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Sonbull [250]
3 years ago
11

Suppose that a sound source is emitting waves uniformly in all directions. If you move to a point twice as far away from the sou

rce, the frequency of the sound will be:________.
a. one-fourth as great.
b. half as great.
c. twice as great.
d. unchanged.
Physics
1 answer:
Helen [10]3 years ago
4 0

Answer:

<em>d. unchanged.</em>

Explanation:

The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.

In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from

v = fλ

that the frequency is tied to the wave, and does not change throughout the waveform.

where v is the speed of the sound wave

f is the frequency

λ is the wavelength of the sound wave.

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1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

PE=mgh

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

KE=\frac{1}{2}mv^2

where v is the speed of the object

When the ball is sitting on the top of the building, we have

  • h=40 m, therefore the potential energy is not zero
  • v=0, since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

h=20 m

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

E=PE+KE=const.

At the top of the building,

E=PE_{top}

While halfway through the fall,

PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}

And the mechanical energy is

E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}

which means

KE_{half}=\frac{E}{2}

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

  • The height of the ball relative to the ground is now zero: h=0. This means that the potential energy of the ball is zero: PE=0
  • The kinetic  energy, instead, is not zero: in fact, the ball has gained speed during the fall, so v\neq 0, and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

4)

The potential energy of the ball as it sits on top of the building is given by

PE=mgh

where:

m = 2 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

h = 40 m is the height of the building, where the ball is located

Substituting the values, we find the potential energy of the ball at the top of the building:

PE=(2)(9.8)(40)=784 J

5)

The potential energy of the ball as it is halfway through the fall is given by

PE=mgh

where:

m = 2 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

h = 20 m is the height of the ball relative to the ground

Substituting the values, we find the potential energy of the ball halfway through the fall:

PE=(2)(9.8)(20)=392 J

6)

The kinetic energy of the ball halfway through the fall is given by

KE=\frac{1}{2}mv^2

where

m = 2 kg is the mass of the ball

v = 19.8 m/s is the speed of the ball when it is halfway through the  fall

Substituting the values into the equation, we find the kinetic energy of the ball when it is halfway through the fall:

KE=\frac{1}{2}(2)(19.8)^2=392 J

We notice that halfway through the fall, half of the initial potential energy has converted into kinetic energy.

7)

The kinetic energy of the ball just before hitting the ground is given by

KE=\frac{1}{2}mv^2

where:

m = 2 kg is the mass of the ball

v = 28 m/s is the speed of the ball just before hitting the ground

Substituting the values into the equation, we find the kinetic energy of the ball just before hitting the ground:

KE=\frac{1}{2}(2)(28)^2=784 J

We notice that when the ball is about to hit the ground, all the potential energy has converted into kinetic energy.

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A teacher told a learner to react benzene (CH) with chlorine (Cl₂) to
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<h3>How to determine the theoretical yield of chlorobenzene, C₆H₅Cl</h3>

From the question given, the following data were obtained

  • Actual yield = 100 g
  • Percentage yield = 65%
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Percentage yield = (Actual / Theoretical) × 100

65% = 100 / Theoretical

0.65 = Actual / Theoretical

Cross multiply

0.65 × Theoretical = 100

Divide both sides by 0.65

Theoretical = 100 / 0.65

Theoretical yield = 153.85 g

<h3>How to determine the mass of benzene, C₆H₆ needed</h3>

Balanced equation

C₆H₆ + Cl₂ → C₆H₅Cl + HCl

Molar mass of C₆H₆ = 78 g/mol

Mass of C₆H₆ from the balanced equation = 1 × 78 = 78 g

Molar mass of C₆H₅Cl = 112.5 g

Mass of C₆H₅Cl from the balanced equation = 1 × 112.5 = 112.5 g

SUMMARY

From the balanced equation above,

112.5 g of C₆H₅Cl were obtained from 78 g of C₆H₆

Therefore,

153.85 g of C₆H₅Cl will be produced from = (153.85 × 78) / 112.5 = 106.67 g of C₆H₆

Thus, the minimum amount of benzene, C₆H₆ needed for the reaction is 106.67 g

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