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polet [3.4K]
2 years ago
12

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

points along the line connecting the two sources.
a) At what distance from source A is there constructiveinterference between points A and B?
b) At what distances from source A is there destructiveinterference between points A and B?
Physics
1 answer:
alexgriva [62]2 years ago
3 0

Answer

given,

distance between A and B is equal to 5 m

let x be the distance of loud speaker from the source A

For constructive interference, path difference = mλ

x - (5 - x ) = 0           , for m = 0

2 x - 5 = 0

2 x = 5

  x = 2.5 cm

For destructive interference

|x - (5 - x )| = (2 m + 1)\dfrac{\lambda}{2}

m = 0

|2x - 5| = \dfrac{\lambda}{2}

|2x - 5| = \dfrac{6}{2}

\pm(2x - 5) = 3

x = 4 , 1 cm

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Answer:

15.065ft

Explanation:

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By definition the drag force is expressed as:

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Where

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For second Newton's Law the Force is also defined as,

F=ma=m\frac{dV}{dt}

Equating both equations we have:

m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A

m(dV)=-\frac{1}{2}\rho C_d A (dt)

\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)

Integrating

\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)

-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t

Here,

V_f = 60mph = 26.82m/s

V_i = 120.7m/s

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C_d = 0.3

\Delta t=7s

Replacing:

\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)

-0.029 = -5.4997r^2

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Answer:

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