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yawa3891 [41]
3 years ago
13

A sound source is located somewhere along the x-axis. Experiments show that the same wave front simultaneously reaches listeners

at x = -7.0 m and x = +3.0 m.part a) What is the x-coordinate of the source?part b) A third listener is positioned along the positive y-axis. What is her y-coordinate if the same wave front reaches her at the same instant it does the first two listeners?
Physics
1 answer:
galina1969 [7]3 years ago
7 0

Answer:

Explanation:

As the source is situated on x - axis , it must be situated in between the two listeners .

So the x coordinate of source is

(-7 + 3 )/2

= - 2 m

The equation of the wave- front will be that o a circle having centre at (-2,0)

and radius = distance between -2 and 3 , that is 5 m

equation of circle

=( x+2 )² + y² = 25

It cuts y axis when x = 0

Putting x = 0

4 + y² = 25

y² = 21

y = + √21 , or - √21

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An insect 5.00 mm tall is placed 20.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ
jenyasd209 [6]

Answer:

a) i = -9.63 cm ,    h ’= .0.24075 cm   erect

b)  i = 259.74 cm ,

Explanation:

For this exercise let's start by finding the focal length of the lens

               1 / f = (n-1) (1 / R₁ - 1 / R₂)

                1 / f = (1.70 -1)) 1 / ∞ - 1/13)

                1 / f = 0.0538

                 f = - 18.57 cm

Now we can use the constructor equation

             1 / f = 1 / o + 1 / i

             1 / i = 1 / f - 1 / o

              1 / i = -1 / 18.57 -1/20

               1 / i = -0.1038 cm

               I = -9.63 cm

For the height of the

image let's use magnification

                 m = h '/ h = - i / o

                  h ’= -h i / o

                  h ’= - 0.5 (-9.63) / 20

                  h ’= .0.24075 cm

b) we invert the lens

The focal length is

             1 / f = (1.70 -1) (1/13 - 1 / int)

              1 / f = 0.0538

             f = 18.57 cm

             1 / i = 1 / f -1 / o

             1 / I = 1 / 18.57 - 1/20

             1 / I = 3.85 10-3

             i = 259.74 cm

     

            h ’= - 0.5 259.74 / 20

             h ’= 6.4935 cm

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speed of sound is 343 Ms at 20 degrees Celsius. The frequency heard from the sound is 256 Hz. what is the sounds wavelength?
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S= 343m/s
F=256Hz

WL= 343ms/256-1
WL=V/F

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D=1/2at^2 <br> solve for a
nignag [31]

Answer:

a = 2d / t²

Explanation:

d = ½ at²

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A 200kg ball on the end of string is swung in horizontal circle with radius of 0.5m . The ball makes revolution every 2second th
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Answer:\dfrac{\pi}{2} ms^{-1}

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Let s be the speed of the ball.

s=\frac{d}{T}=\frac{\pi }{2}ms^{-1}

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