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alex41 [277]
3 years ago
12

Two fire trucks have sirens that emit waves of the same frequency. As the fire trucks approach a person, the person hears a high

er frequency from truck X than from truck Y.
1. Which of the following statements about truck X can be correctly inferred from this information?
A) It is traveling faster than truck Y.
B) It is closer to the person than truck Y.
C) It is speeding up, and truck Y is slowing down.
Physics
1 answer:
Mumz [18]3 years ago
5 0

Answer:

Only option A is correct

Explanation:

From the concept of Doppler effect, only speed matters. Thus, the faster a vehicle is moving, the closer together the sound waves get compressed and the higher the frequency. For example, for a very fast vehicle traveling at the speed of sound; the compressions are all right on top of each other. Thus, faster speed means closer compressions and higher frequencies. Hence, option only option A must be true because X is a higher frequency and so it must be going faster. The distance to the person will affect the volume but will not the pitch so Option B is not correct. Option C too is not correct because It doesn’t matter whether you are speeding up or slowing down, it only matters who is going faster. For example, from option c concept, if truck X was going 10 m/h and speeding up while truck Y was going 50 mph and slowing down, it would not meet the requirement that X has a higher frequency because only actual speed matters, not what is happening to that speed. Thus only option A is the correct answer

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An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99
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<u>Answer:</u> The average atomic mass of the given element is 20.169 amu.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i     .....(1)

We are given:

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Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

  • For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

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Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882  

Putting values in equation 1, we get:

\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]

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