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Svetlanka [38]
3 years ago
6

By what factor does the sound intensity increase if the sound intensity level increases from 60 db to 61 db?

Physics
1 answer:
tiny-mole [99]3 years ago
6 0

Answer:

\frac{I_{2}}{I_{1}} = 10^{0.1}

Explanation:

The sound intensity level β can be mathematically defined as

\beta(dB) = 10log_{10}(\frac{I}{I_{o} } )

where I_{o} is the reference intensity and is equal to 10^{-12} W/m^{2}.

Now,

\beta_{2} - \beta_{1} = 10 log\frac{I_{2} }{I_{o}} -  10 log\frac{I_{1} }{I_{o}}\\62-60= 10log(\frac{\frac{I_{2} }{I_{o} } }{\frac{I_{1} }{I_{o} } } )\\1=10log(\frac{I_{2} }{I_{1} } )\\\\\frac{I_{2} }{I_{1} }=10^{0.1}

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A stone is dropped from the edge of a roof, and hits the ground with a velocity of -180 feet per second. How high (in feet) is t
Neko [114]

Answer:

d = 506.25 ft

Explanation:

As we know by kinematics that

v_f^2 - v_i^2 = 2 a d

here we know that initially the stone is dropped from rest from the edge of the roof

so here initial speed will be zero

now we have

v_i = 0

also the acceleration of the stone is due to gravity which is given as

g = 32 ft/s^2

now we have

v_f = 180 ft/s

so from above equation

180^2 - 0 = 2(32)d

d = 506.25 ft

6 0
3 years ago
To make a given sound seem twice as loud, how should a musician change the intensity of the sound?
Serhud [2]

Answer:

C. Quadruple the intensity

Explanation:

The intensity of the sound is proportional to square of amplitude of the sound.

I ∝ A²

\frac{I_1}{A_1^2} = \frac{I_2}{A_2^2}\\\\I_2 = \frac{I_1A_2^2}{A_1^2}

When the given sound is twice loud as the initial value, then the new amplitude is twice the former.

A₂ = 2A₁

I_2 = \frac{I_1A_2^2}{A_1^2} \\\\I_2 = \frac{I_1(2A_1)^2}{A_1^2} \\\\I_2 = \frac{4I_1A_1^2}{A_1^2}\\\\ I_2 = 4I_1

Thus, to make a given sound seem twice as loud, the musician should Quadruple the intensity

3 0
3 years ago
plz tell fast oktwo paragraphs describing the advantages and disadvantages of living in the town.plz tell​
mars1129 [50]

Answer:

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Many people who reside in small towns enjoy the closeness of knowing their neighbors and the details of neighbors' lives. If a baby is born, a child is ill or a relative dies, people in small towns often want to know about it and be there for you. This can be an advantage, especially if you are lacking a family of your own or enjoy a great deal of socializing. However, if you would rather keep the details of your life to yourself and turn your nose up at idle gossip, small town living might not be for you.

Finances

Jobs can be hard to come by in small towns. You might be required to commute out of town to work but then return to a peaceful environment at the end of a busy day. Housing might be cheaper in small towns, although commuting costs might be higher. Food and child care expenses might be reduced as people are more likely to have gardens and share produce or take turns watching each other's children.

Recreation

One of the disadvantages of a small town is not being close to amusement parks, zoos or museums. However, people in small towns tend to come together during town picnics, festivals or fairs. Depending on your preferences, small towns might provide a relaxed, slower pace when it comes to entertainment compared to the hustle and bustle of a city, where there is constant movement, commotion and entertainment at every corner.

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In a small town, class sizes are usually smaller and teachers and students can get to know each other and their fellow classmates on a deeper level. Many small towns have schools that receive national academic awards or are the highest achieving district in their states. CNN Money ranked Louisville, Colorado as the number one small town in America in 2011 and the school system is academically ranked among the top three in the Denver area. However, some of these towns noted high taxes as a trade-off for the outstanding school systems.

Explanation:

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Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 20
Genrish500 [490]

Answer:

The time it can operate between chargins in minutes is

t=102.8 minutes

Explanation:

Given: m=500kg, r=1.0m, w=200\pi rad/s

a). The rotational kinetic energy

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I=\frac{1}{2}*m*r^2

I=\frac{1}{2}*500kg*(1.0m)^2

I=250 kg*m^2

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K_R=49.348x10^6J

b). The power average 0.8kW un range time can be find

P=\frac{K_R}{t'}

Solve to t'

t=\frac{K_R}{P}

t=\frac{49.348x10^6}{0.8x10^3w}=6168.5s

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3 years ago
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