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3 years ago

15

A tube closed at one end is used to determine the speed of sound in air. The resonances occur every 32 cm when a 530-Hz tuning f

ork is vibrated. What is the velocity of the sound?

1 answer:

soldi70 [24.7K]3 years ago

6 0

Answer:169.6m/s

Explanation:

Wavelength=32cm=0.32m

Frequency=580Hz

Velocity=frequency x wavelength

Velocity=580x0.32

Velocity=169.6m/s

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What happens when light passes through a prism or diffraction grating?

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The light scatters in all directions.

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A 60KG WOMAN IS ON A LADDER 2 M ABOVE THE GROUND. WHAT IS HER POTENTIAL ENERGY

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2 years ago

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8m/s2 . The accel

Zanzabum

Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.

Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m

Now you need the final speed to use it as initial speed of the next part.

Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s

Part B) Free fall

Maximum height, y max ==> Vf = 0

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3 years ago

A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is

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2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as

$a = -\omega^2 A$

Where,

a = Acceleration

A = Amplitude

$\omega$ = Angular velocity

From a reference system in which the downward acceleration is negative due to the force of gravity we will have to

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$-\omega^2 A = -g$

$\omega = \sqrt{\frac{g}{A}}$

From the definition of frequency and angular velocity we have to

$\omega = 2\pi f$

$f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}$

$f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}$

$f = 2.5Hz$

Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz

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3 years ago

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an electric field is the answer

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3 years ago

Read 2 more answers