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Sindrei [870]
3 years ago
5

The volume control on a surround-sound amplifier is adjusted so the sound intensity level at the listening position increases fr

om 23 to 61 dB. What is the ratio of the final sound intensity to the original sound intensity?
Physics
1 answer:
amid [387]3 years ago
5 0

Answer:

Ratio will be \frac{I_2}{I_1}=10^{3.8}=6309.5

Explanation:

We have given that intensity level is increases from 23 dB to 61 dB

We know that level of intensity is given by

i=10log\frac{I}{I_0}

So 23=10log\frac{I_1}{I_0}------eqn 1

And 61=10log\frac{I_2}{I_0}------eqn 2

Subtracting eqn 1 from eqn 2

61-23=10log\frac{I_2}{I_1}

3.8=log\frac{I_2}{I_1}

\frac{I_2}{I_1}=10^{3.8}=6309.5

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Please do number 25! Explain how you got your answer with detail to get Brainliest! Thank you!
Ad libitum [116K]
John weighs 200 pounds.
In order to lift himself up to a higher place, he has to exert force of 200 lbs.

The stairs to the balcony are 20-ft high.
In order to lift himself to the balcony, John has to do
(20 ft) x (200 pounds)  =  4,000 foot-pounds of work.

If he does it in 6.2 seconds, his RATE of doing work is
(4,000 foot-pounds) / (6.2 seconds)  =  645.2 foot-pounds per second.

The rate of doing work is called "power".

(If we were working in the metric system (with SI units),
the force would be in "newtons", the distance would be in "meters",
1 newton-meter of work would be 1 "joule" of work, and
1 joule of work per second would be 1 "watt".
Too bad we're not working with metric units.)

So back to our problem.

John has to do 4,000 foot-pounds of work to lift himself up to the balcony,
and he's able to do it at the rate of 645.2 foot-pounds per second.

Well, 550 foot-pounds per second is called 1 "horsepower".

So as John runs up the steps to the balcony, he's doing the work
at the rate of

           (645.2 foot-pounds/second) / (550 ft-lbs/sec per HP)

=  1.173 Horsepower.  GO JOHN !

(I'll betcha he needs a shower after he does THAT 3 times.)
_______________________________________________

Oh my gosh !  Look at #26 !  There are the metric units I was talking about.

Do you need #26 ?

I'll give you the answers, but I won't go through the explanation,
because I'm doing all this for only 5 points.

a).  5
b).  750 Joules
c).  800 Joules
d).  93.75%

You're welcome.

And #27 is 0.667 m/s .
7 0
3 years ago
If the mass of the object is doubled and the speed is halved then kinetic energy will change by a factor of:
madam [21]

Answer:

kinetic energy will change by a factor of 1/2

Option C) 1/2 is the correct answer

Explanation:

Given the data in the question;

we know that;

Kinetic energy = 1/2.mv²

given that mass of the object is doubled; m1 = 2m

speed is halved; v1 = V/2

Now, New kinetic energy will be; 1/2.m1v1²

we substitute

Kinetic Energy = 1/2 × 2m × (v/2)²

Kinetic Energy = 1/2 × 2m × (v²/4)

Kinetic Energy = 1/2 × m × (v²/2)

Kinetic Energy = 1/2 [ 1/2mv² ]

Kinetic Energy = 1/2 [ KE ]

Therefore; kinetic energy will change by a factor of 1/2

Option C) 1/2 is the correct answer

3 0
2 years ago
A fox fleeing from a hunter encounters a 0.700 m tall fence and attempts to jump it. The fox jumps with an initial velocity of 7.
iris [78.8K]

Answer:0.61 m

Explanation:

Given

Initial velocity (u)=7.40 m/s

launch angle \theta =45^{\circ}

height of fence=0.7 m

Trajectory of Projectile is given by

y=x\tan \theta -\frac{gx^2}{2u^2\cos^2\theta }

For x=2.1 m  

y=2.1\tan (45)-\frac{9.8\times 2.1^2}{2\times 7.4^2\times (\cos 45)^2}

y=2.1-0.789

y=1.31 m

At x=2.1 m Y=1.31

therefore Fox clear fence by a margin =1.31-0.7=0.61 m

7 0
3 years ago
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