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Ah, I see. I was confused about the wording, because I thought "proper time" (t0) was the time measured by the person on earth, but "proper time" is actually the time that is measured by the person that is moving on the spaceship. Thank you so much for your help!

I did know that relativistic time (where observer is moving) should be shorter than proper time (where observer is stationary), but our professor gave us this: proper time = t0 relativistic time = t relativistic time = t0 / sqrt[1 - (v/c)^2], which doesn't give the right relationship... so I can...

Good to know. Our professor gave us the wrong equation in notes. Thanks for your help!

Homework Statement A certain star is 76.6 light-years away. How many years would it take a spacecraft traveling 0.946c to reach that star from Earth, as measured by observers on Earth? How many years would it take to reach that star from Earth, as measured by observers on the spacecraft ...

Ah, I see. I thought that the infared rays were only the ones that landed on level 3, but they can land on level 3, 4, OR 5? However, visible rays ONLY land on level 2 and UV rays ONLY land on level 1, is that correct? Thanks for your help!

Homework Statement How many transitions involving the hydrogen energy levels shown correspond to photons of infrared light? The infrared range is less than 1.8 eV Homework Equations Paschen Lines = Infared wavelengths The Attempt at a Solution I thought that since Paschen lines =...

Homework Statement Two antennas located at points A and B are broadcasting radio waves of frequency 96.0 MHz, perfectly in phase with each other. The two antennas are separated by a distance d=12.40m. An observer, P, is located on the x axis, a distance x=55.0m from antenna A, so that APB...

So I know the difference in the paths is 1.4m, but when I multiply that by the wavelength, (3.125m), and convert to radians it is wrong.

I know that AP = 55m and BP = 56.4m, I just don't know how to find the wavelength of pathlength difference

Homework Statement Two antennas located at points A and B are broadcasting radio waves of frequency 96.0 MHz, perfectly in phase with each other. The two antennas are separated by a distance d=12.40m. An observer, P, is located on the x axis, a distance x=55.0m from antenna A, so that APB...

Ah, just figured it out...you have to use "m" and multiply it by the wavelength.

So...I think I just realized that pathlength difference refers to "m", but what are the units??

Homework Statement Two slits separated by a distance of d = 0.12mm are located at a distance of D = 780mm from a screen. The screen is oriented parallel to the plane of the slits. The slits are illuminated by a coherent light source with a wavelength of lambda = 549E-6 mm. The interference...

Nevermind, I got it myself. I was using the wrong value to convert eV. Thanks anyway

I didn't ignore it, I don't understand where my units are wrong. eV, m = kg If you would tell me which part I'm doing wrong I could try and fix it.

I'm using m = 9.11E-31 eV = 796 796 = 1/2*9.11E-31 v^2 796 = 4.55E-31 v^2 1.748E33 = v^2 v = 4.18E16 Don't know where I'm wrong...

eV = 1/2 mv^2 eV / (1/2m) = v^2 v = sqrt (eV / 1/2m) Am I doing this wrong?

For: 796 = 1/2 * mv^2 (where m = 9.11E-31) I keep getting 4.18E16 for v I don't understand what part I'm doing wrong, I've done it about ten times...

I tried converting eV to joules, which gave me 1.27E-16 joules, which gave me 8.36E6 m/s but that is wrong too. I think the units for mass (kg) are correct, I'm not sure what I have wrong...

If 796 = 1/2 * 9.11E-31 * v^2, I get v to be 4.18E16 m/s which is wrong...i'm not sure what I'm doing wrong...

Homework Statement What is the speed of a 796eV electron? Homework Equations E = hc/lambda E = hv The Attempt at a Solution I'm not sure how to get a value for meters... I know if I take h (plank's constant) and divide the electron, I will get 4.14E-15 ev*s / 796 eV = 5.20E-18...

I didn't know that all virtual images are upright, that is good to know! However, if Part F is true, I still have something else wrong, as I tried: TFTFTT So I'm not sure what other one I have wrong...

I guess my prof said that a diverging lens could possibly form an inverted image, although I'm not sure how. I also already tried that combination: TFTFTT So I must either have another one wrong or something... not sure...

