Closer. The path difference is small when the detector is far away and then increases as d decreases. We'll "get to" m = 1 first.
The intensity of a point source goes like 1/4 \pi r^2 , so if the intensity of the source is I_0, the intensity at the hole should be I_0/4 \pi x^2. Only a...
Sorry, I don't follow: m is just 1.
Yes.
Sure, but the intensity of the source is an overall constant which divides out.
I assume the source is isotropic, so no.
I guess not really, but my idea is that it's so small that it's close enough - otherwise I'm not sure what to do.
Homework Statement
A point source of light (wavelength \lambda = 600 \, \text{nm} ) is located a distance x = 10\,\text{m} away from an opaque screen with a small circular hole of radius b. A very small photodiode is moved on an axis from very far away toward the screen. The first...
Homework Statement
The picture of the circuit is attached; I want to find |V_{A}/V_{J}|. This seems really easy but I haven't done circuit analysis in forever.
Homework Equations
Complex impedances, Z_{C} = 1/i\omega C, Z_{R} = R.
The Attempt at a Solution
First R_{A} and C_{A} are in...
Ah yes, thank you! The dimensions were incorrect.
Applying the correction, we find the same result. Also, no, I didn't account for the changing magnetic field as the cylinder speeds up; doing this with Faraday yields the same answer, so it's probably right. Thanks again to both of you :smile:
Homework Statement
An infinite wire of linear charge density \lambda lies on the z axis. An insulating cylindrical shell of radius R is concentric with the wire and can rotate freely about the z axis. The charge per unit area on the cylinder is \sigma = -\lambda/2\pi R while the mass per unit...
I'm assuming you meant that moment of inertia = torque divided by angular acceleration, because what you have there isn't true (just look at the units.)
I agree with this.
I'm assuming that you're referring to this page. You'll note that T is in fact the kinetic energy of the object and...
The way you've written the fractions is somewhat ambiguous. It's hard to say whether you meant that R_{\text{max}}=\frac{u^2}{g}\frac{1}{\sqrt{1+\frac{2gh}{u^2}}} or R_{\text{max}}=\frac{u^2}{g}\sqrt{1+\frac{2gh}{u^2}} (especially since the term in the radical is dimensionless), but I've...
I'm trying to model the flight of a small spherical object (such as a ping-pong ball) through air at smallish velocities (\approx 5{_{m/s}}) with linear drag so that F_{d}=-bv. The problem is, I can't find a table of linear drag coefficients (b) anywhere; it's always just the normal drag...
You don't actually need t, that's just to confirm that there's only one root to the equation so that the collision is "just barely" avoided. The problem is still to solve this for d, which will give you the answer:
\sqrt{[-v]^2-4[(1/2)(a_{1}+a_{2})(d)]}=0
No, v can be anything. The roots are:
t = \frac{v\pm \sqrt{v^2-2d(a_{1}+a_{2})}}{a_{1}+a_{2}}
Then we picked a d to make the square root 0 so that:
t = \frac{v\pm 0}{a_{1}+a_{2}}
t = \frac{v}{a_{1}+a_{2}}
Simply eliminating one root (ie. the negative root) because a negative time doesn't make any sense is usually a good idea, but here the negativity of an answer is totally arbitrary. You could put the origin anywhere you want and even get both roots to be negative. All you know is that one car...
Okay, fair enough. You should've used (1/2)(a_{1}+a_{2}) instead of just (a_{1}+a_{2}) for the coefficient of t^2 but the idea is there. The main point though to notice is that there are in fact two solutions because of the plus or minus sign. Thus we need to pick our value of d such that...
Note that you've set the positions equal to each other and since the positions are functions of time, this must be a quadratic equation for the times when the positions are equal. Collecting like terms gives:
0 = (1/2)(a_{1}+a_{2})t^2 + (-v)t + d
Now what can you do with this?
:rofl:
I meant the shapes of the graphs, but that's fine too. In fact, the curves are parabolas due to the t^2 terms. There are three possible ways that two parabolas can intersect; once, twice or never. We already know that we want them to intersect once, so what are you going to do with...
Yes, all of that is good! Summarizing what you just said, we have the following:
x_{1}=vt - \frac{1}{2}a_{1}t^2
x_{2}=d + \frac{1}{2}a_{2}t^2
You still need to tell me what they look like though :)
Okay, here's the thing. In physics questions, "just barely avoid" often means that the event would actually happen in real life, but for the purposes of the problem you can consider it as having been avoided. In other words, when you solve this question and find a d, if you actually used that d...
Oops, thanks for catching that typo :redface:
You should be subbing in i\theta into \sin(\theta)=\frac{1}{2i}(e^{i\theta} - e^{i(-\theta)}). In other words, simplify the following:
\sin(i\theta)=\frac{1}{2i}(e^{i(i\theta)} - e^{i(-i\theta)})
And then compare it to the definition of...
Alright, I'm going to show you a little bit of wizardry. We know that
e^{i\theta}=\cos(\theta) + i \sin(\theta) \; \; \; [1]
Now let \theta = -\theta so that:
e^{i(-\theta)}=\cos(-\theta) + i \sin(-\theta)
Using the facts that cosine is even and sine is odd (which you already knew)...
Okay, fair enough.
This isn't true. Try finding \sin(i\theta) first with Euler's formula, and then compare the result to what you were given for \sinh(\theta).
Say that the positions of car 1 and car 2 are x_1 and x_2 respectively. Imagine plotting x_1 and x_2 as a function of time on the same graph. What will the shapes of the curves be? What can you say about the number of times we want the curves to intersect?
Hint: x_1(t=0) = 0 and x_2(t=0) = d
You can't just assume that \omega = \sqrt{k/M}. This would be true if the mass was just sliding with no friction, but it will also have some rotational energy since it's a cylinder.
I would proceed by writing down the total energy in terms of the angle \theta of displacement, and then...
Okay. So you basically just have to define dm in terms of some small change in theta and some small change in r preemptively so you can just integrate them both up right away.
A is the area of the disk formed by integrating up all the rings
But what if I want to write it all out in one double integral? That must be possible...