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3 years ago

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

2 answers:

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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A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra

lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

$m_{1}=5.8 kg$ is the mass of the bowling ball

$V_{1}=4.34 m/s$ is the velocity of the bowling ball

$m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg$ is the mass of the ping-pong ball

$V_{2}$ is the velocity of the ping-pong ball

Now, the momentum $p_{1}$ of the bowling ball is:

$p_{1}=m_{1}V_{1}$ (1)

$p_{1}=(5.8 kg)(4.34 m/s)$

$p_{1}=25.172 kg m/s$ (2)

And the momentum $p_{2}$ of the ping-pong ball is:

$p_{2}=m_{2}V_{2}$ (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

$p_{1}=p_{2}$ (4)

$m_{1}V_{1}=m_{2}V_{2}$ (5)

Isolating $V_{2}$ :

$V_{2}=\frac{m_{1}V_{1}}{m_{2}}$ (6)

$V_{2}=\frac{25.172 kg m/s}{0.002214 kg}$ (7)

Finally:

$V_{2}=11369.46 m/s$

6 0

3 years ago

For two objects with masses 100 kg and 500 kg to be 20 meters apart , find the gravity between them

Vikki [24]

Answer:

8,3375*10^-9 N

Explanation:

4 0

3 years ago

Scientists and astronomers have found that in galaxies with central black holes, there are also large star formations near those

sweet-ann [11.9K]

Answer:

Explanation:

nothing to do with black holes creating star or related

5 0

3 years ago

Read 2 more answers

Based on the information in the table, which two elements are most likely in the same group, and why?

velikii [3]

Answer: bismuth and nitrogen, because they have the same number of valence electrons

Explanation:

Elements are distributed in groups and periods in a periodic table.

Elements that belong to same groups will show similar chemical properties because they have same number of valence electrons.

The number of valence electrons in Bismuth and nitrogen are 5 and thus thus they will show similar chemical properties and thus belong to the same group.

The atomic masses of elements in a group will differ drastically.

The group number has got nothing to be the isolation year.

Thus bismuth and nitrogen belong to same group because they have the same number of valence electrons

3 0

3 years ago

Read 2 more answers

1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

$\tau=Fd$