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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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artcher [175]

Answer:

We conclude that the mass of a rock with a force of 500 N and an acceleration of 75 m/s² is 6.7 kg.

Hence, option D is correct.

Explanation:

Given

  • Force F = 500 N
  • Acceleration a = 75 m/s²

To determine

Mass m = ?

Important Tip:

  • The mass of a rock can be found using the formula F = ma

Using the formula

F = ma

where

  • F is the force (N)
  • m is the mass (kg)
  • a is the acceleration (m/s²)

now substituting F = 500, and a = 75 m/s² in the formula

F = ma

500 = m(75)

switch sides

m\left(75\right)=500

Divide both sides by 75

\frac{m\cdot \:75}{75}=\frac{500}{75}

simplify

m=\frac{20}{3}

m=6.7 kg

Therefore, we conclude that the mass of a rock with a force of 500 N and an acceleration of 75 m/s² is 6.7 kg.

Hence, option D is correct.

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3 years ago
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Answer:

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Explanation:

Given that,

Charge 1, q_1=3.11\ \mu C=3.11\times 10^{-6}\ C

The distance between charges, r = 0.241 m

Force experienced by particle 1, F = 3.44 N

We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :

F=k\dfrac{q_1q_2}{r^2}

q_2=\dfrac{Fr^2}{kq_1}

q_2=\dfrac{3.44\times (0.241)^2}{9\times 10^9\times 3.11\times 10^{-6}}

q_2=7.13\times 10^{-6}\ C

or

q_2=7.13\ \mu C

So, the magnitude of electric charge 2 is q_2=7.13\ \mu C. Since, the force is attractive then the magnitude of charge 2 must be negative.

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Having _________is the most obvious difference between a eukaryotic and a prokaryotic cell, but there are other differences as w
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