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3 years ago

Why do reflecting telescopes usually have a secondary mirror in addition to a primary mirror?

1 answer:

8 0

Once the starlight or moonlight comes down the tube and hits the big main mirror at the bottom, it's reflected, and heads back up the tube, along the same way it came in. But if you're looking through an eyepiece, the eyepiece is on the SIDE of the tube. So you need another small mirror, to catch the focused light before it goes up out of the tube, and send it sideways to the eyepiece.

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A wave traveling in the positive x-direction with a frequency of 50.0 Hz is shown in the figure below. Find the following values

Klio2033 [76]

Answer:

Explanation:

a. The amplitude is the measure of the height of the wave from the midline to the top of the wave or the midline to the bottom of the wave (called crests). The midline then divides the whole height in half. Thus, the amplitude of this wave is 9.0 cm.

b. Wavelength is measured from the highest point of one wave to the highest point of the next wave (or from the lowest point of one wave to the lowest point of the next wave, since they are the same). The wavelength of this wave then is 20.0 cm. or $\lambda=20.0cm$

c. The period, or T, of a wave is found in the equation

$f=\frac{1}{T}$ were f is the frequency of the wave. We were given the frequency, so we plug that in and solve for T:

$50.0=\frac{1}{T}$ so

$T=\frac{1}{50.0}$ and

T = .0200 seconds to the correct number of sig fig's (50.0 has 3 sig fig's in it)

d. The speed of the wave is found in the equation

$f=\frac{v}{\lambda}$ and since we already have the frequency and we solved for the wavelength already, filling in:

$50.0=\frac{v}{20.0}$ and

v = 50.0(20.0) so

v = 1.00 × 10³ m/s

And there you go!

5 0

2 years ago

HELPP ME IN PHYSICS +15 POINTS!! But a right answer not links or I’ll report. -_-

frozen [14]

The answer shud be 31 . 14m

8 0

2 years ago

Use the graph below to answer the following question: if average acceleration is calculated using the equation, “ change in velo

sergiy2304 [10]

Answer:

$a=9\ cm/s^2$

Explanation:

Average Acceleration

Acceleration is a physical magnitude defined as the change of velocity over time. When we have experimental data, we can compute it by calculating the slope of the line in velocity vs time graph.

Note: We cannot see if the time axis is numbered in increments of 1 second, and we'll assume that.

When $t_2=4\ sec$ , the graph shows a value of $v_2=36\ cm/s$

When $t_1=0\ sec$ , the object is at rest, $v_1=0$

We compute the average acceleration as

$\displaystyle a=\frac{v_2-v_1}{t_2-t_1}$

$\displaystyle a=\frac{36\ cm/s-0\ cm/s}{4\ sec-0\ sec}$

$\displaystyle a=\frac{36\ cm/s}{4\ s}$

$\boxed{a=9\ cm/s^2}$

6 0

2 years ago

A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p

alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

$F=qvB$

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

$qvB = \frac{mv^2}{r}$

which can be rewritten as

$v=\frac{qB}{mr}$

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

$\frac{2\pi r}{T}=\frac{qB}{mr}$

So, we get:

$T=\frac{2\pi m r^2}{qB}$

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) $a. f/10$

The frequency of revolution of a particle in uniform circular motion is

$f=\frac{1}{T}$

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

$f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}$

6 0

2 years ago

The mass of jupiter is 300 times the mass of the earth. Jupiter orbits the sun with Tjupiter = 11.9 yr in an orbit with Rjupiter

mihalych1998 [28]

Answer:

c) 11.9 yr

Explanation:

The orbital period is proportional to r^(3/2) and does not depend on the satellite's mass. Any object at Jupiter position will have the same orbital period regardless of mass.

By keppler's law we know that

T^2= r^3

T= orbital time period

r= mean distance of the planet from the Sun.

clearly, The orbital period does not depend on the satellite's mass

there, the correct answer will be c= 11.9 yr.

5 0

3 years ago