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noname [10]
4 years ago
10

A constant tone is being applied to a speaker. The voltage across the speaker is 5 volts. The voltage across the speaker is incr

eased to 15 volts in order to increase the sound level. What is the decibel gain, rounded to the nearest decibel?
Physics
2 answers:
nexus9112 [7]4 years ago
8 0

Answer:

10 do gain

Explanation:

The decibel formula for voltage: db = 20*log(%28%28E1%29%29%2F%28%28E2%29%29)

:

E1 = 15 v; E2 = 5v

:

db = 20*log(%2815%29%29%2F%285%29%29)

:

db = 20*log(3)

:

db = 20 * .47712

:

9.54 db ~ 10 db gain

lubasha [3.4K]4 years ago
5 0

Answer:

decibel gain is 10 db

Explanation:

given data

voltage v1 =  5 volts

voltage v2 = 15 volts

to find out

decibel gain

solution

we have given that 5 volt voltage that is increase to 15 volts

so

we know here decibel gain formula that is

decibel gain = 20 log(v2/v1)   ....................1

put here value of v1 and v2  in equation 1

decibel gain = 20 log(15/5)

decibel gain = 20 log3

decibel gain = 20 × 0.4771

decibel gain = 9.54 db = approx 10 db

so decibel gain is 10 db

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5 0
4 years ago
A wheel of radius 30 cm is rotating at a rate of 2.0 revolutions every 0.080 s. (A) through what angle, in radians, does the whe
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2revs in 0.08s
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6 0
3 years ago
Long sand ridges oriented at right angles to the wind are called _____ dunes
S_A_V [24]

The answer is:

Transverse dune

The explanation:

Transverse dune : is abundant barchan dunes It  may merge into barchanoid ridges, which then grade into linear .

The transverse dunes is called that because they lie transverse, or across, the wind direction, with the wind blowing perpendicular to the ridge crest.

It is large, very asymmetrical, elongated dune lying at right angles 90° to the prevailing wind direction.

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3 0
3 years ago
Read 2 more answers
A car accelerates uniformly from rest to a speed of 7.6 m/s in 4.6 s. How far does the car travel in that time?
uysha [10]

Answer:

The car will travel a distance of 17.45 meters.

Explanation:

Given:

Initial velocity (V_i) = 0

Final velocity (V_f) = 7.6 m/s

Time taken = 4.6 s

Acceleration = (Final velocity - Initial Velocity )/time

a=\frac{(V_f-Vi)}{t}= \frac{7.6-0}{4.6}=1.65\ ms^-2

We have to calculate total distance traveled by the car.

Let the distance traveled be 'd'

Equation of motion:

d=V_i(t)+\frac{at^2}{2}

Plugging the values.

⇒d=V_i(t)+\frac{at^2}{2}

⇒d=0+\frac{1.65*(4.6)^2}{2}

⇒d=17.45\ m

The car will travel a distance of 17.45 meters for the above case.

4 0
3 years ago
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