Digital bits on a 12.0-cm diameter audio CD are encoded along an outward spirali

Question

Digital bits on a 12.0-cm diameter audio CD are encoded along an outward spiraling path that starts at radius R1=2.5cm and finishes at radius R2=5.8cm. The distance between the centers of neighboring spiral-windings is 1.6m(=1.6×106m). Part A Determine the total length of the spiraling path. [Hint: Imagine "unwinding" the spiral into a straight path of width 1.6m, and note that the original spiral and the straight path both occupy the same area.] Express your answer to two significant figures and include the appropriate units. Digital bits on a 12.0-cm diameter audio CD are encoded along an outward spiraling path that starts at radius R1=2.5cm and finishes at radius R2=5.8cm. The distance between the centers of neighboring spiral-windings is 1.6m(=1.6×106m). Part A Determine the total length of the spiraling path. [Hint: Imagine "unwinding" the spiral into a straight path of width 1.6m, and note that the original spiral and the straight path both occupy the same area.] Express your answer to two significant figures and include the appropriate units. Digital bits on a 12.0-cm diameter audio CD are encoded along an outward spiraling path that starts at radius R1=2.5cm and finishes at radius R2=5.8cm. The distance between the centers of neighboring spiral-windings is 1.6m(=1.6×106m). Part A Determine the total length of the spiraling path. [Hint: Imagine "unwinding" the spiral into a straight path of width 1.6m, and note that the original spiral and the straight path both occupy the same area.] Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

so the area wound = (R22 - R12)

the area unwound = L*w

so L = *((5.8*10-2)2 -(2.5*10-2 )2 )/1.6*10-6 =5378 m

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