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never [62]
3 years ago
9

A sound source of frequency f moves with constant velocity (less than the speed of sound) through a medium that is at rest. A st

ationary observer hears a sound whose frequency is appreciably different from f because:________
a. interference effects set up a standing-wave pattern that alters the frequency
b. the wavelength established in the medium is not the same as it would be if the source were at rest.
c. the equation that relates velocity of propagation, frequency, and wavelength of a sound traveling through a medium does not apply in this situation
d. the sound wave travels through the medium with a velocity different from that which it would have if the source were at rest
Physics
1 answer:
ANTONII [103]3 years ago
7 0

Answer:

The correct answer is d

Explanation:

In this exercise they ask us which statement is correct, for this we plan the solution of the problem, this is a Doppler effect problem, it is the frequency change due to the relative speed between the emitter and the receiver of sound.

The expression for the Doppler effect of a moving source is  

f ’= (v / (v- + v_s) f  

From this expression we see that if the speed the sound source is different from zero feels a change in the  frequency.

The correct answer is d

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60

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A fire breaks out and increases the Kelvin temperature of a cylinder of compressed gas by a factor of 2.4. What is the final pre
kherson [118]

Answer:

P(final) is 2.4 times P(initial).

Explanation:

Here we can assume that the cylinder did not break and it's volume and number of moles of gas present in the cylinder remains constant.

Given the temperature increases by a factor of 2.4. Let us assume that the initial temperature be T_{1} and the final temperature be T_{2}.

Given that T_{2}=2.4\times T_{1}

Now we know the ideal gas equation is PV=nRT

here V=constant , n=constant , R=gas constant(which is constant).

\frac{P}{T}=constant

\frac{P_{1} }{T_{1}}=\frac{P_{2} }{T_{2} }

P_{2}=(\frac{T_{2} }{T_{1} } )P_{1}

P_{2}=(\frac{2.4T_{1} }{T_{1} } )P_{1}

P_{2}=2.4\times P_{1}

7 0
3 years ago
Which is required for
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Answer:

You need (B) Two substances in direct contact with one another to have conduction.

I hope this helped at all.

5 0
3 years ago
A combination of two identical resistors connected in series has an equivalent resistance of 12. ohms. What is the equivalent re
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Answer:

R1 + R2 = R = 12 for resistors in series - so R1 = R2 if they are identical

2 R1 = 12         and R1 = R2 = 6 ohms

1 / R = 1 / R1 + 1 / R2     for resistors in parallel

R = R1 * R2 / (R1 + R2) = 6 * 6 / (6 + 6) = 3

The equivalent resistance would be 3 ohms if connected in parallel

7 0
2 years ago
Most automobiles have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made o
Colt1911 [192]

Answer:

There is a loss of fluid in the  container of 0.475L

Explanation:

To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.

The formula that describes this thermal expansion process is given by:

\Delta V = \beta V_0 \Delta T

Where,

\Delta V =Change in volume

V_0 =Initial Volume

\Delta T = Change in temperature

\beta = coefficient of volume expansion (Coefficient of copper and of the liquid for this case)

There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,

\Delta V_T = \Delta V_l - \Delta V_c

Where,

\Delta V_l= Change in the volume of liquid

\Delta V_c= Change in the volume of copper

Then replacing with the previous equation we have:

\Delta V = \beta_l V_0 \Delta T- \beta_c V_0 \Delta T

\Delta V = (\beta_l-\beta_c)V_0\Delta T

Our values are given as,

Thermal expansion coefficient for copper and the liquid to 20°C is

\beta_c = 51*10^{-6}/\°C

\beta_l = 400*10^{-6}/\°C

V_0 = 16L

\Delta T = (95\°C-10\°C)

Replacing we have that,

\Delta V = (\beta_l-\beta_c)V_0\Delta T

\Delta V = (400*10^{-6}/\°C-51*10^{-6}/\°C)(16L)(95\°C-10\°C)

\Delta V = 0.475L

Therefore there is a loss of fluid in the container of 0.475L

6 0
3 years ago
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