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Snowcat [4.5K]
3 years ago
11

The ceiling of your lecture hall is probably covered with acoustic tile, which has small holes separated by about 5.9 mm. Using

light with a wavelength of 527 nm, how far could you be from this tile and still resolve these holes? The diameter of your eye is about 5 mm.( )mCould you resolve these holes better with red light or with violet light? (You have only 1 try.)1. violet.2. red
Physics
1 answer:
timurjin [86]3 years ago
5 0

Answer:

45.88297 m

Violet

Explanation:

x = Gap between holes = 5.9 mm

\lambda = Wavelength = 527 nm

D = Diameter of eye = 5 mm

L= Distance of observer from holes

From Rayleigh criteria we have the relation

\frac{x}{L}=1.22\frac{\lambda}{D}\\\Rightarrow L=\frac{xD}{1.22\lambda}\\\Rightarrow L=\frac{5.9\times 10^{-3}\times 5\times 10^{-3}}{1.22\times 527\times 10^{-9}}\\\Rightarrow L=45.88297\ m

A person could be 45.88297 m from the tile and still resolve the holes

Resolving them better means increasing the distance between the observer and the holes. It can be seen here that the distance is inversely proportional to the wavelength. Violet has a lower wavelength than red so, violet light would resolve the holes better.

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Explanation:

Because the temperature and the radiation are not correlated, they're not represented as functions of each other, they're represented as independent variables thus using graph 5 you cannot figure out how one affect another

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2 years ago
Each year, an average person in the United States is exposed to a radiation level of _____.
Anna11 [10]

Answer:

The correct answer is 0,2 rems

Explanation:

People are exposed to natural sources of radiation all the time. According to recent estimates, the average person in the United States receives an effective dose of approximately 3 mSv per year of natural radiation, which is equivalent to 0.3 rems. This amount includes cosmic radiation from outer space and is average because it varies depending on the region people are in.

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Have a nice day!

3 0
3 years ago
3, A 4kg block is pushed 2m at an acceleration of 0.2m/s square up a vertical wall by a constant force F applied at an angle of
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<h3>Work done by the applied force</h3>

The work done by the applied force is calculated as follows;

W = Fd

F - Ff = ma

where;

  • F is applied force
  • Ff is frictional force

Fcos(37) - μmgsin(37) = ma

Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)

0.799F - 7.077 = 0.8

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W = 9.86 x 2 x cos(37)

W = 15.75 J

Thus, the work done by the applied force on the block against the frictional force is 15.75 J.

Learn more about work done here: brainly.com/question/25573309

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