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levacccp
2 months ago
11

A tube with a cap on one end, but open at the other end, produces a standing wave whose fundamental frequency is 130.8 Hz. The s

peed of sound is 343 m/s. a. If the cap is removed , what is the new fundamental frequency?
Physics
1 answer:
Softa [3K]2 months ago
8 0

Answer:

A. 261.6 Hz

B. 0.656 m

Explanation:

A.

When a tube is open on one side and closed on the other, the fundamental frequency is given by:

F1 = V/4*L

Where,

F1 = fundamental frequency

V = speed of sound

L = length of the tube

If the tube is open at both ends:

F'1 = V/2*L

Where:

F'1 = new fundamental frequency

Thus,

F'1 = 2 * F1

= 2 * 130.8

= 261.6 Hz

B.

F1 = V/4*L

Or

F'1 = V/2*L

Given:

V = 343 m/s

F1 = 130.8

L = 343/(4 * 130.8)

= 0.656 m

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1 month ago
An organ pipe open at both ends has a radius of 4.0 cm and a length of 6.0 m. what is the frequency (in hz) of the third harmoni
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When air is forced into the open pipe,

L = \frac{nλ}{2}

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It is important to note that n=1 corresponds to the fundamental frequency, n=2 corresponds to the first harmonic, and so forth.

Thus, the third harmonic will be for n=4

With L=6m and n=4, solving for λ yields:

λ=\frac{(2)*(6)}{4} =3m

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1 month ago
Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells
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Answer:

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b) The count of blood cells is  N_t=1.04*10^{13}

Explanation:

From the problem statement, we learn that

         The blood volume is  V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3

         The density of blood is  \rho_b = 1060 kg/m^3

         % of blood which consists of cells is  = 45.0%

        the % of blood that is  plasma is  = 55.0%

        density of blood cells is  \rho_d = 1125kg/m^3

         % of cells that are white is  = 1%

        % of cells that are red is  = 99%

         The red blood cell diameter is  = 7.5 \mu m = 7.5*10^{-6}m

         The red blood cell radius is  = \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m

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               m = \rho_b * V_b

Substituting values

            m = 1060 * 0.00546

               m= 5.7876kg

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                         = 0.45 * 5,7876

                         = 2.60442 kg

The volume of cells is V_c = \frac{m_c}{\rho_d}

                                      = \frac{2.60442}{1125}

                                      = 2.315 *10^{-3} m^3

The white blood cells volume is V_w = 1% of the cells volume

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                                                       = 2.315*10^{-5}m^3

The volume of a single cell is V_s = 4 \pi r^3

                                                                        = 4*(3.142) * (3.75*10^{-6})^3

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                                                    = 1.037*10^{13}

The total white blood cell count is   =\frac{V_w}{V_s}

                                                          = \frac{2.315 * 10^{-5}}{2.21*10^{-16}}

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The overall number of blood cells is  N_t= 1.037*10^{13} + 1.04*10^{11}

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Answer:

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I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}

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Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf

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