

This implies that m = 1

y = 1 × 589×10^-9 × 1/0.00025

y = 589×10^-9/0.00025

y = 2,356,000 × 10^-9

y = 2.356 × 10^-3 m = 2.356mm

For the second dark fringe, this implies that m = 2

y = 1 × 2 × 589×10^-9/0.00025

y = 1178 × 10^-9 /0.00025

y = 4,712,000 × 10^-9

y = 4.712 × 10^-3 m = 4.712mm

8 0
3 years ago
A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.
fgiga [73]

The vertical velocity of the projectile upon returning to its original is 17. 74 m/s

<h3>How to determine the vertical velocity</h3>

Using the formula:

Vertical velocity component , Vy = V * sin(α)

Where

V = initial velocity = 36. 6 m/s

α = angle of projectile = 29°

Substitute into the formula

Vy = 36. 6 * sin ( 29°)

Vy = 36. 6 * 0. 4848

Vy = 17. 74 m/s

Thus, the vertical velocity of the projectile upon returning to its original is 17. 74 m/s

Learn more about vertical velocity here:

brainly.com/question/24949996

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8 0
1 year ago
A balloon is expanded to the same volume as that of a human head. Do an order-of-magnitude estimate of the volume of this balloo
cestrela7 [59]

Answer:

Volume of balloon =  1000 cm^3

Explanation:

 The head of a normal person can be assumed as a sphere with radius 10 cm.

 Volume of sphere =\frac{4}{3} \pi r^3, where r is the radius.

 We have approximate radius = 10 cm.

  Approximate volume of head =\frac{4}{3} \pi r^3=\frac{4}{3} *\pi* 10^3=4188cm^3

 In the given options the closest value to the approximate volume is 1000 cm^3.

 So, volume of head = Volume of balloon =  1000 cm^3

4 0
2 years ago
Nasa’s goal is to help industry reduce emissions from aircraft by how much by 2050 compared to 2005?
lys-0071 [83]

Answer:

50%

Explanation:

Commercial aviation is responsible for 2% of global carbon emission. In the year 2009, the members of IATA (International Air Transport Association) had drawn a pledge to:

1. Halve the carbon emission due to aircraft by 2050, relative to emission of 2005

2. To make growth of industry carbon neutral by 2020

3. To cut CO₂ emission by 1.5% per year till 2020

To achieve these targets, a four pillar action plan was created.

8 0
3 years ago
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