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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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Force F → = (−8.0 N)iˆ + (6.0 N)jˆ acts on a particle with position vector r → = (3.0 m)iˆ + (4.0 m)jˆ. What are (a) the torque
disa [49]

Answer with Explanation:

We are given that

F=-8\hat{i}+6\hat{j}

r=3\hat{i}+4\hat{j}

a.We have to find the torque on the particle about the origin.

We know that

Torque=\tau=r\times F=\begin{vmatrix}i&j&k\\3&4&0\\-8&6&0\end{vmatrix}

By using the formula

\tau=50\hat{k}

b.\mid \tau\mid =\mid F\mid \mid r\mid sin\theta

\mid F\mid=\sqrt{(-8)^2+(6)^2}=10

\mid r\mid=\sqrt{3^2+4^2}=5

\mid \tau\mid=\sqrt{(-50)^2}=50

Substitute the values then we get

50=10\times 5 sin\theta

sin\theta=\frac{50}{50}=1

sin\theta=sin90^{\circ}

Because sin90^{\circ}=1

\theta=90^{\circ}

3 0
3 years ago
This chemical equation represents a ______________ reaction.
Ede4ka [16]

Answer:

you didnt put anything

Explanation:

7 0
3 years ago
Read 2 more answers
An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average mag
Afina-wow [57]

Answer:

The induced emf in the loop is 7.35\times 10^{-4}\ V

Explanation:

Given that,

Length of the wire, L = 1.22 m

It changes its shape is changed from square to circular. Then the side of square be its circumference, 4a = L

4a = 1.22

a = 0.305 m

Area of square, A=a^2=(0.305)^2=0.0930\ m^2

Circumference of the loop,

C=2\pi r=L\\\\r=\dfrac{L}{2\pi}\\\\r=\dfrac{1.22}{2\pi}=0.194\ m

Area of circle,

A'=\pi r^2\\A'=\pi (0.194)^2\\\\A'=0.118\ m^2

The induced emf is given by :

\epsilon=\dfrac{\d\phi}{dt}\\\\\epsilon=\dfrac{\d(BA)}{dt}\\\\\epsilon=B\dfrac{A'-A}{t}\\\\\epsilon=0.125 \times \dfrac{0.118-0.0930}{4.25}\\\\\epsilon=7.35\times 10^{-4}\ V

So, the induced emf in the loop is 7.35\times 10^{-4}\ V

8 0
3 years ago
A ball rolls off a desk at a speed of 3 m/s and lands .40 seconds later. How far from the base of the desk does the ball land?
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Is the velocity constant? Is there any friction?

3 meters per second

then after 40 seconds it must 3*40 = 120 meters

120 meters or 0.12 km if you will

7 0
3 years ago
Which type of listening response includes the use of head nods, facial expressions, and short utterances such as "uh-huh" that s
BARSIC [14]
This type of listening response is called back-channel signal. This allows the speaker to know that the listener is attentive or willing to engage a conversation between them. It is shown through short utterances, facial expressions, head nods and others. 
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3 years ago
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