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Vaselesa [24]
4 years ago
6

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 remains the same, but the intensity decreases
B. the frequency decreases and the intensity increases
C. the frequency and intensity increase
D. the frequency decreases and the intensity decreases
Physics
1 answer:
Olin [163]4 years ago
3 0

Answer:

The correct answer is option 'd': The frequency decreases and the intensity of the sound decreases.

Explanation:

1) <u>Effect on Frequency </u>

According to Doppler's effect of sound we have

for a source of sound moving away from the observer the relation between the observed and the original frequency is given by

f_{app}=\frac{c-v_{rec}}{c+v_{s}}\times f_{original}

where

c = speed of sound in air

v_{rec} is the velocity of observer of sound

v_{s} is the velocity of source of sound

f_{o} is the original frequency of sound

As we see the ratio is less than 1 thus the frequency of sound that the observer receives is less than that of source.

2) <u>Effect on Intensity:</u>

At a distance 'r' from source emitting a wave of Power 'P' is given by

I=\frac{P}{4\pi r^{2}}

As we see on increasing 'r' intensity of sound decreases.

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Answer:

4600 ft.Ibs

Explanation:

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8 0
4 years ago
A sample of gold has a volume of 2 cm3 and a mass of 38.6 grams. What would be the density, and three other properties of the sa
lukranit [14]

The density of the gold is calculated to be "19,300 kg/m³".

<u>Explanation:</u>

Given:

Volume = 2cm³

Mass = 38.6 grams.

To Find:

Density of the gold = ?

Solution:

Density is obtained by dividing mass of the sample by its volume and it is given in the units of kg/m³.

Mass in grams is converted into kg as,

1 g = 0.001 kg

38. 6 g = \frac{38.6}{1000} = 0.0386 kg

Now we have to convert cm³ to m³ as,

1 cm³ = 10⁻⁶ m³

2 cm³ = 2 × 10⁻⁶ m³

So Density = \frac{0.0386 k g}{2 \times 10^{-6} m^{3}}=19,300 \mathrm{kg} / \mathrm{m}^{3}

Physical properties of Gold:

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3 0
3 years ago
Light rays in a material with index of refrection 1.29 1.29 can undergo total internal reflection when they strike the interface
deff fn [24]

Answer:

The second material's index of refraction is 1.17.

Explanation:

Given that,

Refractive index of the material, n = 1.29

Critical angle is 65.9 degrees.

We need to find the second material's index of refraction. We know that at critical angle of incidence, angle of refraction is equal to 90 degrees. Using Snell's law as:

n_1\sin \theta_c=n_2\sin (90)\\\\n_2=n_1\sin \theta_c\\\\n_2=1.29\times \sin (65.9)\\\\n_2=1.17

So, the second material's index of refraction is 1.17.

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devlian [24]

The answer is "the product of the object's moment of inertia and the object's angular velocity.

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How do the interactions between the two atoms change when the atoms get closer than the balanced point?
ValentinkaMS [17]

Answer:

The strong attraction of each shared electron to both nuclei stabilizes the system, and the potential energy decreases as the bond distance decreases. If the atoms continue to approach each other, the positive charges in the two nuclei begin to repel each other, and the potential energy increases.

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