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slega [8]
3 years ago
14

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

Physics
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.
5 0
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
Read 2 more answers
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