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sdas [7]
3 years ago
15

A speeder is pulling directly away and increasing his distance from a police car that is moving at 24 m/s with respect to the gr

ound. The radar gun in the police car emits an electromagnetic wave with a frequency of6.7 × 10 9 Hz. The wave reflects from the speeder’s car and returns to the police car, where its frequency is measured to be 370 Hz less than the emitted frequency. Find the speeder’s speed with respect to the ground.
Physics
1 answer:
iogann1982 [59]3 years ago
8 0
2,062,305 2,062,305 <span>2,062,305</span>
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A) 4.7 cm

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sin \theta=\frac{n \lambda}{d}

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In this problem,

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B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

5 0
3 years ago
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