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3 years ago

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A tube with a cap on one end, but open at the other end, has a fundamental frequency of 130.8 Hz. The speed of sound is 343 m/s

(a) If the cap is removed, what is the new fundamental frequency of the tube

1 answer:

sergey [27]3 years ago

8 0

Answer:

Y = V / f where Y equals wavelength

4 Y1 = V / f1 for a closed pipe the wavelength is 1/4 the length of the pipe

2 Y2 = V / f2 for the open pipe the wavelength is 1/2 the length of the pipe

Y1 / Y2 = 2 = f2 / f1 dividing equations

f2 = 2 f1

the new fundamental frequency is 2 * 130.8 = 261.6

(The new wavelength is 1/2 the original wavelength so the frequency must double to produce the same speed.

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A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

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Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :

$V_{2}=V_{1}+at$ (6)

$V_{2}^{2}=V_{1}^{2}+2ad$ (7)

Where:

$a$ is the acceleration

$d=0.01 m$ is the length the ball was compressed

$t$ is the time

Finding $a$ from (7):

$a=\frac{V_{2}^{2}-V_{1}^{2}}{2d}$ (8)

$a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)}$ (9)

$a=90.25 m/s^{2}$ (10)

Substituting (10) in (6):

$57 m/s=-38 m/s+(90.25 m/s^{2})t$ (11)

Finding $t$ :

$t=1.052 s$ (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force $F$ is proportional to the variation of momentum $\Delta p$ in time $\Delta t$ :

$F=\frac{\Delta p}{\Delta t}$ (13)

$F=\frac{65.075 kgm/s}{1.052 s}$ (14)

Finally:

$F=61.82 N$

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An 1800-kg truck pulls a 710-kg trailer away from a stoplight with an acceleration of 1.20 m/s2 . Part A What is the net force e

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Answer:

F = 852 N

Explanation:

We apply Newton's second law to the trailer :

F = m*a Formula (1)

F : net force exerted by the truck on the trailer Newtons (N)

m : mass of the trailer in kilograms (kg)

a : acceleration of the trailer in meters over second square (m/s²)

Data

a=1.20 m/s² : acceleration of the trailer

m=710 kg : mass of the trailer

We replace data in the Formula (1) to calculate the net force exerted by the truck on the trailer

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Particles q1, 92, and q3 are in a straight line.

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The net force on q₃ will be 17.51 N. The net force is the algebraic sum of the two forces on the pleading q₃

<h3 /><h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The force,by the charge q₁ on the q₃;

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The force,by the charge q₂ on the q₃;

$\rm F_{32}} = \frac{Kq_2q_3}{r^2} \\\\ \rm F_{32}} = \frac{9 \times 10^9 \times 25.5 \times 10^{-6}\times \times -74.9\times 10^{-6}}{(0.300)^2} \\\\ F_{32}} =-19.09 \ N$

The net force is the sum of the two forces;

$\rm F_{net}=F_{32}+F_{31}\\\\\ \rm F_{net}=36.6-19.9 \\\\ \rm F_{net}=17.51 \ N$

Hence, the net force on q₃ will be 17.51 N.

To learn more about Columb's law, refer to the link;

brainly.com/question/1616890

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