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Bond [772]
2 years ago
9

Consider a cylindrical pipe with both ends open and fundamental resonance frequency 150 Hz. If this particular pipe has a length

of L, what the wavelengths of three lowest modes produced by this pipe? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a. L,2L, 3L b. 2L, L, 2L/3 с. 2L/3, L, 2L d. L, L/2, L/3 e. 4L, 24, 4L/3 f. 4L, 4L/3, 4L/5 g. Neither of above.
Physics
1 answer:
Naya [18.7K]2 years ago
8 0

Answer:

answers, the correct one is b

Explanation:

For this problem we use the fact that since the ends are open at these points we have a maximum,

fundamental L =λ/2 λ = 2L

second harmonic L = λ λ = 2L / 2

third harmonic L = 3 λ/2 λ = 2L / 3

n harmonic lam = 2L / n

 with n an integer

therefore the length of the first three harmonics are: 2L, L, 2L / 3

when examining the different answers, the correct one is b

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Answer:

The time taken in years is   x = 125 \  years

Explanation:

From the question we are told that

   The  speed is  v  = 540.00 \ km /hr  =  \frac{540 *1000}{3600} =  150\  m/s

    The distance from the sun to Pluto is  d =  5.9*10^{9} \  k m =  5.9*10^9 * 1000 =  5.9*10^{12} \  m

Generally the time  taken is mathematically represented as

     t =  \frac{d}{v}

=>   t = \frac{5.9*10^{11}}{150}

=>   t =  3.933*10^{9}

Converting to years

   1 year  \to  3.154*10^7 \  s

    x \  years  \to 3.933*10^{9}

=>  x = \frac{ 3.933*10^{9} * 1 }{ 3.154 *10^7}

=>    x = 125 \  years

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What is the specific heat of water? A. 1.00 J/gºC
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Answer:

B 4.18J\g degree C is the specific heat of water

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The roller-coaster car shown in fig. 6-41 (h1 = 30 m, h2 = 12 m, h3 = 20 m), is dragged up to point 1 where it is released from
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<span>Since there is no friction, conservation of energy gives change in energy is zero Change in energy = 0 Change in KE + Change in PE = 0 1/2 x m x (vf^2 - vi^2) + m x g x (hf-hi) = 0 1/2 x (vf^2 - vi^2) + g x (hf-hi) = 0 (vf^2 - vi^2) = 2 x g x (hi - hf) Since it starts from rest vi = 0 Vf = squareroot of (2 x g x (hi - hf)) For h1, no hf Vf = squareroot of (2 x g x (hi - hf)) Vf = squareroot of (2 x 9.81 x 30) Vf = squareroot of 588.6 Vf = 24.26 For h2 Vf = squareroot of (2 x 9.81 x (30 – 12)) Vf = squareroot of (9.81 x 36) Vf = squareroot of 353.16 Vf = 18.79 For h3 Vf = squareroot of (2 x 9.81 x (30 – 20)) Vf = squareroot of (20 x 9.81) Vf = 18.79</span>
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Two basketballs of equal mass are rolling toward each other at constant velocities. The first basketball (B1) has a velocity of
slamgirl [31]

v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)

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where,

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v'₂ = final velocity of ball 2

The final velocity of the balls after head on elastic collision would be

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Substituting the velocities in the equation

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If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.

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