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drek231 [11]
3 years ago
15

If the angular magnification of an astronomical telescope is 42 and the diameter of the objective is 72 mm, what is the minimum

diameter of the eyepiece required to collect all the light entering the objective from a distant point source located on the telescope axis?
Physics
1 answer:
andriy [413]3 years ago
4 0

Answer:

Minimum diameter=1.714

Explanation:

Angular magnification=fob/fey=42

fob=40*fey

Hence minimum diameter=diameter of objective/angular magnification

minimum diameter=72/42=1.714

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A sinusoidal wave travels along a string. if the time for a particular point to move from maximum displacement to zero displacem
Lisa [10]
The time it takes for a point to reach 0 displacement from maximum displacement is \frac{\pi}{2} of a cycle. Thus, one period (2\pi) will take T=0.17*4=0.68s.

The frequency is simply f=\frac{1}{T}=\frac{1}{0.68}=1.47Hz.

The speed is given by: v=f\lambda=1.47*1.4 = 2.06m/s
7 0
3 years ago
Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.47 T. When viewed edge-on, the first surfac
alexgriva [62]

Answer:

a. A = 0.0859 m^2

b. A = 0.0178 m^2

Explanation:

Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.47 T. When viewed edge-on, the first surface is tilted at an angle of from the horizontal, and a net magnetic flux of 8.4 103 Wb passes through it. The same net magnetic flux passes through the second surface. (a) Determine the area of the first surface. (b) Find the smallest possible value for the area of the second surface.

take note that the question has not specified th angle which the surface is tilted so i assume the angle is at 12^{0} to the horizontal

flux = BAcos(\alpha)

B=magnetic flux in Weber

A=area of the flat surface in m^2

\alpha=the angle to the horizontal

a) 8.4 x10^-3= (.47)Acos(78)

alpha has to be the angle from the normal and not the horizontal so 90-12=78,

 8.4 x10^-3

/(.47)cos(78)

A = 0.0859 m^2

b) If flux remains the same then for it to be the smallest possible area it needs to be perpendicular to the magnetic field so alpha would be 0.

8.4 x10^-3 = (.47)Acos(0)

A = 0.0178 m^2

7 0
4 years ago
What are lowland areas on the moon
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Craters flat land or rock
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A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro
Mkey [24]

Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia =  1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

7 0
4 years ago
How do you find the number of valence electrons in one atom of magnesium?
Iteru [2.4K]
Magnesium is element 12 in the periodic table and has 2 valance electrons.
7 0
4 years ago
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