Two radio antennas are 10 km apart on a north-south axis on a seacoast. The ante

Question

Two radio antennas are 10 km apart on a north-south axis on a seacoast. The antennas broadcast identical AM radio signals, in phase, at a frequency of 1.6 MHz. A steamship, 200 km offshore, travels due north at a speed of 15 km/hr and passes east of the antennas. A radio on board the ship is tuned to the broadcast frequency. The reception of the radio signal on the ship is a maximum at a given instant. The time interval until the next occurrence of maximum reception is closest to:

Explanation / Answer

here,

L = distance of ship from shore ( 2 * 10^5 m )

d = separation of sources ( 10^4 m )

let the distance from one loud to the next loud be x

frequency , f = 1.6 * 10^6 Hz

wavelength = 3 * 10^8 / 1.6*10^6

wavelength = 187.5 m

The equation is

n * wavelength = x* d / L

n = 1

1 * 187.5 = x*10^4 /( 2*10^5)

x = 3750 m

speed of ship . v = 15 km/h

v = 4.17 m/s

Then the time , t = x / v

t = 3750 /4.17

t = 900 s

t = 15 min

the time interval until the next occurrence of maximum reception is closest to 15 min

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.