Recent content by jayjay713

yes! The answer is 601 nm. If you take the wavelength and times it by the index of refraction of water (1.33), you get 150.29. Multiply that by 4 and get ~601 nm. Don't know how that fits into the formula of 2T = wavelength / something (in example its over 4 or 2, don't know how to get that...

Homework Statement A thin layer of water (n=1.33) is surrounded by air what wavelength of light will not be transmitted through the water at the point where the water is 113 nm thick? The Attempt at a Solution - In phase - Path difference of 2T. - Constructive Interference I just...

but if I used that formula, wouldn't it give me a speed relative to the earth? Or would it be relative to the quasar? :O

A quasar is moving away from the Earth with a speed of 0.850C. It emits a proton that eventually reaches earth, and is traveling at a speed of 0.519C relative to the earth. How fast is the proton moving relative to the quasar? Is this answer as simple as it seems? is the answer simply...

i don't know what it means? it's on the formula sheet there is an example using it. all i want to know is the speed of the proton relative to the quasar... QUASAR<-0.85c-<QUASAR "shoots" PROTON>0.519>PROTON <EARTH> the velocities are relative to earth. THERE! I DREW IT OUT :) Does...

it's his stinkin isu for grade 12 physics. we are given a booklet with a few examples but its really complicated stuff.

A quasar is moving AWAY from the Earth with a speed of 0.850C. Itemits a proton that eventually reaches the earth, and is traveling at a speed of 0.519C relative to the EARTH. how fast is the proton moving RELATIVE to the QUASAR? my attempt: u = u' +v / (1 + u'v/c^2) u = 0.578c RELATIVE to...

Homework Statement a quasar is moving away from Earth with a speed 0f 0.850c. It emits a proton that eventually reaches Earth with a speed of 0.519c relative to the earth. How fast is the proton moving relative to the QUASAR. The Attempt at a Solution u = u' + v / 1 + u'(v) / c^2 u...

oh okay. so b would be equal to 1 as well because of the postulate. thank you for your help!

How did he determine the mass of an oil drop and the charge of the electron?

0.1 is the sweet spot. It is very similar to the oil drip experiment!

3.5 1.9 0.9 5.4 2.8 -2.1 8.2 0.7 8.9 see what I mean? absolute value algorithm ends with 0.6 0.6 how do i finish it and get the mass? is 0.6 grams the mass?:S

Interesting that you say it reminds you of Millikan's oil drip experiment. How so?

YIIIIIIIIIIP i did it without the negatives and i got 0.9, 0.3, 0.9 .. which equals 0.6, 0.6. BUT if i use negatives for the last step it would be 0.6 - (-0.6) = 1.2 g! is that a good guess? or should i stick with 0.6?

i got -198.8! in the 10th difference haha what... i did it on excel as well, same answer! what am i doing wrong?