studentteacher

What's a good way to teach freshman high-school students how to balance this: __ KMnO4 + __ HCl --> ___ KCl + __MnCl2 + __H2O + __Cl2

If object distance is positive it is in front of the lens. If image distance is positive, the image is behind the lens and real. (But, our text doesn't give a coherent definition of "image" or "front" or "behind." I assume "front" means where the light is coming from and "behind" means where the lens sends the light. How can an image be in front of a lens? Or an object behind it?) In this case. the object (source of light) is the slide, and the image is "the screen". Since they are on opposite sides of the lens, they have the same sign, and the sign in the equation doesn't require modification. I guess.

I don't understand this part. How can it be correct to write -1 * (x) = x ? My problem-solving was the same as yours, except I used di/do = - 49, because that's what the equation says to use.

I can't get a sensible answer. I think the problem is that it is wrong to state "object-distance + image-distance = 320 cm." because one of the distances is going to be negative. Using substitution, I get that do = -6.67 and di then = 326.67, which is impossible. I don't really understand presenting the problem this way. There must be a general relationship for distance between image and object, instead distance between object & image and lens.

I think I'm just blanking on the algebra. Problem: A slide projector (old book) needs to project a 98 cm high image of a 2cm tall slide. The screen is 320 cm from the slide. What focal length does the lens need? How far should the lens be from the slide? So, image-height = 98 cm. object-height = 2 cm. object-distance + image-distance = 320 cm. Formulas are 1/do + 1/di = 1/ f and hi/ho = - di/do. Should be able to find f, but can't.....

I am really just not getting this. here is a homework problem. A film of oil (n=1.25) floats on water (n=1.33). What is the thinnest film that produces a strong green reflection (wavelength = 500 nm). And,2 about anti-reflection coatings on glasses. A 90-nm thick coating is applied to a lens. What must be the coating's index of refraction to be most effective at 480 nm? Assume the coating's index of refraction is less than that of the lens. If the index of refraction of the coating is 1.38, what's the minimum thickness the coating should be to be most effective at 480 nm? None of the formulas given above include a term for index of refraction. Is the idea that you figure ut which of the first two equations for delta-d to use, based on the index of refraction of the materials, and then substitute into one of the second two based on whether you're working with constructive or destructive interference? I think in the first hw problem, both waves of shifted, since the air-oil-water sequence is always from low n to higher n. So, solve the equation 2t=500/1.25. For the second, again both waves are shifted, but we're looking at destructive interference. So, 2*90=0.5*480/n ... and n = 1.33

Hi thanks for the reply. Is the unit of delta-d distance? If the ray is perpendicular to the surface, is dealt-d just the same as 2t?

I'm a little confused by what these equations are saying: ∆d = 2t (phase shift for neither or both waves) ∆d = 2t + 0.5 * wavelength (phase shift for only one wave) ∆d = m*wavelength (constructive) ∆d = (m + 0.5)*wavelength (destructive) I think t is the thickness of the film, but then what is ∆d?

We're given the mass of the block, the mass of the bullet, the spring constant, and the amplitude. So, I used the spring constant and amplitude to find total energy. And then set the kinetic energy of the bullet before the collision equal to the total energy, and solved for v. I don't understand why that was wrong, or what is right.

So, I was tutoring a student and we got to a question of a type I've seen before. A block attached to a spring is resting on a frictionless horizontal surface. A bullet is fired into it. You're told the amplitude of the oscillation and the spring constant. So, you find the total energy. You get the right answer. Another part of this question is to find the speed of the bullet before it hit the block. So, you set 1/2 mv^2 equal to the total energy found above, and solve for v. But somehow, I kept getting the wrong answer, and this was very embarrassing in front of my student. Is there a trick here?

You travel 3 m north in 1s, and 4 m east 2 s. Displacement: 5 m in 3s. Average veolcity is 5/3 m/s. or V north = 3 m/s, and v east = 2 m/s. Average velocity for trip is (allegedly) 3.6 m/s. Obviously this wrong, but conceptually I'm not quite sure how to explain why it doesn't work.