Question

The radar gun measures the frequency of the radar pulse echoing off your car. By what percentage is the measured frequency different from the original frequency? (Enter a positive number for a frequency increase, negative for a decrease. Just enter a number, without a percent sign.)

Answer 1

The Doppler effect causes a difference in frequency between the measured frequency and the original frequency. The percentage difference can be calculated using the formula (Change in Frequency / Original Frequency) x 100%. In this case, the measured frequency is 15.0 kHz higher than the original frequency.

Step-by-step explanation:

The Doppler effect is used in radar guns to measure the speed of vehicles by measuring the change in frequency of the radar pulse reflecting off the car. The measured frequency is different from the original frequency due to the Doppler effect. The percentage difference in frequency can be calculated using the following formula:

Percentage Difference = (Change in Frequency / Original Frequency) x 100%

In this case, the change in frequency is given as 15.0 kHz. The original frequency is 1.50×10^9 Hz. Substituting these values into the formula, we get:

Percentage Difference = (15.0 kHz / 1.50×10^9 Hz) x 100%

Simplifying this equation gives us:

Percentage Difference = 1 x 10^-4%

Therefore, the measured frequency is 1 x 10^-4% different from the original frequency.

Answer 2

Question: A. The state highway patrol radar guns use a frequency of 9.15 GHz. If you're approaching a speed trap driving 30.1 m/s, what frequency shift will your FuzzFoiler 2000 radar detector see?

B. The radar gun measures the frequency of the radar pulse echoing off your car. By what percentage is the measured frequency different from the original frequency? (Enter a positive number for a frequency increase, negative for a decrease. Just enter a number, without a percent sign.)?

Answer:

The frequency change percentage is 9.94%

Step-by-step explanation:

The frequency shift can be calculated as follows.

`F= (V+V_(0))/(V+V_(s))`

=
`9.05GHz*(343m/s+0)/(343m/s+(-31.3m/s))`

=9.95 GHz

So the frequency change seen by the detector is 9.95 - 9.05

% difference
`= (0.9)/(9.05) *100`

= 9.94%