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2 years ago

How do we use sounds in ways other than just listening?

Physics

2 answers:

Whitepunk [10]2 years ago

7 0

Answer:

talking and we can make our own sound

Explanation:

Leviafan [203]2 years ago

6 0

<h2>How do we use sounds in ways other than just listening?</h2>

Sound could be an element for detecting people or entities, This is very useful for blind people. Sounds could also heal your chakra or make you feel like dancing.

<em>Hope I helped. :)</em>

<em />

<em>- Valenteer</em>

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Which vector has an x-component with a length of 3?<br> А. С.<br> B. d<br> C. a<br> D. b

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Answer:

B.d

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3 years ago

Read 2 more answers

2. Name two specific contributions Bohr made to our understanding of atomic

pickupchik [31]

Answer:

Bohr's greatest contribution to modern physics was the atomic model. ... Bohr was the first to discover that electrons travel in separate orbits around the nucleus and that the number of electrons in the outer orbit determines the properties of an element

Explanation:

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3 years ago

Two of the three tuning forks have known frequencies. When a 610 Hz fork and the unknown fork are struck together, four beats pe

RUDIKE [14]

-- In combination with 610 Hz, the beat frequency is 4 Hz.

So the unknown frequency is either (610+4) = 614 Hz
or else (610-4) = 606 Hz.

In combination with 605 Hz, the beat frequency will be
either (614-605) = 9 Hz or else (606-605) = 1 Hz.

-- In actuality, when combined with the 605 Hz, the beat
frequency is too high to count accurately. That must be
the 9 Hz rather than the 1 Hz.

So the unknown is (605+9) = 614 Hz.

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3 years ago

In your own words, answer the following 3 questions about an object experiencing red shift? 1. Describe the motion of an object

german

1. When an object is moving away from us, the light from the object is known as redshift, and when an object is moving towards us, the light from the object is known as blueshift.

2. A wavelength increases in size, and its frequency, and energy decrease.

3. The frequency of a wave increases, and its wavelength decreases.

Redshift is an important term for astronomers. The term can be taken literally. The wavelengths of light are stretched and perceived as shifting toward the red portion of the spectrum. The same thing happens to sound waves when the source moves relative to the observer.

As the wave frequency decreases, the wavelength increases as long as the wave velocity remains constant. If the wave speed stays the same as the frequency decreases, it means that fewer wave peaks or troughs pass through a given point in a given time period. The number of complete wavelengths in a given unit of time is called frequency. Frequency and energy decrease with increasing wavelength.

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1 year ago

provides some pertinent background for this problem. A pendulum is constructed from a thin, rigid, and uniform rod with a small

gavmur [86]

Answer:

the period of the physical pendulum is 0.498 s

Explanation:

Given the data in the question;

$T_{simple$ = 0.61 s

we know that, the relationship between T and angular frequency is;

T = 2π/ω ---------- let this be equation 1

Also, the angular frequency of physical pendulum is;

ω = √(mgL / $I$ ) ------ let this equation 2

where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and $I$ is moment of inertia of rod.

Now, moment of inertia of thin uniform rod D is;

$I$ = $\frac{1}{3}$ mD²

since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.

we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/ $I$ ) OR T = 2π√( $I$ /mgL)

so we can use $I$ = $\frac{1}{3}$ mD² for moment of inertia of the rod

Since center of gravity of the uniform rod lies at the center of rod

so that L = $\frac{1}{2}$ D.

now, substituting these equations, the period becomes;

T = 2π/√( $I$ /mgL) OR T = $2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } }$ OR T = 2π√(2D/3g ) ----- equation 3

length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

ω $_{simple$ = 2π/ $T_{simple$ OR ω $_{simple$ = √(g/D) OR ω $_{simple$ = 2π√( D/g )

so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) × $T_{simple$

we substitute in value of $T_{simple$

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s

8 0

3 years ago