A police is directed towards thief both having mutually perpendicular velocities initially but police is always directed towards thief. We need to find time when police catches the thief
There's a trick for doing these types of questions, but I can't remember it ...
I'll have to go away and think about it. Monday mornings are very busy for me and in a few minutes I have to start work and I'll be working for several hours.
@Jasmine there are many to solve this particular, problem as you started with a projectile problem, so it would be easy for us to go through that. First Let fake mod, show his efforts.
Let the ball thrown from. The ground has two component, $vcosx$, $vsinx$, .
Horizontal component of both projectile will be same throughout journey.
So ball which was thrown from top of a tower with velocity u.
Now let say they collide at time t.
So we get distance =speed. Time.
So. At this time. Their horizontal component are equal so $u=vcosx$.
Now I you can find the time, by putting this value in the tow equation.
I got a question similar to police thief,. Which you was asking
Sorry typo were.
@Jasmine we can discuss it when you want.
@Jasmine ah, sorry, in projectile, we if the height of tower let 10m then vertical distance traveled by 1 which is on ground is let say x, and other h-x.
Or you can use the relative like u-vcosx will be the velocity h be height, t=h/(u-vocsx)
So I know that two particles can be entangled in a quantum way, but is it possible that more than two particles be entangled in a quantum way? Most descriptions provide with two-particles cases, so I wonder. (It's hard to think of three particles entangled in spin, or so.)
I think the problem is that until you do QM at university it's hard to understand what entanglement actually means. So you are trying to describe something you haven't learned about.
I suspect what it means is you adjust $V_{bb}$ to make $I_c = 2mA$ i.e. half the saturation value. Then you calculate the value of $V_i$ that just saturates.
@AdvilSell If you look at the output voltage $V_c$ then it can vary from 1V to 5V. It can't go below 1V because that's the BE voltage drop, and it can't go above 5V because that's the max supply voltage. OK so far?
@AdvilSell well, OK, it does satisfy that equation while the other answers don't. But 3.5V would oversaturate the transistor so I don't understand why they say that.
$V_{be}$ is a regular forward biased PN junction, so it has the usual voltage drop across it. The question says 1V though I would have thought 0.6V was more typical for a silicon transistor.
But anyhow, $V_{cb}$ is a reverse biased junction that only allows current to flow due to the carriers created by the base current, and I think $V_{cb}$ can fall to effectively zero when the transistor is saturated.
That's effectively what I used when I said $I_cR_c = V_{cc} - V_{be}$.
The Hertzsprung–Russell diagram, abbreviated as H–R diagram, HR diagram or HRD, is a scatter plot of stars showing the relationship between the stars' absolute magnitudes or luminosities versus their stellar classifications or effective temperatures. More simply, it plots a star's luminosity (brightness) against its temperature (color).
The diagram was created circa 1910 by Ejnar Hertzsprung and Henry Norris Russell and represents a major step towards an understanding of stellar evolution.
The related color–magnitude diagram (CMD) plots the apparent magnitudes of stars against their color, usually...
The brightness is plotted on the y axis and it varies by many orders of magnitude.
there should be some cutoff magnitude. I don't know exactly what's the best cutoff, but I would guess somewhere around 4.0 to 5.0 apparent magnitude. If the cutoff is 4.0, then that gives us 3 tiers to look at (1 to 2, 2 to 3, 3 to 4). Are each of those tiers uniformly distributed?
The apparent magnitude is how bright the star looks to us here on Earth, so it's a function of two factors: 1. how bright the star is i.e. the absolute magnitude 2. how far away the star is
And just to be confusing higher magnitudes, i.e. more positive numbers, are dimmer.
The brightest star as seen from Earth is Sirius with a magnitude of -1.46.
The dimmest star visible with the naked eye has a magnitude of +6.5, but far dimmer stars can be seen with telescopes.
The dimmest objects visible are very distant galaxies with an apparent magnitude of about +30.