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
vovikov84 [41]
3 years ago
7

Technological advances are now making it possible to link visible-light telescopes so that they can achieve the same angular res

olution as a single telescope over 300 meters in size. Part A What is the angular resolution (diffraction limit) of such a system of telescopes for observations at a wavelength of 500 nanometers?
Physics
2 answers:
leonid [27]3 years ago
4 0

Answer: The angle of resolution =47.05°

Explanation:

The angle of resolution represents the resolve power and precision of optical instruments such as the eye, camera and e en a microscope. It is given by the equation:

Sin A = 1.220(W÷D)

Sin A= 1.220(300÷500)

Sin A = 1.220(0.6)

Sin A = 0.732

A= Sin^-1 0.732

A= 47.05°

devlian [24]3 years ago
3 0

Answer:

Explanation:

θ = 1.22 × (λ/D)

where,

θ is the angular resolution (radians) λ is the wavelength of light

= 500 nm

= 0.5 × 10^-6 m

D is the diameter of the lens' aperture.

= 300 m

θ = 1.22 × (0.5/300)

= 0.00203 arcsecond.

You might be interested in
According to our present theory of solar system formation, why were solid planetesimals able to grow larger in the outer solar s
nirvana33 [79]

Answer:

According to our present theory of solar system formation, why were solid planetesimals able to grow larger in the outer solar system than in the inner solar system? Because only metal and rock could condense in the inner solar system, while ice also condensed in the outer solar system.

Explanation:

5 0
2 years ago
What is the speed of a giraffe that has a
nevsk [136]
4 m/s because it is THR fastest voice
6 0
2 years ago
A student drops a 0.4kg ball from a height a of 49m above the ground. Neglect drag. Answer each of the following questions about
Elan Coil [88]

• The only force acting on the ball, and thus the net force, is due to its own weight, with magnitude

<em>w</em> = <em>m</em> <em>g</em> = (0.4 kg) (9.8 m/s²) ≈ 3.9 N

pointing downward.

• The ball is in free fall, so its acceleration is <em>g</em> = 9.8 m/s² (also pointing downward).

• The ball is dropped from a height of 49 m, so its height <em>y</em> at time <em>t</em> is

<em>y</em> = 49 m - 1/2 <em>g</em> <em>t</em> ²

Set <em>y</em> = 0 and solve for <em>t</em> :

0 = 49 m - 1/2 <em>g t</em> ²

<em>t</em> ² = (98 m) / <em>g</em>

<em>t</em> = √((98 m) / <em>g</em>) = √(10) s ≈ 3.2 s

• The ball's velocity <em>v</em> at time <em>t</em> is

<em>v</em> = - <em>g t</em>

so that at the time found previously, the ball will have attained a velocity of

<em>v</em> = - <em>g</em> (3.2 s) ≈ -31 m/s

and thus a <em>speed</em> of about 31 m/s.

4 0
3 years ago
The formula v = √ 2.3 r models the maximum safe speed, v , in miles per hour, at which a car can travel on a curved road with ra
Archy [21]

Answer:

562 miles per hour.

Explanation:

As given in the question, the formula for the maximum speed on a curved road is

v=\sqrt{2.3} r

Given value of r=370 feet

So the maximum safe speed will be

v=\sqrt{2.3} \times 370 = 1.52\times 370 = 562.4 miles per hour.

Rounding off to the nearest whole number we get the maximum safe speed at the curved road is 562 miles per hour.

6 0
4 years ago
Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f
vampirchik [111]

Answer:

\large \boxed{\text{30 rev/s}}

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

 I₁ = 3.5    kg·m²s⁻¹

ω₁ = 6.0    rev·s⁻¹

 I₂ = 0.70 kg·m²s⁻¹

Calculation:

\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}

8 0
4 years ago
Other questions:
  • I need an answer for 1 and 2 pleaseeee
    10·1 answer
  • Juan draws a free-body diagram of an object that is in dynamic equilibrium moving to the left. Which labels correctly complete t
    15·2 answers
  • The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This sp
    13·1 answer
  • Which of the following statements about the resistsance of a wire is correct? Select THREE answers.
    11·1 answer
  • पाल
    5·1 answer
  • Turner’s treadmill starts with a velocity of
    6·2 answers
  • How many atoms are in a cell
    13·1 answer
  • When we look at a star
    14·1 answer
  • Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
    14·1 answer
  • Can you answer these questions would mean so much grades are due next week and im just really busy will give 35 points
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!