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
Sloan [31]
2 years ago
10

The resolution of a telescope is ultimately limited by the diameter of its objective lens or mirror. A typical amateur astronome

r's telescope may have a 6.09 in diameter mirror. What is the minimum angular separation (in arc seconds) of two stars that can be resolved with a 6.09 in scope
Physics
1 answer:
slava [35]2 years ago
3 0

Answer:

 θ =  3.19  arc second

Explanation:

The resolution of a telescope is given by the rayleigh criterion, which establishes that two objects are separated if the principal maximum of diffraction of one of them coincides with the first minimum of diffraction of the second object, based on this the solution is given by the first diffraction minimum, the a slit is

        a sin θ = m λ

with m = 1

in the case of circular apertures the equation must be found in polar coordinates, therefore a numerical constant is introduced

        a sin θ = 1.22 λ

Angles are measured in radians and in these experiments they are small

        sin θ = θ

       θ= 1.22  λ  / a

in this case a = 6.09 in, the wavelength is wrong = 550 10⁻⁹ m which is the maximum resolution of the human eye

l

et's reduce the magnitudes to the SI system

        d = 6.09‘  2.54 10⁻-2 m / 1 inch = 15.4686 10-2 m

let's calculate

       θ = 1.22 550 10-9 / 15.468 10-2

       θ = 15.5 10⁻⁶ rad

       rad = 2.06 105 s

       θ = 15.5 10⁻⁶ rad  2.06 105s/ 1 rad

       θ =  3.19   s

     

You might be interested in
What is the momentum of a 0.1-kg mass moving with a speed of 5 m/s
Andrej [43]
This question is wrong because in momontum we will write acceleration instead of speed.     suppose acceleration is 5m/s2 then 
 P= ma 
                then put values
   
4 0
3 years ago
You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, w
Vesna [10]

Answer:

\ \text{m/s}

Explanation:

u_1 = Velocity of one lump = 3x+3y-3z

u_2 = Velocity of the other lump = -4x+0y-4z

m = Mass of each lump = 30\ \text{g}

The collision is perfectly inelastic as the lumps stick to each other so we have the relation

mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=\ \text{m/s}

The velocity of the stuck-together lump just after the collision is \ \text{m/s}.

4 0
2 years ago
A particular balloon can be stretched to a maximum surface area of 1257 cm2. The balloon is filled with 3.1 L of helium gas at a
chubhunter [2.5K]

Answer:

The ballon will brust at

<em>Pmax = 518 Torr ≈ 0.687 Atm </em>

<em />

<em />

Explanation:

Hello!

To solve this problem we are going to use the ideal gass law

PV = nRT

Where n (number of moles) and R are constants (in the present case)

Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} --- (*)

Our initial state is:

P1 = 754 torr

V1 = 3.1 L

T1 = 294 K

If we consider the final state at which the ballon will explode, then:

P2 = Pmax

V2 = Vmax

T2 = 273 K

We also know that the maximum surface area is: 1257 cm^2

If we consider a spherical ballon, we can obtain the maximum radius:

R_{max} = \sqrt{\frac{A_{max}}{4 \pi}}

Rmax = 10.001 cm

Therefore, the max volume will be:

V_{max} = \frac{4}{3} \pi R_{max}^3

Vmax = 4 190.05 cm^3 = 4.19 L

Now, from (*)

P_{max} = P_1 \frac{V_1T_2}{V_2T_1}

Therefore:

Pmax= P1 * (0.687)

That is:

Pmax = 518 Torr

6 0
3 years ago
!!!!HURRY 30 POINTS!!!!
Nataly_w [17]
The anwser is C flooding
3 0
3 years ago
A 5 kg block is being pushed horizontally across a level surface at a constant velocity. What is the magnitude of the Normal for
luda_lava [24]

Answer:

50 N.

Explanation:

On top of a horizontal surface, the normal force acting on an object is equivalent to the force of gravity acting on the object. That is:

\displaystyle \begin{aligned} F_N = F_g & = ma \\ & = mg\end{aligned}

The mass of the block is 5 kg and the given force due to gravity is 10 N/kg. Substitute and evaluate:

\displaystyle F_N = F_g = (5\text{ kg})(10 \text{ N/kg}) = 50 \text{ N}

In conclusion, the normal force acting on the block is 50 N.

5 0
2 years ago
Other questions:
  • An experimenter adds 970 J of heat to 1.75 mol of an ideal gas to heat it from 10.0∘C to 25.0∘C at constant pressure. The gas do
    6·1 answer
  • Kevin goes bowling. Whenever he bowls the ball, he transfers energy from his hand to the bowling ball. The amount of energy befo
    9·1 answer
  • How much mass dose a vehicle have if it needs 29000 Newton’s of force to accelerate it at a rate of 4m/s
    9·1 answer
  • If the mass of a material is 41 grams and the volume of the material is 5 cm^3, what would the density of the material be? answe
    15·2 answers
  • A runner starts at point A runs around a 1 mile track and finishes the run back at point A. Which of the following statement is
    15·1 answer
  • Consider a 10 gram sample of a liquid with specific heat 2 J/gK. By the addition of 400 J, the liquid increases its temperature
    6·1 answer
  • Rod A and rod B are cylindrical rods made of the same metal. amd they differ only in size. Rod B has double the length and doubl
    7·1 answer
  • If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is
    15·1 answer
  • A car is traveling 50 mph. It leaves Boston, Massachusetts at 5 a.m. and is
    7·1 answer
  • Does gravitational potential energy increase when getting closer to planet universal gravitation law.
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!