Question

A freight version of Boeing's 747 airliner has 2.7

Answer 1

Let's assume that the 747 can hold each CD, in its case, without space loss (for instance, think of packing oranges into a box...they don't exactly stack well)

2.7 * 10^4 cu.ft. = Volume of 747
10.8 * 10^1 cu.in = Volume of CD and case

1 cubic foot is equivalent to 1728 cubic inches

(2.7 * 10^4 * 1.728 * 10^3) / (10.8) =
0.432 * 10^7 CDs
4.32 * 10^6 CDs

4,320,000 CDs on 1 jet

a)
Each CD has 650 MB, so:

4.32 * 10^6 * 650 =
2808 * 10^6 MB =
2.808 * 10^9 MB

2,808,000,000 Megabytes of storage

b)
First, let's convert Megabytes to bits

1,048,576 (bytes/MB) * 8 (bits/byte) = 8,388,608 (bits/MB)

Now, the plane stores 2.808 * 10^9 MB

2.808 * 10^9 (MB) * 8.388608 * 10^6 (bits/MB) =
23.555211264 * 10^15 bits

A modem transfers 45 * 1024 bits per second

23.555211264 * 10^15 / (45 * 1024) =
511180800000 =
5.111808 * 10^11 seconds

(roughly 16200 years)


c)
Okay, we have to transmit 2,808,000,000 Megabytes in 12 hours (43200 seconds)

2.808 * 10^9 MB =
2.808 * 10^9 * 1024^2 Bytes =
2.808 * 10^9 * 1024^2 * 8 Bits =
23555211.264 * 10^9 Bits =
2.3555211264 * 10^16 Bits

We have to transmit

2.3555211264 * 10^16 Bits in 43200 Seconds

(2.3555211264 / 4.32) * (10^16 / 10^4) b/s =
0.54525952 * 10^12 b/s =
5,4525952 * 10^11 b/s

1 T3 line can transfer 45 Megabits per second

45 * 8 * 1,048,576 =
377487360 =
3.7748736 * 10^8 b/s

(5.4525952 / 3.7748736) * (10^11 / 10^8) =
1.4444444444444444444444444444444 * 10^3 =
1444.444.....

It would take 1445 T3 connections to transmit all of the information stored in the CDs in a 12 hour period