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3 years ago

What are the characteristics of the image based on the values?Check all that apply

Physics

2 answers:

Ahat [919]3 years ago

8 0

inverted, real, behind the lens is the answer

kifflom [539]3 years ago

7 0

Answer:

Inverted

Real

Behind the lens

Explanation:

As we know that

$d_o = 16 cm$

$d_i = 16 cm$

also we know that

$h_o = 4 cm$

$h_i = -4 cm$

now we know that magnification is given by

$M = \frac{d_i}{d_o} = \frac{h_i}{h_o}$

so we can say here that

M = 1

so here image size will be same as object size and it must be real and inverted as magnification is negative in sign

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3 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]

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

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

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

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

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

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$f_n=n f_1$ (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:

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