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Korvikt [17]
3 years ago
7

Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid

er points along the line connecting the two sources.Required:a. At what distance from source A is there constructive interference between points A and B?b. At what distances from source A is there destructive interference between points A and B?
Physics
1 answer:
Korolek [52]3 years ago
7 0

Answer:

a

    z= 2.5 \ m

b

   z =  (1 \ m ,  4 \ m )

Explanation:

From the question we are told that

     Their distance apart is  d =  5.00 \ m

      The  wavelength of each source wave \lambda =  6.0 \ m

Let the distance from source A  where the construct interference occurred be z

Generally the path difference for constructive interference is

              z - (d-z) =  m \lambda

Now given that we are considering just the straight line (i.e  points along the line connecting the two sources ) then the order of the maxima m =  0

  so

        z - (5-z) =  0

=>     2 z - 5 =  0

=>     z= 2.5 \ m

Generally the path difference for destructive  interference is

           |z-(d-z)| = (2m + 1)\frac{\lambda}{2}

=>         |2z - d |= (0 + 1)\frac{\lambda}{2}

=>        |2z - d| =\frac{\lambda}{2}

substituting values

          |2z - 5| =\frac{6}{2}

=>      z  =  \frac{5 \pm 3}{2}

So  

      z =  \frac{5 + 3}{2}

      z =  4\ m

and

      z =  \frac{ 5 -3 }{2}

=>   z =  1 \ m

=>    z =  (1 \ m ,  4 \ m )

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Explanation:

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  mg(h_{f} -h_{o})  = \frac{1}{2} kx^{2}

Where k = spring constant, g = 9.8 \frac{m}{s^{2} }

  k = \frac{2mg(h_{f} -h_{o} )}{x^{2} }

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the amount of surface area of the block contact with the surface is 2.03*10^-2*m2 what is the average pressure exerted on the su
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Complete question:

A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa

Answer:

The average pressure exerted on the surface by the block is 9655.17 Pa

Explanation:

Given;

density of the lead, ρ =  1.13 x 10⁴ kg/m³

mass of the lead block, m = 20 kg

surface area of the area of the block, A = 2.03 x 10⁻² m²

Determine the force exerted on the surface by the block due to its weight;

F = mg

F = 20 x 9.8

F = 196 N

Determine the pressure exerted on the surface by the block

P = F / A

where;

P is the pressure

P = 196 / (2.03 x 10⁻²)

P = 9655.17 N/m²

P = 9655.17 Pa

Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa

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3 years ago
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Answer:

c. is more than that of the fluid.

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let s1= specific heat of solid and s2= specific heat of liquid

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Yo sup??

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