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Korvikt [17]
3 years ago
8

By what factor must the amplitude of a sound wave be increased in order to increase the intensity by a factor of 9?a. 9 b. 2 c.

4 d. 3
Physics
2 answers:
jolli1 [7]3 years ago
6 0

Answer:

3 times option (d) is correct

Explanation:

Initial intensity = Io

final intensity = 9 Io

initial amplitude = A

Final amplitude = ?

let the final amplitude is A'.

Intensity of a wave is directly proportional to the square of the amplitude of the wave.

I ∝ A²

So, Io ∝ A²   ... (1)

9Io ∝ A'²     .... (2)

Divide (2) by (1)

9 = A'²/A²

A' = 3 A

So, the amplitude becomes 3 times.

Marta_Voda [28]3 years ago
4 0

Answer:

option D

Explanation:

given,

intensity\ \alpha \ (Amplitude)^2

increase the intensity by factor of 9

    I₁ = I₀

    I₂ = 9 I₀

now,

\dfrac{I_1}{I_2}=\dfrac{A_1^2}{A_2^2}

\dfrac{I_0}{9I_0}=(\dfrac{A_1}{A_2})^2

(\dfrac{A_1}{A_2})^2=\dfrac{1}{9}

\dfrac{A_1}{A_2}=\dfrac{1}{3}

      A₂ = 3 A₁

hence, amplitude increase with the factor of 3

so, the correct answer is option D

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Is pressure water a acid , basic or natural substance
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7 0
4 years ago
The crowd at a concert lifts a singer to a height of 2.2 m. The crowd uses a total of force of 600 N. How much work has the crow
nordsb [41]

Answer:

<h2>1320 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 600 × 2.2

We have the final answer as

<h3>1320 J</h3>

Hope this helps you

7 0
3 years ago
A ball is thrown horizontally from the top of a building 54 m high. The ball strikes the ground at a point 35 m horizontally awa
Ivanshal [37]

Answer:

V=34.2 m/s

Explanation:

Given that

Height , h= 54 m

Horizontal distance , x = 35 m

Given that , the ball is thrown horizontally , therefore the initial vertical velocity will be zero.

In vertical direction :

We know that

V^2_y=U^2_y+2 g h

Now by putting the values in the above equation we got

V^2_y=U^2_y+2 g h

V^2_y=0^2+2\times 9.81 \times 54

Assume g= 9.81 m/s^2

Thus

V^2_y=1059.48

V_y=\sqrt{1059.48}\ m/s

V_y=32.54 m/s

We also know that

V_y=U_y+ g\times t

32.54=9.81\times t

t=\dfrac{32.54}{9.81}=3.31\ s

In horizontal direction :

x=U_x\times t

U_x=\dfrac{35}{3.31}=10.54\ m/s

Thus the resultant velocity

V=\sqrt{V^2_y+U^2_x}

V=\sqrt{32.54^2+10.54^2}=34.2\ m/s

V=34.2 m/s

Therefore the velocity will be 34.2 m/s.

5 0
4 years ago
A 2500‐kg vehicle traveling at 25 m/s can be stopped by gently applying the breaks for 20 seconds. What is the average force sup
katen-ka-za [31]
Momentum = (mass) x (speed)

Change in momentum = (force) x (time)

The initial momentum is (mass) x (speed) = 2500x 25 = 62,500 kg-m/s.

Since you want to <u>stop</u> the vehicle, that number is also the required <em>change</em>
in momentum ... you want the vehicle to wind up with zero momentum.

62,500 = (force) x (time) = 20 x force

Divide each side by 20 :

force = 62,500 / 20 = <em>3,125 newtons </em>
3 0
3 years ago
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