(II) On an audio compact disc (CD), digital bits of information are encoded sequentially along a spiral path. Each bit occupies about 0.28 μ m. А CD player’s readout laser scans along the spiral’s sequence of bits at a constant speed of about 1.2 m/s as the CD spins. ( a ) Determine the number N of digital bits that a CD player reads every second. ( b ) The audio information is sent to each of the two loudspeakers 44,100 times per second. Each of these samplings requires 16 bits and so one would (at first glance) think the required bit rate for a CD player is N 0 = 2 ( 44 , 100 samplings second ) ( 16 bits sampling ) = 1.4 × 10 6 bits second , where the 2 is for the 2 loudspeakers (the 2 stereo channels). Note that N 0 is less than the number N of bits actually read per second by a CD player. The excess number of bits (= N − N 0 ) is needed for encoding and error-correction. What percentage of the bits on a CD are dedicated to encoding and error-correction?
(II) On an audio compact disc (CD), digital bits of information are encoded sequentially along a spiral path. Each bit occupies about 0.28 μ m. А CD player’s readout laser scans along the spiral’s sequence of bits at a constant speed of about 1.2 m/s as the CD spins. ( a ) Determine the number N of digital bits that a CD player reads every second. ( b ) The audio information is sent to each of the two loudspeakers 44,100 times per second. Each of these samplings requires 16 bits and so one would (at first glance) think the required bit rate for a CD player is N 0 = 2 ( 44 , 100 samplings second ) ( 16 bits sampling ) = 1.4 × 10 6 bits second , where the 2 is for the 2 loudspeakers (the 2 stereo channels). Note that N 0 is less than the number N of bits actually read per second by a CD player. The excess number of bits (= N − N 0 ) is needed for encoding and error-correction. What percentage of the bits on a CD are dedicated to encoding and error-correction?
(II) On an audio compact disc (CD), digital bits of information are encoded sequentially along a spiral path. Each bit occupies about 0.28 μm. А CD player’s readout laser scans along the spiral’s sequence of bits at a constant speed of about 1.2 m/s as the CD spins. (a) Determine the number N of digital bits that a CD player reads every second. (b) The audio information is sent to each of the two loudspeakers 44,100 times per second. Each of these samplings requires 16 bits and so one would (at first glance) think the required bit rate for a CD player is
N
0
=
2
(
44
,
100
samplings
second
)
(
16
bits
sampling
)
=
1.4
×
10
6
bits
second
,
where the 2 is for the 2 loudspeakers (the 2 stereo channels). Note that N0 is less than the number N of bits actually read per second by a CD player. The excess number of bits (= N − N0) is needed for encoding and error-correction. What percentage of the bits on a CD are dedicated to encoding and error-correction?
Section 4.5 Tangential and Radial AccelerationFigure represents the total acceleration of a particle moving clockwise in a circle of radius 2.50 m at a certain instant of time. For that instant, find (a) the radial acceleration of the particle, (b) the speed of the particle, and (c) its tangential acceleration.
In the answer, it gives a formula f-ff=m(0) and afterwards simplies into 120 - ff = 0 and further into ff=120N. My question is how do you not have to divide by 0 to finish the equation if that makes sense. To simply further why are we able to ignore the m(0) portion?
A particle moves at constant speed in a circle of radius 2.06 cm. If the particle makes four revolutions every second, determine the magnitude of its acceleration
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