1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
shutvik [7]
3 years ago
13

The pupil of a cat's eye narrows to a slit width of 0.5 mm in daylight. What is the angular resolution of the cat's eye in dayli

ght (wavelength = 500 nm)?
A. 0.01 rads
B. 10-5 rads
C. 10-3 rads
D. 10-4 rads
Physics
1 answer:
Elenna [48]3 years ago
4 0

Answer:

C. 10⁻³ rads

Explanation:

Here, we shall use Rayleigh's Criterion to find out the angular resolution of Cat's eye during day light. Rayleigh's Criterion is written as follows:

θ = λ/a

where,

θ = angular resolution of Cat's eye = ?

λ = wavelength = 500 nm = 5 x 10⁻⁷ m

a = slit width of eye = 0.5 mm = 5 x 10⁻⁴ m

Therefore,

θ = (5 x 10⁻⁷ m/5 x 10⁻⁴ m)

Therefore,

θ = 0.001

θ = Sin⁻¹(0.001)

θ = 0.001 rad = 1 x 10⁻³ rad

Hence, the correct answer is:

<u>C. 10⁻³ rads</u>

You might be interested in
Help with these please someone?
DIA [1.3K]

Answer:

8. 2.75·10^-4 s^-1

9. No, too much of the carbon-14 would have decayed for radiation to be detected.

Explanation:

8. The half-life of 42 minutes is 2520 seconds, so you have ...

1/2 = e^(-λt) = e^(-(2520 s)λ)

ln(1/2) = -(2520 s)λ

-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1

___

9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...

6.5·10^7/5.73·10^3 ≈ 11344

half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.

7 0
3 years ago
The wave function of a particle is exp(i(kx-omegat)), where x is distance, t is time, and k and co are positive real numbers. Th
Step2247 [10]

Answer:

Momentum, p=\hbar k

Explanation:

The wave function of a particle is given by :

y=exp[i(kx-\omega t)]...............(1)

Where

x is the distance travelled

t is the time taken

k is the propagation constant

\omega is the angular frequency

The relation between the momentum and wavelength is given by :

p=\dfrac{h}{\lambda}............(2)

From equation (1),

k=\dfrac{2\pi}{\lambda}

\lambda=\dfrac{2\pi}{k}

Use above equation in equation (2) as :

p=\dfrac{h k}{2\pi }

Since, \dfrac{h}{2\pi}=\hbar

p=\hbar k

So, the x-component of the momentum of the particle is \hbar k. Hence, this is the required solution.

8 0
3 years ago
An object at 27°C has its temperature increased to 37°C.
Firlakuza [10]

Answer:

b. 14

Explanation:

T_{i} = Initial temperature = 27 °C = 27 + 273 = 300 K

T_{f} = Final temperature = 37 °C = 37 + 273 = 310 K

P_{i} = Initial Power radiated by the object

P_{f} = Final Power radiated by the object

We know that the power radiated is directly proportional to fourth power of the temperature. hence

\frac{P_{f}}{P_{i}} = \frac{T_{f}^{4} }{T_{i}^{4} }\\\frac{P_{f}}{P_{i}} = \frac{(310)^{4} }{(300)^{4} }\\\frac{P_{f}}{P_{i}} = 1.14\\P_{f} = (1.14) P_{i}

Percentage increase in power is given as

\frac{(P_{f} - P_{i})\times100}{P_{i}} \\\frac{((1.14) P_{i} - P_{i})\times100}{P_{i}} \\14

3 0
3 years ago
A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E} = 708.03 N/C

Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C

5 0
3 years ago
PLEASE ANSWER BOTH QUICK!! 50 POINTS!!! PLEASE DON'T IGNORE!!
RUDIKE [14]

Answer:

A. B. Pls Give Branliest

Explanation:

3 0
3 years ago
Read 2 more answers
Other questions:
  • Which of the following is NOT an example of general characteristic of expository writing? Select all that apply.. .
    12·2 answers
  • The speed of an electromagnetic wave is a constant, 3.0 × 108 m/s. The wavelength of a wave is 0.3 meters. What is the frequency
    15·2 answers
  • A Fathom is an old depth measure equal to 6 Feet How deep in meters is a 5_ fathom deep water channel​ show working out
    8·1 answer
  • For questions 22 – 24, write an equation for the reaction of hydrogen chloride and sodium sulfide to produce hydrogen sulfide wi
    12·2 answers
  • Invader Zim’s spaceship is sitting at rest in outer space. The ship then accelerates at a uniform rate of 12 m/s2 for 10 seconds
    9·1 answer
  • What to do if the patient stops breaching during Fainting ?​
    9·2 answers
  • Two children hang by their hands from the same tree branch. The branch is straight, and grows out from the tree trunk at an angl
    6·1 answer
  • A boat is moving along the water. The forward thrust from the engines is 35,000 N. Water resistance is 20,000 N and air resistan
    10·1 answer
  • PLS HELP WILL MARK BRAINLIEST IF RIGHT NEED IMMEDIATELY PLEASEEEEE
    8·2 answers
  • The name of C (S) + o2 (g) CO2 (g)
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!