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egoroff_w [7]
3 years ago
7

25 POINTS FOR CORRECT ANSWER

Physics
2 answers:
castortr0y [4]3 years ago
5 0
No.  I do not agree with Stefan.  Quite the contrary.  I disagree
with his description of "<span>angle of incidence" as the angle between
the surface of the mirror and the incoming ray. 

The correct description of "angle of incidence" is </span><span>the angle between
the NORMAL TO the surface of the mirror and the incoming ray. 

Thus, the true angle of incidence is the complement of the angle that
Stefan calculates or measures.</span>
Dvinal [7]3 years ago
3 0

I do not agree With stefan hope this helps


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An open pipe of length 0.58 m vibrates in the third harmonic with a frequency of 939Hz. What is the speed of sound through the a
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363 m/s is the speed of sound through the air in the pipe.

Answer: Option B

<u>Explanation:</u>

The formula used to calculate the wavelength given as below,

      Wavelength (\lambda)=\frac{\text { wave velocity }(v)}{\text { frequency }(f)}

      v=\lambda \times f   --------> eq. 1

In power system, harmonics define by positive integers of the fundamental frequency. So the third order harmonic is a multiple of the third fundamental frequency. Each harmonic creates an additional node and an opposite node, as well as an additional half wave within the string.

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         L=\frac{\lambda}{2}

For second harmonic, n =2

         L=\frac{2 \lambda}{2}=\lambda

For third harmonic, n =3

         L=\frac{3 \lambda}{2}

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Here given f = 939 Hz, L = 0.58 m...And, substitute eq 2 in eq 1 and values, we get

   v=\frac{2 \times 0.58 \times 989}{3}=\frac{1089.24}{3}=363.08 \mathrm{m} / \mathrm{s}

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julia-pushkina [17]

1)

\begin{gathered} E1=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_1v_2^2 \\ where: \\ m_1=m_2=0.25kg \\ v_1=7m/s \\ v_2=-7m/s \\ so: \\ E1=\frac{1}{2}0.25(7^2)+\frac{1}{2}0.25(7^2) \\ E1=6.125+6.125 \\ E1=12.25J \end{gathered}

Answer:

d. 12.25J

------------------------

2)

According to the conservation of energy:

\begin{gathered} E1=E2 \\ so: \\ E2=12.25J \end{gathered}

Answer:

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3)

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Answer:

d. 0 kg∙m/s

-----------------------------------------

4)

Using conservation of momentum:

\begin{gathered} P1=P2 \\ so: \\ P2=0 \end{gathered}

Answer:

b. 0 kg•m/s

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