# Staysys

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9

1. ## Looking Back Through Time

Ahh, yes you are correct sorry about that. So it should be changed to say that the mirror would have to be at least as big as the radius of the Earth.
2. ## Looking Back Through Time

If you want to see the entire Earth then it would have to be a mirror at least as big as Earth itself. It's one big misunderstanding about mirrors that if you get farther away you will have a wider field of vision, but that's not true. You will see the same amount of stuff inside the mirror no matter how far away it is, so if you are just using a handmirror you'll only be able to see your own face (if it's perfectly perpendicular to your vision). But assuming you could make a mirror larger than the diameter of the Earth and you could launch it out into space so that after it is 100 lightyears from Earth its orientation will be perfectly pointing back at Earth, then 100 years after THAT event you will be able to see an image of Earth in the past. NOTE: It won't be 200 years because the mirror probably took longer than 100 years to get 100 lightyears away. But you could do calculations to make it so that you could see 200 years into the past if that is what you want. The problem is the Earth will be moving in a complex orbit relative to the mirror, so the mirror won't always be facing the Earth, maybe if it was rotating about all its axis you would be able to get images every period of its rotation.
3. ## Does 'outside of time and space' have any logical meaning whatsoever?

You don't know that. However, something existing "outside" of space does not make sense. It is a difficulty of semantics, because what (some) people probably have in mind when they say it probably does actually make sense. But if you take it literally at face value it does not. Because "outside" refers to a place, and "places" to us require space. It would probably be more accurate for them to say that "it exists, but not in space or time", and that is a statement which currently may or may not be possible.
4. ## Velocity of melting icecream

oh right, I don't know how I missed that. I guess because I arrived at the same answer OP said it was supposed to be. But, ya I think you are right about using a cylinder then. I'm not sure if the OP is even still checking his thread though so I won't bother getting an actual number right now though.
5. ## Velocity of melting icecream

umm.. did you see anything wrong with the solution I posted in my first post?
6. ## Velocity of melting icecream

I don't think that works. If you can just treat it as a falling cylinder then after about 10 seconds the chocolate would be flowing out at a speed of near 100m/s. That clearly doesn't happen, because the chocolate slowly oozes it's way to the hole opening and then falls out at a low speed, it isn't being accelerated at freefall toward the hole.
7. ## Velocity of melting icecream

You can just use conservation of energy (kinetic + potential = constant) (1/2)mv^2 + mgh = constant dividing by volume this change the mass terms to density: (1/2)pv^2 + pgh = constant This should be constant at all places in the fluid so we consider the very top of the fluid as the first point. Here v = 0. The second point we consider is the hole at the bottom of the cone where h = 0. so we can now set the equation equal at the two points: pgh = (1/2)pv^2 and just solve for v: v^2 = 2gh v = sqrt(2gh) ========================================== ========================================== If you google this it seems everyone tells you to use Bernoulli's equation, which looks very similar to what I had: (1/2)pv^2 + pgh = constant except with an additional P on the left side for the pressure at that point, and they say P is the same at both points so it cancels anyway.... But, honestly it seems wrong to me to use Bernoulli's equation for this. I don't see why P would be the same at both points, and also Bernoulli's equation is only supposed to be valid for two points along the same stream line, which doesn't seem to work in this case. If anyone can explain to me why Bernoulli's equation would work here please do! Otherwise I'll just assume it's not appropriate.
8. ## Rectangle tiles resize algorithm

Well if you know the width, height, and starting coordinate of each rectangle then you can find the resized values as follows: Say these are your variables -------------------------------- totalWidth = (width of entire rectangle) totalHeight = (height of entire rectangle) totalWidthNew = (width to resize entire rectangle to) totalHeightNew = (height to resize entire rectangle to) width = (initial width of the Nth small rectangle) height = (initial height of the Nth small rectangle) x = (initial x coordinate of Nth small rectangle) y = (initial y coordinate of Nth small rectangle) Then you just recalculate width, height, x, and y by doing: newWidth = (width / totalWidth) * totalWidthNew newHeight = (height / totalHeight) * totalHeightNew newX = (x / totalWidth) * totalWidthNew newY = (y / totalHeight) * totalHeightNew Just reassign width, height, x, and y for each of the current values of each small rectangle, and recalculate their newWidth, newHeight, newX, and newY values like above. Please ask if anything is unclear. NOTE: I just came up with all of this off the top of my head so I haven't tested it, but I am pretty confident it will work like you want.
9. ## Doppler Effect In Sound

Just remember that the frequency is only changing relative to the observer, so you can use the normal doppler effect equations even for an accelerating body, just plug in the instantaneous velocity of the observer at the moment he is measuring the frequency and the original frequency that was emmitted by the source and that will tell you what the accelerating observer sees.
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