A television picture in the United States is composed of 525 "lines" wich sweep across the face of the picture tube. These lines represent the path of an electron beam wich sweeps out these 525 lines every 1/30 of a second. If the horizontal dimension of the picture is 20 inches (50.8cm), how fast must the beam travel? (The beam must sweep back across the screen to start a new line, but assume that the time for this re sweep.) express your answer in ft/sec and miles/hr (or m/sec and km/hr).
2006-06-28 18:36:26 · 2 answers · asked by best 1 in Science & Mathematics ➔ Physics
Try this--I used to be pretty good at math story problems: First of all, how far must the beam travel, all told? 525 lines * 50.8 cm/line = 26,670 cm or 266.7m. Then, you have 266.7m / (1/30)s = 8,001m/s, or about 8*10^3 m/s; 8*10^3 m/s / (1000m/km) / (60s/hr) = .1334km/hr --- I think.
2006-06-28 18:48:08 · answer #1 · answered by Mark 3 · 0⤊ 0⤋
Another way to solve: the beam sweeps 525 lines in 1/30 second, then it sweeps one line in 1/(30*525) sec. In that time it moves 20", so the beam sweep velocity is 20/(30*525) in/sec. or 20/(12*30*525) ft/sec
2006-06-28 19:19:43 · answer #2 · answered by gp4rts 7 · 0⤊ 0⤋