Could someone help me out on this. I don\'t even know why I am doing this to beg
ID: 3549134 • Letter: C
Question
Could someone help me out on this. I don't even know why I am doing this to begin with, but I am confused.
Please show your work and could explain how you got your answers. Thank You
Many computers use one byte (8 bits) of data for each letter of the alphabet. There are 44 million words in the Encyclopedia Britannica. What is the bit density (bits/in2) of the head of a pin if the entire encyclopedia is printed on it? Assume the average word is five letters long. What is the byte density? What is the area of a single bit in nm2? A CD-ROM has a storage density of 46 megabytes/in2 and a DVD has a storage density of 329 megabytes/in2. Is the pinhead better or worse than these two storage media? How much better or worse?Explanation / Answer
Assuming the head of a pin is 1/16th of an inch in diameter, the area is pi*(1/32)^2.
Since each word is average 5 letters long, and there are 44 million words, there is a total of 220000000 bytes. Multiply that by 8, and you get 1.76 billion bits total over the head of the pin. There fore the bit density is #bits/area = 1.76billion bits/(pi(1/32)^2)) = approx 574 trillion bits per inch^2
Byte density is bit density divided by 8
c) inch^2 = (2.54cm^2) = (25.4e-6nm)^2. So you need to divide 574billion by (25.4e-6)^2
d) The pin head is WAY better. 574 trillion bits per inch^2 = about 75 trillion bytes per inch^2, or about 70 terabytes per inch^2