Answers to any of the following questions about digital storage?
I've searched for the methods to solve these problems but can not find anything. I have no idea how to solve any of these problems, if anyone can explain how to solve the problems below your help will be greatly appreciated!! Thanks so much in advance 
1) Estimate the length of the spiral carrying pits and lands on an ordinary CD of diameter 12 cm. Assume the adjacent parts of the spiral are 1600 nm apart. Explain your method, listing the assumptions you have made.
2) It takes 4.0 minutes to record a song on a CD at a rate of 44 100 words per second. A word has a length of 32 bits. How many bytes does the song take?
3) Calculate the storage capacity of an 80-minute CD that was recorded at a rate of 44 100 bits per second using two channels of 16-bit words each.
4)a) Explain how digital data is stored on a CD
b) Explain how interference of light can be used to distinguish a pit from a land
c) A CD player uses a laser of wavelength 680 nm. What is an appropriate wavelength?
1) At the end of the spiral, the diameter is 12 cm, or the circumference is 38 cm. At the other end, the beginning, the diameter is about 4 cm at a guess, so the circumference is 13 cm. Take the average of the two as 25 cm.
In the distance between a radius of 2 cm and a radius of 6 cm, there is 4 cm. Divided by 1600 nm that is 25000 groves. Multiply that by average length of 25cm to get 625000 cm.
I could get more exact numbers by reading online the red-book spec for CDs
b) 44100 words/sec x 32bits/word x 1 byte/8 bits = 176400 bytes.
but 44100 words/sec seems wrong, I’d have to check.
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I will go for number 1. Consider each track as a circle. You know how many circles there are because you know how many nm have data tracks. Just find the length of the middle track (the average) and multiply that by the number of circles total. This should be a very good estimate of the length of the spiral. My sense is that this will be a very large number because there are many circles.
2. 1/8 bytes/bit * 32 bits/word * 44100 words/second * 60 seconds/minute * 4.0 minutes
3. Channels and word size are redundant.
44100 bits/second * 60 seconds/minute * 80 minutes
4. check the internet, it has nice pictures