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3 years ago

Explain what happens to the pitch of a cell phone ringing when the amplitude of a sound wave increases.

Physics

2 answers:

lys-0071 [83]3 years ago

4 0

the pitch of the cell phone ringing will remains same when the amplitude of a sound wave increases.

The pitch of a sound wave depends on its frequency. more frequency means higher pitch and less frequency means lower pitch. The amplitude of wave is the maximum displacement of the particles of the medium from their mean positions. Increase in amplitude will increase the loudness and not the pitch. So the pitch remains same as the amplitude of sound wave is increased.

rewona [7]3 years ago

3 0

As the amplitude of a sound wave increases the pitch of the ringing would be much higher (like if you were to inhale helium.. just with a phone)

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Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

The radial acceleration is $a_c = 0.9574 m/s^2$

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

3 0

3 years ago

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Darya [45]

The answer to that is 3

7 0

3 years ago

Can a object have have zero velocity and nonzero acceleration?

vladimir1956 [14]

Answer:

yes

Explanation:

3 0

3 years ago

The tops of the towers of the Golden Gate Bridge in San franciscoo , are 227 m above the water. Suppose a worker drops a 655 g w

yaroslaw [1]

Answer:

Kinetic energy is 1425.11 J.

Explanation:

Given:

Mass of the wrench is, $m=655\ g=0.655\ kg$

Height of fall is, $h=227\ m$

Force of resistance is, $F=0.141\ N$

Now, the total energy at the top is equal to the potential energy of the wrench at the top since the kinetic energy at the top is 0.

Now, potential energy at the top is given as:

$U=mgh\\U=0.655\times 9.8\times 227\\U=1457.113\ J$

Now, the potential energy at the top is converted to kinetic energy at the bottom and some energy is wasted in overcoming the resistance force by air.

Potential Energy = Kinetic energy + Energy to overcome resistance.

⇒ Kinetic energy = Potential Energy - Energy to overcome resistance.

Energy to overcome resistance force is the work done by the wrench against the resistance force and is given as:

$W_{res}=Fh=0.141\times 227=32.007\ J$

Therefore, Kinetic energy at the bottom is given as:

$K=U-W_{res}\\\\K=1457.113\ J - 32.007\ J\\\\K=1425.106\approx1425.11\ J$

Hence, the kinetic energy of the wrench be when it hits the water is 1425.11 J.

7 0

3 years ago

A material used for electrical wiring must have

lapo4ka [179]

A certain amount of Conductivity

5 0

3 years ago