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Bumek [7]
3 years ago
12

Question #5: Explain how a reflecting telescope is different from a refracting telescope. List the two different types of reflec

ting telescopes given in your text.
Physics
1 answer:
Anni [7]3 years ago
6 0

The main component in a reflecting telescope is a mirror where the light will bounce off and is then focused into a smaller area. In contrast, a refracting telescope uses lenses that focus the light as it travels towards the other end.

Two different types of reflecting telescopes are:

1.Cassegrain reflector

2.Newtonian telescope

Explanation:

  • The distinction between the two is in how they manipulate the incoming light in order to magnify the image. The main component in a reflecting telescope is a mirror where the light will bounce off and is then focused into a smaller area.
  • Key advantage of reflecting telescopes is how big you can make them. With lenses, the maximum size is limited to about one meter, largely because of the problems stated above as well as the skyrocketing costs.
  • The Newtonian telescope, also called the Newtonian reflector, is a type of reflecting telescope invented Sir Isaac Newton, using a concave primary mirror and a flat diagonal secondary mirror. The Newtonian telescope's simple design has made it very popular with amateur telescope makers.
  • The Cassegrain reflector is a combination of a primary concave mirror and a secondary convex mirror, often used in optical telescopes and radio antennas, the main characteristic being that the optical path folds back onto itself, relative to the optical system's primary mirror entrance aperture.

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<span>The factors that are used to determine power are:
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5 0
3 years ago
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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 5.00 s, it rotates 13.9 rad. Du
alisha [4.7K]

Answer:

(a) Angular acceleration is 1.112 rad/s².

(b) Average angular velocity is 2.78 rad/s .

Explanation:

The equation of motion in Rotational kinematics is:

θ = θ₀ + 0.5αt²

Here θ is angular displacement at time t, θ₀ is angular displacement at time t=0, t is time and α is constant angular acceleration.

(a) According to the problem, θ is 13.9 rad, θ₀ is zero as it is at rest and t is 5 s. Put these values in the above equation:

13.9 = 0 + 0.5α(5)²

α = 1.112 rad/s²

(b) The equation of average angular velocity is:

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ω = 2.78 rad/s

3 0
3 years ago
How far can a sound wave travel in 90 seconds when the ambient air temperature is 10 C?
Ksju [112]

Answer:

s = 30330.7 m = 30.33 km

Explanation:

First we need to calculate the speed of sound at the given temperature. For this purpose we use the following formula:

v = v₀√[T/273 k]

where,

v = speed of sound at given temperature = ?

v₀ = speed of sound at 0°C = 331 m/s

T = Given Temperature = 10°C + 273 = 283 k

Therefore,

v = (331 m/s)√[283 k/273 k]

v = 337 m/s

Now, we use the following formula to calculate the distance traveled  by sound:

s = vt

where,

s = distance traveled = ?

t = time taken = 90 s

Therefore,

s = (337 m/s)(90 s)

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3 years ago
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Answer:

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Explanation:

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So the relation between the time and the velocity is linear. If we plot the velocity in function of time, the plot is a line, and the acceleration is the slope of this line.

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3 years ago
Acceleration usually has the symbol a. It is a vector. What is the correct way
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Explanation:

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