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Allushta [10]
3 years ago
11

These graphs show measurements of the density of air as different sound waves pass a single point.

Physics
2 answers:
inessss [21]3 years ago
7 0

Answer: The graph with the highest density of air.

Explanation: The graphs are missing, but il try to explain this problem anyways.

Soundwaves are mechanical waves, that need a medium to move (commonly, the medium is the air)

As those mechanical waves affect the air, the density of air in some areas increase and in others decrease, and when that "wave of air" impacts a receptor, like your ear, you receive the information of the sound, like pitch and intensity.

As more air impacts the receptor, more "loud" is the noise, and as you may know, density = mass/volume, so as biggest is the density of air in a point, this means that we have more mass of air at that point, which is directly related to the intensity or loudness of the sound wave at that point

Ymorist [56]3 years ago
5 0

Answer: D!

It is the option with the greatest amplitude.

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Answer:

m = 3 kg

The mass m is 3 kg

Explanation:

From the equations of motion;

s = 0.5(u+v)t

Making t thr subject of formula;

t = 2s/(u+v)

t = time taken

s = distance travelled during deceleration = 62.5 m

u = initial speed = 25 m/s

v = final velocity = 0

Substituting the given values;

t = (2×62.5)/(25+0)

t = 5

Since, t = 5 the acceleration during this period is;

acceleration a = ∆v/t = (v-u)/t

a = (25)/5

a = 5 m/s^2

Force F = mass × acceleration

F = ma

Making m the subject of formula;

m = F/a

net force F = 15.0N

Substituting the values

m = 15/5

m = 3 kg

The mass m is 3 kg

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1) Arrange the following spectral regions in order of increasing energy: infrared, microwave, ultraviolet, visible.
Anton [14]

Answer:

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Option (A) is correct.

Explanation:

Given:

Arrange the following spectral regions in order of increasing energy: infrared, microwave, ultraviolet, visible.

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So ultraviolet has higher energy and microwave has lower energy.

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