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sineoko [7]
3 years ago
13

  If a string vibrates at the fundamental frequency of 528 Hz and also produces an overtone with a frequency of 1,056 Hz, this o

vertone is the 
*A. *second harmonic.
*B. *first harmonic.
*C. *fourth harmonic.
*D. *third harmonic.

14.   The _______ of a sound wave is defined as the amount of energy passing through a unit area of the wave front in a unit of time. 
*A. *amplitude
*B. *compression
*C. *intensity
*D. *frequency
Physics
1 answer:
Leto [7]3 years ago
6 0

your answer is *A. *second harmonic.

i just took the test

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How can liquid pressure phenomena be used in daily life. Give one example with explanation of working​?
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Answer:

An example in which liquid pressure phenomena can be used in daily life is in Water blasting

Explanation:

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There are different varieties of water blasting, including;

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What energy output objects work with the turbine?
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Answer:

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6 0
2 years ago
Why does the sample on a microscope slide need to be very thin?
ziro4ka [17]

B. so light can shine through it from below.

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3 years ago
Read 2 more answers
For thermal equilibrium at temperature Tan appropriate measure of energy is kT where k is Boltzmann's constant. Convert the foll
Schach [20]

Answer:

1 cm⁻¹ =1.44K  1 ev = 1.16 10⁴ K

Explanation:

The relationship between temperature and thermal energy is

     E = K T

The relationship of the speed of light

    c =λ f = f / ν          1/λ= ν

The Planck equation is

          E = h f

Let's start the transformations

     c = f λ = f / ν        

     f = c ν

     E = h f

     E = h c ν

     E = KT

     h c ν = K T

     T = h c ν  / K =( h c / K) ν

Let's replace the constants

     h = 6.63 10⁻³⁴ J s

     c = 3 10⁸ m / s

     K = 1.38  10⁻²³ J / K

 

     v = 1 cm-1 (100 cm / 1 m) = 10² m-1

   

     T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 1 10²

     A = h c / K = 1,441 10⁻²

     T =  1.44K

     ν = 103 cm⁻¹ = 103 10² m

     T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 103 10²

     T = 148K

1 Rydberg = 1.097 10 7 m

As we saw at the beginning the λ=1 / v

     T = (h c / K) 1 /λ

     T = 1,441 10⁻²  1 / 1,097 10⁷

     T = 1.3 10⁻⁹ K

    E = 1Ev (1.6 10⁻¹⁹ J /1 eV) = 1.6 10⁻¹⁹ J

    E = KT

    T = E/K

    T = 1.6 10⁻¹⁹ /1.38 10⁻²³

    T = 1.16 10⁴ K

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3 years ago
Ideally, the resistance of an ammeter should be:
pav-90 [236]
Ideally the resistance should be ZERO
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