Question

: How long does it take for the first bit to arrive from the transmitter to the receiver over a 17 Mbps channel of length 1000 km. Assume speed of light is 2.0x10 8 m/s in the channel. Give your answer in ms. @Physics

Answer 1

2.0x10^8 m/s *correction

Answer 2

let me just say that I got 5 ms and was wrong. homework console said it was 5x10^-1 ms..

Answer 3

I haven't done these before either but I'll give it a shot:
First, I don't think the 17 Mbps value is even a factor. I believe what the question is really asking is how fast a bit can arrive over a 1000km cable at the speed of light given

To answer that I'd use the velocity formula which is \[V = \frac{distance}{time}\]which can be rearranged to get\[t = \frac{d}{v}\]so it's:\[t=\frac{1000000}{2 x 10^8}\]\[t = .005s\]

Answer 4

yep. that was also my solution.. we know that 0.005 s is 5 ms too, right? idk about the homework console haha. :))

Answer 5

it keeps on saying that it's 5x10^-1 lol.

Answer 6

yes, it's 5 milliseconds...which could also be written as 5 x 10^-3

Answer 7

If it was 5 x 10^-1, that would be a half a second...which doesn't even make sense :)

Answer 8

haha i guess it's a glitch. thanks, i just wanted to confirm that I'm not alone haha.

Answer 9

one more left! wanna try it out?

Answer 10

lemme look

Answer 11

Ok, I see how their answer is right I think for this one.

Answer 12

this might help

Answer 13

Well that changes things...one min :)

Answer 14

lol used d/c+ L/R and got the same thing, L/R is too small to change things around anyway too

Answer 15

Ok, let's redo this one using the equation given:
\[\frac{1000000}{2 x 10^8} + \frac{1}{17000000}=5 * 10^{-4}s\]
\[0.0005s *1000 = 5 x 10^{-1}\]

Answer 16

Now let's do the other one :)

Answer 17

oh wow, thanks haha. i think i got the other one, please confirm.

Answer 18

ok

Answer 19

hey, after checking the calculation again im still getting 0.005. even after using your calculation..

Answer 20

I agree, the bandwidth doesn't matter.\[1000km \times \frac{1000m}{1km} \times \frac{1s}{2 \times 10^{8}m} \times \frac{1000ms}{1s}\]That works out to be 5ms. However, if the original problem said 100km then their answer would be right. (The other problem you posted used 100km, so that's why I suggested it.)

Answer 21

@windsylph, The formulas you posted are good and all, but if you learn dimensional analysis then you won't even need them. Start with the quantity they give you, then multiply by conversion factors (equal to 1), making sure the units cancel, to convert their quantity to the right units (dimension). If you have to add two quantities, convert them to the same units first. These problems almost solve themselves once you start identifying the correct conversion factors.

Answer 22

Haha thanks, I almost always use dimensional analysis for my homework problems in this class, but I didn't figure out at first that I had to add the two times to get the correct answer.