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neonofarm [45]
3 years ago
15

Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the buzz

er over the side while the balloon is rising at constant speed, what can you conclude about the sound you hear as the buzzer falls toward the ground?
(A) The frequency and intensity increase
(B) The frequency decrease and intensity increase
(C) The frequency decrease and intensity decrease
(D) The frequency remains the same, but the intensity decreases.
Physics
1 answer:
amid [387]3 years ago
7 0

Answer:

(C) The frequency decrease and intensity decrease

Explanation:

The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source, or the wave source is moving relative to the observer, or both.

if the observer and the source move away from each other as is the case for this problem, the wavelength heard by the observer is bigger.

The frequency is the inverse from the wavelength, so the frequency heard will increase.

The sound intensity depends inversely on the area in which the sound propagates. When the buzzer is close, the area is from a small sphere, but as the buzzer moves further away, the wave area will be from a larger sphere and therefore the intensity will decrease.

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It traveled 4 centimeters.

Explanation:

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In this graph we have the following:

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3 years ago
Sound waves can be modeled by the equation of the form y=20sin(3t+theta). Determine what type of interference results when sound
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2 years ago
A real power supply can be modeled as an ideal EMF of 60 Volts in series with an internal resistance. The voltage across the ter
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Answer:

5 ohms

Explanation:

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Now, we know that, the voltage drop in the battery is given as:

V_d=Ir

Therefore, the voltage across the terminals of the battery is given as:

V= E-V_d\\\\V=E-Ir

Now, rewriting in terms of 'r', we get:

Ir=E-V\\\\r=\frac{E-V}{I}

Plug in the given values and solve for 'r'. This gives,

r=\frac{60-40}{4}\\\\r=\frac{20}{4}\\\\r=5\ ohms

Therefore, the internal resistance of the battery is 5 ohms.

5 0
3 years ago
What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?
Daniel [21]

To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

F=\frac{kq_{1}q_{2}  }{r^{2} }

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Given  q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C and  r=8 cm=8\times10^{-2} m.

Substituting these values in above formula we get,

F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9}  N\\\\F=8.8\times10^{-4}N

Thus, the repulsive force between two pith balls is 8.8\times10^{-4}N.

7 0
3 years ago
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