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Helen [10]
3 years ago
7

A small omnidirectional stereo speaker produces waves in all directions that have an intensity of 8.00 at a distance of 4.00 fro

m the speaker.
At what rate does this speaker produce energy?

What is the intensity of this sound 9.50 from the speaker?

What is the total amount of energy received each second by the walls (including windows and doors) of the room in which this speaker is located?
Physics
1 answer:
slavikrds [6]3 years ago
5 0

Answer:

A. We have that radius r = 4.00m intensity I = 8.00 W/m^

total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W

b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2

c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J

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It is a privilege, because not everyone might have resources for training, and therefore no driving skills.
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vodomira [7]

Answer:

the true about n i b i n is they are Commissioner

8 0
3 years ago
A softball of mass 0.220 kg that is moving with a speed of 5.5 m/s (in the positive direction) collides head-on and elastically
Elanso [62]

Answer:

The velocity and mass of the target ball are 1.6 m/s and 1.29 kg.

Explanation:

Given that,

Mass of softball = 0.220 kg

Speed = 5.5 m/s

(a). We need to calculate the velocity of the target ball

Using conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

0.220\times5.5+m_{2}\times0=0.220\times(-3.9)+m_{2}v_{2}

1.21=-0.858+m_{2}v_{2}

m_{2}v_{2}=2.068....(I)

The velocity approach is equal to the separation of velocity

u_{1}-u_{2}=v_{2}-v_{1}

5.5-0=v_{2}-(-3.9)

v_{2}=1.6\ m/s

(b). We need to calculate the mass of the target ball

Now, Put the value of v₂ in equation (I)

m_{2}\times1.6=2.068

m_{2}=\dfrac{2.068}{1.6}

m_{2}=1.29\ kg

Hence, The velocity and mass of the target ball are 1.6 m/s and 1.29 kg.

3 0
3 years ago
a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
3 years ago
While in a stream 39 cm deep, they look down into the water and see a craw fish at the bottom. How deep does the stream appear t
Helga [31]

Answer:

The  depth of stream to the student is  d_1  =  0.2932 \  m

Explanation:

From the question we are told that

   The actual  depth of the stream is d =  39 \ cm  =  0.39 \ m  

    The  refractive index of the water is  n =  1.33

Generally the apparent depth of the stream is mathematically represented as

         d_1  =  \frac{d}{1.33}

substituting values

        d_1  =  \frac{ 0.39}{1.33}

        d_1  =  0.2932 \  m

7 0
4 years ago
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