hw22 - = 225 196 92 p = 105 . 6 p 32 P 36 Compare 940 on...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 21 Fall, 2004 Solution to HW-22 32 P25 Estimate the average power output of the sun, given that about 1350 W/m 2 reaches the upper atmosphere of the earth. The power from the sun radiates uniformly outward through a spherical surface whose radius R is the earth-sun distance 149 . 6 × 10 6 km, and the Fux at radius R is given as S = 1350 W/m 2 . Hence the total power output is P total = S × 4 πR 2 = (1350 W / m 2 )4 π (149 . 6 × 10 9 m) 2 =3 . 8 × 10 26 W 32 P34 An amateur radio operator wishes to build a re- ceiver that can tune a range from 14.0 MHz to 15.0 MHz. A variable capacitor has a minimum capacitance of 92 p±. (a) What is the required value of the inductance? (b) What is the maximum capacitance used on the variable capacitor? The resonant frequency ω of the tuner will satisfy ω 2 = 1 /LC ,so L =1 2 C . C min = 92 p± will correspond to the maximum frequency, so L = 1 (2 π 15 . 0 × 10 6 rad / s) 2 × 92 p± =1 . 22 µ H . C is inversely proportional to the square of the frequency ( C =1 2 L ), so decreasing the frequency from 15.0 to 14.0 MHz increases C by a factor of (15 / 14) 2 : C max
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = 225 196 92 p = 105 . 6 p 32 P 36 Compare 940 on the AM dial to 94 on the M dial. Which has the longer wavelength, and by what factor is it larger? 940 AM corresponds to f = 940 kHz, and 94 M is 94 MHz. The wavelengths follow from = c/f : 940 = 3 . 10 8 940 10 3 = 319 m 94 = 3 . 10 8 94 10 6 = 3 . 19 m The AM wavelength is longer by a factor of 100. 32 P 40 A radio voice signal from the Apollo crew on the moon is beamed to a listening crowd from a radio speaker. If you are standing 50 m from the loudspeaker, what is the total time lag between when you hear the sound and when the sound left the moon? Let the earth-moon distance be D = 384 10 6 m, the terrestrial distance be d = 50 m, and the speed of sound in air be v = 348 m/s. Then the total time t is t = D c + d v = 384 10 6 3 10 8 + 50 348 = 1 . 28 s + 0 . 14 s = 1 . 42 s November 6, 2004...
View Full Document

Ask a homework question - tutors are online