Homework Statement Starting with a real object, which of the following statements are TRUE about the image formed by a single lens? (Choose ALL correct answers, e.g. B, AC, CDF.) A) For a converging lens an object has to be placed between the focal length and the lens in order to form a...

right. I'm an idiot. thanks for your help!

yep, I'm awful at trig so that isn't surprising. haha. i'm still not getting it right though: if i use that identity, i get: 1 * 2 sin theta cos theta = 1.51 * sin theta If I divide the left by sin theta, the sin theta cancels out (right?), leaving me: 2 cos theta = 1.51 which gives me a...

Homework Statement A beam of light in air strikes a piece of glass (n=1.51) and is partially reflected and partially refracted. Find the angle of incidence if the angle of reflection is twice the angle of refraction. Homework Equations n1sintheta1 = n2sintheta2 The Attempt at a...

I went through it a couple times and got the same thing. Thanks so much for your help!

when I try to solve that, I get: x = (h^2 + d^2 + h^2n^2) / n^2 or x = (h + d + hn) / n

ok, i see how you got to here: 1 + (h/x)^2= (1+( h/d^2) )n^2 but not I'm still not sure how to solve for x?

ok just tried to do it again and I'm completely lost.

shoot, now when I did it again, I got x/n = d, so x = nd I think x = nd is correct...

would solving that result in x = d/n? That is what I got. Here is how I got it: (x^2 + h^2) / x^2n^2 = (d^2 + h^2) / d^2 1/n^2 + h^2/x^2 = 1 + h^2/d^2 1/n^2x^2 = 1/d^2 nx = d x = d/n Is that right? Thanks for your help on this I appreciate it by the way

sorry, I see what I did - I forgot the n in the top equation before. so now I'm still confused as to how to solve for x...

I must have done something wrong, because x can't equal d. x is less than d because of the index of refraction (in the diagram). Here is how I got the formula: the theta for the apparent angle of where the fish appears is I called theta apparent the theta for the angle of where the fish...

I did apply Snell's law, and I did tak the refraction of light into account. that is how I got the equation (x is the distance of the actual fish, where d is the distance of the virtual fish). I just don't know how to solve the algebra, can you help me with that?

OK - I managed to find a formula to solve for X, but now I'm having trouble with the algebra: [ x / sqrt(x^2 + h^2) ] = [ d / sqrt(d^2 + h^2) ] Can anyone help me remember how to get rid of the sqrt of x on the bottom of the left hand side? If I multiply by the sq rt of the ( ) I will have to...

Homework Statement After finishing your physics classes with the firm conviction that the questions were strange and out of touch with reality, you decide to take up spear fishing. While wading through water (of index of refraction=n and depth h) you spot a fish that appears to be a...

ah, i see. that makes it simpler, i often make the problems harder than they need to be. thanks for your help!

yep, that's a lot easier. oops. thanks!

Ah, I see. Thank you!

Homework Statement A 1.10m long FM antenna is oriented parallel to the electric field of an EM wave. How large must the electric field be to produce a 1.28 mV voltage between the ends of the antenna? Homework Equations u = EoErms^2 I = uc Intensity = Power / area The Attempt at a...

Homework Statement A laser beam enters a 10.0cm thick glass window at an angle of 30 degrees from the normal. The index of refraction of the glass is 1.46. At what angle from the normal does the beam travel through the glass? Homework Equations sin theta 1 / sin theta 2 = n1 / n2 = v1 /...

Homework Statement A super nova releases 1.9E45 J of energy over a 10 day period. It is 1560 ly from earth. A detector, facing the star, is a disk of radius 7cm. How much energy reaches the detector? (The detector is in orbit, so it can face the star for the entire 10-day period) Homework...

No...current has to go from hi to low, or positive to negative. So there is no current.

Sorry, I should have scanned the problem earlier...I attached it here - can you see if it makes more sense to you now? Thanks! The only part I don't understand is the current on the middle block (and hence, the force on the middle block). I get the first and third parts. Thanks!

I see! Thank you for your help!

Then I have no clue what equation to use. I know the B fields oppose each other on the outside of each wire, but when you say, "Can you find a set of radii from the two wires that makes the combination go to zero" I guess I don't know how to solve for two different variables at once. Don't worry...

I know the EMF is zero when the wire box is in the constant square magnetic field, that is why I assumed current is zero. I just don't really know how to prove it, other than there is no change in magnetic field, so there is no induced current. That seems to make sense.

I still don't understand it, sorry. Thanks for trying. I'll have to go get help tomorrow. I got r = .004 m but that was wrong.