All categories

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.

You might be interested in

you see a lightning bolt in the sky. You hear a clap of thunder 3 seconds latter. The speed of 330 m/s. How far away was the lig

Naddika [18.5K]

The i got answer is 990m

8 0

3 years ago

1. Is the sequence geometric? If so, identify the common ratio.

frozen [14]

1) yes;2 6*2=12 12*2=24 24*2=48
2)Next Term (or nth term) = ar^n-1

a = first term, i.e. 5
r = common ratio i.e. 3 (as 15/5=3 and 45/15=3 
n = .. 
as you already have 1st , 2nd and 3rd terms
substituting now 
T4= ar^n-1 
= 5*3^4-1 
= 5*3^3 
= 5*27 
T4 = 135
T5= ar^n-1
= 5*3^5-1 
= 5*3^4 
= 5*81 
T5 = 405

6 0

4 years ago

One of the harmonic frequencies of tube A with two open ends is 576 Hz. The next-highest harmonic frequency is 648 Hz. (a) What

balu736 [363]

(a) 288 Hz

The difference between any two harmonics of an open-end tube is equal to the fundamental frequency, $f_1$ (first harmonic):

$f_{n+1}-f_n = f_1$ (1)

In this problem, we are told the frequencies of two successive harmonics:

$f_n = 576 Hz\\f_{n+1}=648 Hz$

So the fundamental frequency is:

$f_1 = 648 Hz-576 Hz=72 Hz$

Now we know that one of the the harmonics is $f_n=216 Hz$ , so its next highest harmonic will have a frequency of

$f_{n+1}=f_n+f_1 = 216 Hz+72 Hz=288 Hz$

(b) n=4

The frequency of the nth-harmonic is an integer multiple of the fundamental frequency:

$f_n=n f_1$ (2)

Since we know $f_n = 288 Hz$ , we can solve (2) to find the number n of this harmonic:

$n=\frac{f_n}{f_1}=\frac{288 Hz}{72 Hz}=4$

(c) 4445 Hz

For a closed pipe (only one end is open), the situation is a bit different, because only odd harmonics are allowed. This means that the frequency of the nth-harmonic is an odd-integer multiple of the fundamental frequency:

$f_n=(2n+1) f_1$ (2)

so, the difference between any two harmonics tube is equal to:

$f_{n+1}-f_n = (2(n+1)+1)f_1-(2n+1)f_1=(2n+3)f_1-(2n+1)f_1=2f_1$ (1)

In this problem, we are told the frequencies of two successive harmonics:

$f_n = 4699 Hz\\f_{n+1}=4953 Hz$

So, according to (1), the fundamental frequency is equal to half of this difference:

$f_1 = \frac{4953 Hz-4699 Hz}{2}=127 Hz$

Now we know that one of the harmonics is $f_n=4191 Hz$ , so its next highest harmonic will have a frequency of

$f_{n+1}=f_n+2f_1 = 4191 Hz+254 Hz=4445 Hz$

(d) n=17

We said that the frequency of the nth-harmonic is equal to an odd-integer multiple of the fundamental frequency:

$f_n=(2n+1) f_1$ (2)

Since we know $f_n = 4445 Hz$ , we can solve (2) to find the number n of this harmonic:

$n=\frac{1}{2}(\frac{f_n}{f_1}-1)=\frac{1}{2}(\frac{4445 Hz}{127 Hz}-1)=17$

7 0

3 years ago

a solenoid of length 1.00 cm1.00 cm and radius 0.550 cm0.550 cm has 4545 turns. if the wire of the solenoid has 1.95 amps1.95 am

steposvetlana [31]

The magnitude of the magnetic field inside the solenoid is 4.56 X 10-3 T

A magnetic subject is a vector subject that describes the magnetic effect on shifting electric fees, electric currents, and magnetic materials. A shifting rate in a magnetic subject experiences a force perpendicular to its own speed and to the magnetic area.

A magnetic area is a vector area that describes the magnetic have an impact on transferring electric powered prices, electric powered currents, and magnetic substances. A moving rate in a magnetic area studies a force perpendicular to its very own pace and to the magnetic subject.

Calculation

B = UNI/L

B = (4pi X 10-7)(25)(1.45)/(.01)

B = 4.56 X 10-3 T

Learn more about magnetic field here - brainly.com/question/26051825

#SPJ4

4 0

1 year ago

Suppose that you measure a galaxy's redshift, and from the redshift you determine that its recession velocity is 30,000 (3×104 k

Liula [17]

Get the homework, burn it, and run!!!

3 0

4 years ago