2007-12-02 13:58:31 · 4 answers · asked by Richard P 1 in Science & Mathematics ➔ Engineering
radius = 13"
44 ft / sec = 528 inches / sec
circumference = 2 x r x π
circumference = 2 x 13 x π = 81.68 inches per revolution
528 inches / sec ÷ 81.68 inches / rev = 6.43 revs / sec
6.43 revs / sec x 60 seconds / min = 388 RPM's.
2007-12-02 14:45:07 · answer #1 · answered by Thomas C 6 · 2⤊ 0⤋
Circumference of wheel = 2 X (pi) X radius
Circ. = 2 X 3.14159 X 13in = 81.68 inches or 6.807 ft.
Revolutions per second = 44/6.807 = 6.454 rev./sec.
Revolutions per minute = 6.454 X 60 = 387.8 rev./min.
2007-12-02 22:28:38 · answer #2 · answered by vacafan 2 · 1⤊ 0⤋
2*pi*13 = 81.68 in. / 12 in./ft = 6.8 ft. = 1 rev.
44 / 6.8 = 6.47 rev. / sec.
6.47 * 60 sec/min. = 388 rpm
2007-12-02 22:43:44 · answer #3 · answered by Hate the liars and the Lies 7 · 1⤊ 0⤋
2,436.92 RPM
T(Uc(V)))
12(44ft/sec)=528in/sec
60(528in/sec)=31,680in/min
(31,680in/min)/(13in)= 2,436.92RPM
2007-12-02 22:02:24 · answer #4 · answered by james R 2 · 0⤊ 5⤋