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guapka [62]
3 years ago
9

A source of sound is kept in a jar in a vacuum. Air is slowly introduced in to the jar. What happens to the sound coming out of

the jar? A The loudness of the sound increases gradually. B The loudness of the source decreases gradually. C The sound remains unchanged. D The pitch of the sound increases gradually. E
The pitch of the sound decreases gradually.
Physics
1 answer:
timama [110]3 years ago
3 0
<span>The loudness of the sound increases gradually as the air is slowly introduced in to the jar. This is because sound needs a physical medium and in a vacuum there is none. The air provides that medium and as it is introduced, the transfer of sound energy increases</span>
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C. volume of water and temperature

Explanation:

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A lightning bolt with 13 kA strikes an object for 14 μ s. How much charge is deposited on the object?
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Answer:

0.182C

Explanation:

Using Q= It

= 13x10^3 . 14x10^-6

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3 years ago
Ivan drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 7 hours. When Ivan drove h
lesya692 [45]

Answer:252 miles

Explanation:

Given

During his way to mountain it took 7 hr to drive

and during his return trip it took 4 hr to return

Let x be the distance between home and mountain

average speed for return is  27 miles per hour faster than his former trip

let v be the speed on his way to mountain thus v+27 is his return speed

thus 7=\frac{x}{v}----1

for return trip

4=\frac{x}{v+27}-----2

divide  1 & 2

\frac{7}{4}=\frac{x\cdot (v+27)}{v\cdot x}

7v=4v+4\cdot 27

3v=4\cdot 27

v=36 mph

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7 0
3 years ago
A body falls from the top of the tower and during the last second of its fall it fall through 23mvfind height of tower.
DanielleElmas [232]

Answer:

39.7 m

Explanation:

First, we conside only the last second of fall of the body. We can apply the following suvat equation:

s=ut+\frac{1}{2}at^2

where, taking downward as positive direction:

s = 23 m is the displacement of the body

t = 1 s is the time interval considered

a=g=9.8 m/s^2 is the acceleration

u is the velocity of the body at the beginning of that second

Solving for u, we find:

ut=s-\frac{1}{2}at^2\\u=\frac{s}{t}-\frac{1}{2}at=\frac{23}{1}-\frac{1}{2}(9.8)(1)=18.1 m/s

Now we can call this velocity that we found v,

v = 18 m/s

And we can now consider the first part of the fall, where we can apply the following suvat equation:

v^2-u^2 = 2as'

where

v = 18 m/s

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s' is the displacement of the body before the last second

Solving for s',

s'=\frac{v^2-u^2}{2a}=\frac{18.1^2-0}{2(9.8)}=16.7 m

Therefore, the total heigth of the building is the sum of s and s':

h = s + s' = 23 m + 16.7 m = 39.7 m

7 0
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nordsb [41]
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5 0
3 years ago
Read 2 more answers
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