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ikadub [295]
4 years ago
15

Frequency is the ____________ that move past a point during a specific amount of time. Frequency is measured in ____________ , a

nd is classified as high, medium, or low. Frequency is interpreted as the ____________ of a sound. Intensity refers to the ____________ of a sound and is measured in ____________ . Louder sounds ____________ the rate of nerve signals relayed to the brain.
Physics
1 answer:
yulyashka [42]4 years ago
6 0

Answer:

Frequency is <u>the number of waves</u> that move past a point during a specific amount of time. Frequency is measured in <u>Hertz</u>, and is classified as high, medium, or low. Frequency is interpreted as the <u>pitch</u> of a sound. Intensity refers to the <u>loudness</u> of a sound and is measured in <u>decibels</u>. Louder sounds <u>increase</u> the rate of nerve signals relayed to the brain.

Explanation:

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We will use the formula:
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4 years ago
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A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does
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Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right) (1)

Where:

y(t) - Position of the mass as a function of time, measured in meters.

A - Amplitude, measured in meters.

k - Spring constant, measured in newtons per meter.

m - Mass of the block, measured in kilograms.

t - Time, measured in seconds.

\phi - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

k = \frac{m\cdot g}{\Delta y} (2)

Where:

g - Gravitational acceleration, measured in meters per square second.

\Delta y - Deformation of the spring due to gravity, measured in meters.

If we know that m=1.65\,kg, g = 9.807\,\frac{m}{s^{2}} and \Delta y = 0.260\,m, then the spring constant is:

k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}

k = 62.237\,\frac{N}{m}

If we know that A = 0.130\,m, k = 62.237\,\frac{N}{m}, m=1.65\,kg, x(t) = 0\,m and \phi = 0\,rad, then (1) is reduced into this form:

0.130\cdot \cos (6.142\cdot t)=0 (1)

And now we solve for t. Given that cosine is a periodic function, we are only interested in the least value of t such that mass reaches equilibrium position. Then:

\cos (6.142\cdot t) = 0

6.142\cdot t = \cos^{-1} 0

t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s

t \approx 0.255\,s

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

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3 years ago
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Answer:

true

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Correct answer choice is :


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