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3 years ago

14

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

2 answers:

fiasKO [112]3 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]3 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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When is the electric flux on a section of a closed surface positive?.

rodikova [14]

When the electric field pierces outward through the section.

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2 years ago

One person is in a pool and is diving to a depth of 2.3m, and another person is diving to a depth of 3m. What pressure does each

Vlad [161]

Explanation:

The static pressure is P = ρgh, where ρ is the density of the fluid and h is the depth.

For the first person:

P = (1000 kg/m³) (9.8 m/s²) (2.3 m)

P = 22,500 Pa

For the second person:

P = (1000 kg/m³) (9.8 m/s²) (3 m)

P = 29,400 Pa

6 0

3 years ago

A 225-kg object and a 525-kg object are separated by 3.80 m. (a) Find the magnitude of the net gravitational force exerted by th

pav-90 [236]

Answer: $F_{net}=3.383\times 10^{-7}\ N$

Explanation:

Given

Mass of first object $m_1=225\ kg$

Mass of second object $m_2=525\ kg$

Distance between them $d=3.8\ m$

$m_3=61\ kg$ object is placed between them

So force exerted by $m_1$ on $m_3$

$F_{13}=\frac{Gm_1m_3}{1.9^2}$

$F_{13}=\frac{6.674\times 10^{-11}(225\times 61)}{1.9^2}$

$F_{13}=2.5374141274×10^{−7}\ N$

Force exerted by $m_2\ on\ m_3$

$F_{23}=\frac{Gm_2m_3}{1.9^2}$

$F_{23}=\frac{6.674\times 10^{-11}(525\times 61)}{1.9^2}$

$F_{23}=5.920632964\times 10^{-7}\ N$

So net force on $m_3$ is

$F_{net}=F_{23}-F_{13}$

$F_{net}=5.920632964\times 10^{-7}-2.5374141274\times 10^{-7}$

$F_{net}=3.383\times 10^{-7}\ N$

i.e. net force is towards $m_2$

(b)For net force to be zero on $m_3$ , suppose

So force exerted by $m_1$ and $m_2$ must be equal

$F_{13}=F_{23}$

$\Rightarrow \frac{Gm_1m_3}{x^2}=\frac{Gm_2m_3}{(3.8-x)^2}$

$\Rightarrow \frac{m_1}{x^2}=\frac{m_2}{(3.8-x)^2}$

$\Rightarrow (\frac{3.8-x}{x})^2=\frac{m_2}{m_1}$

$\Rightarrow \frac{3.8-x}{x}=\sqrt{\frac{525}{225}}$

$\Rightarrow 3.8-x=1.52752x$

$\Rightarrow 3.8=2.52x$

$\Rightarrow x=1.507\ m$

4 0

3 years ago

An airplane is flying in a horizontal circle at a speed of 92.1 m/s. The 55.6 kg pilot does not want his radial acceleration to

Fudgin [204]

Answer:

r = 102.43 m

Explanation:

Newton's second law for this case is

F = ma

Where the acceleration is centripetal

a = v² / r

r = v² / a

They indicate that the radial acceleration is 8.45 g

r = v² / 8.45 g.

r = 92.1² / 8.45 9.8

r = 102.43 m

7 0

3 years ago

2. When a fire truck moves away from you, you hear the pitch of the siren go down. This is

Yanka [14]

the Doppler effect. (I don't know how to explain it lol)

8 0

3 years ago

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