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2 years ago

14

What properties of a sound wave determine volume, pitch, and timbre?

2 answers:

fiasKO [112]2 years ago

8 0

Sound waves are changes in pressure generated by vibrating molecules. The physical characteristics of sound waves influence the three psychological features of sound: loudness, pitch, and timbre. Loudness depends on the amplitude,or height, of sound waves. The greater the amplitude, the louder the sound perceived.

IgorLugansk [536]2 years ago

6 0

Amplitude is the biggest answer Because it dictates how loud or soft the sound is for the volume. For pitch it is gonna he loudness and time that are the properties because that addresses the frequency. For timbre it is the wave form, the bigger the wave the bigger the overtones, or harmonics are present.

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A force of 20.0 N is applied to a 3.00 kg object for 4.00 seconds. Calculate the impulse experienced by the object.

GenaCL600 [577]

Answer:

Impulse = 80Ns

Explanation:

Given the following data;

Mass = 3kg

Force = 20N

Time = 4 seconds

To find the impulse experienced by the object;

Impulse = force * time

Impulse = 20*4

Impulse = 80Ns

Therefore, the impulse experienced by the object is 80 Newton-seconds.

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2 years ago

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stellarik [79]

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2 years ago

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A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15Â° below the

Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, $(\theta)$ = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ $F_x=Fcos(\theta)$ and ⇒ $F_y=Fsin (\theta)$

⇒ $F_x=15cos(15)$ ⇒ $F_y=15sin (15)$

⇒ Applying concept of forces.

⇒ $\sum F_x=F_n_e_t =F-f$

⇒ $F_n_e_t =F-f$

⇒ $ma =F-f$ <em> ...Newtons second law Fnet = ma</em>

⇒ $a =\frac{F-f}{m}$

⇒ Plugging the values.

⇒ $a =\frac{15cos(15)-0}{40}$ <em>...f is the friction which is zero here.</em>

⇒ $a =\frac{14.48}{40}$

⇒ $a=0.362\ ms^-^2$

Magnitude of the acceleration of the crate is 0.362 m/s^2.

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2 years ago

If a green light shines on a red toy, the toy will look:

disa [49]

Not 100% sire but I think it'd be Yellow since we see red and green light together as Yellow

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2 years ago

A car starts from rest, speeds up with constant acceleration, and travels 400 meters in 10 seconds. What was the acceleration of

PIT_PIT [208]

Answer:

Acceleration of the car will be $a=8m/sec^2$

Explanation:

We have given that car starts from rest so initial velocity of the car u = 0 m/sec

And car traveled 400 m in 10 sec

So distance traveled by car s = 400 m

Time taken to compete this distance t = 10 sec

We have to find the acceleration of the car

From second equation of motion we know that $s=ut+\frac{1}{2}at^2$

So $400=0\times 10+\frac{1}{2}\times a\times 10^2$

$a=8m/sec^2$

So acceleration of the car will be $a=8m/sec^2$

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2 years ago