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Anit [1.1K]
3 years ago
13

a sound intensity of 70 dB is observed a distance of d from the source of sound. what sound intensity would be observed at a dis

tance 2d from the source
Physics
1 answer:
elixir [45]3 years ago
5 0

Answer:

Correct option is

B

16dB

Sound level at A is ⊥o⊥B

If the intensity of sound at A 4 cm away from the source is I

A

then.

⊥o=10log

I

o

I

A

Where I

o

=10

−12

w m

−2

⇒⊥=log

10

−12

I

A

⇒

10

−12

I

A

=10

⇒I

A

=10

−11

wm

−2

If the source (s) is emmitting the sound energy of power P

I

A

=

4×(4)

2

P

⇒P=10

−11

×64π w

⇒ Intensity of sound (I

B

) ar point B,2 m

I

B

=

4π(2)

2

P

I

B

=

16π

64π×10

−11

10 m

−2

⇒I

B

=4×10

−11

wm

−2

⇒ Sound level at 2m=10log(

I

I

B )

=10log(10 −12

4×10 −11)

=10log40

=16.02

Hence Sound level at 2m from sourceis 16 dB

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Answer:

decreases

Explanation:

Remeber:

There is always inverse relation between frequency and wavelength.

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We know that

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3 years ago
Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
shusha [124]

Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

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- Insert limit i.e a/y = 0

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Hence the Electric Field is off a point charge of magnitude 3q.

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