While a CD player is reading the outer edge of an audio CD the disc spins at approximately 3.33 revolutions per second. The radius of a CD is 0.06 meters and the data transfer rate is 1.2 million bits per second (a bit is the smallest piece of information one can have - it is either a one or a zero). How many bits are stored per meter on the outer rim of a CD to the nearest thousand?
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Nice one (y)
thanks
bits per meter*
thanx
nice
hmmmm i forgot taking circumference
Here is the series of equation you must solve to get the answer
3 . 3 3 1 2 0 0 0 0 0 = 3 6 0 . 3 6 0
2 ∗ 0 . 0 6 ∗ π 3 6 0 . 3 6 0 = 9 5 5 8 5 5 . 5 4 4 0 9 5 4 6 7 4 8 2
When you round this quotient to the nearest thousand you get 956000.
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The circumference of the CD is 2 π × 0 . 0 6 = 0 . 1 2 π . Multiplying this by 3 . 3 3 gives the number of meters passed per second: 0 . 3 9 9 6 π m/s. Now we divide 1.2 million bits by this number to get the number of bits per meter: 1 2 0 0 0 0 0 ÷ ( 0 . 3 9 9 6 π ) ≈ 9 5 6 0 0 0 bits per second.