# Help me with my Physics!

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what i would do is create a triangle pyramid of the ballon oragami with the egg being in the certer one, that way you have air cushions on all sides regardless of how it dropped, id put a pic up but i hate my photoshop skills

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My dad had a good idea - make a sort of sphere around the egg, supported by pillars from the egg to the sphere.

The problem is that they won't stop the acceleration of the egg once the sphere hits the ground - the sphere will collapse due to the flimsiness of the paper.

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what i would do is create a triangle pyramid of the ballon oragami with the egg being in the certer one, that way you have air cushions on all sides regardless of how it dropped, id put a pic up but i hate my photoshop skills
Balloon origami?
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i cant believe i still remember how to make those

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Parachutes not allowed. I had thought of a cone as well, but it needs to last through several drops.

I thought of making one with little paper protrusions all along the outside kind of making it look like a spiked ball.

Don't think the paper would be strong enough to make it work though.

Can't use the five sheets of paper to make five cones and just hope that you only get tested on five drops?

I just can't see anything made of paper being reused that way. Even if you make shredded paper padding, after each drop it'll pad less and less.

I seriously doubt that there'd be that much sideways pressure on the egg from the cone, but what do I know? I'm a chemistry major.

I think the paper/tape would break before the egg anyways. I mean, have you made an omlette? Those little guys are tough!

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Bah, all I did in high school physics was light up a few lights and wave a few ropes around. The cone idea is a good start but as mentioned, the thing has to be dropped multiple times (presumably without modification between drops). Stuffing the tip with paper as Bahamut mentioned may work but as Kaiger said, it might make the structure more solid. Using shredded paper instead won't work for the second drop because the initial drop would have compressed the whole tip into a brick wall basically. Having 5 individual cones for each test wouldn't work either because the crumpling effect would be severely reduced if the tip isn't properly reinforced.

If you can at least touch your design after every drop, you could panel beat the design back into shape. Yeh, the external structure would be weakened but if you had shredded paper reinforcing the tip of the cone, you could get those back into a loose mess instead of a brick wall.

Playing around with air pressure with only 5 sheets of paper and some tape won't work too well either (you could give the damn thing wings if you weren't limited). A parachute wouldn't work at 1m, not enough air resistance so don't even consider that idea (you're not allowed to anyway, so it doesn't matter).

No chance you could tamper with the egg eh? Soaking the whole thing in vinegar would do wonders...or enchance the exploding effect when it hits the floor. Too bad you can't turn normal paper into bubble wrap either. If only Macgyver were part of the OCR forums, he'd save your egg with bits of shoelace and some fluff off the back of your shirt.

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I can touch the design after each drop.

Hmph. I will have to sleep on this.

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protest, egg drops are inhumane against unborn chickens. or tell them you're religion forbids it. they'll have to give you an 'a'

or say screw it and roll the egg in a paper, and chuck it at someone you really dislike. followed by 'damn, i guess my design didn't work'

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protest, egg drops are inhumane against unborn chickens. or tell them you're religion forbids it. they'll have to give you an 'a'

or say screw it and roll the egg in a paper, and chuck it at someone you really dislike. followed by 'damn, i guess my design didn't work'

I kind of need an A.

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i believe the balloons are also called waterbombs

http://www.things2make.com/Things2make_files/Instructions%20over%205/waterbombs.htm

just blow up the bombs and don't add water, itll be like an air cushion, make 5, put the egg in the center and tape the others so that it looks like a 3 sided pyramid, that way, any way it falls it has support, not sure if the phsyics will work out, but it has been 5 years since i took that course

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I have a simple idea. Make something like a paper cube, but angle the vertical sides a little bit outwards so they give in slightly when the egg lands on it.

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I have a physics question too actually.

A 200 gram piece of metal is 30 degrees Celsius and is put into a beaker of boiling water, how much heat energy does the metal absorb?

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Oh i did this and won the competition at my school

you make a big cylinder, make a coen and put it titled in the cylinder

make little cylinders aaround the side for stabalization and then cut the rest of your paper into shreds and put it between the cone and the cylinder and some inside the cone too

works like a charm

boil the egg?

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Black_Lit

This is for the metal question. I'm pretty sure that it depends on how length of time it's in there, its material properties and it's dimensions. However looking through my old heat transfer book I found that for a convective heat transfer which is what you have, the genral formula would be something like this q"=h(Ts-Tinfinity). q" is the heat flux transfered into the body, h is the convective heat transfer coefficient, Ts is the surface temperature of the object, and Tinfinity is the temperature of the fluid that its placed into. Normally the convective heat transfer coefficient is a bitch to calculate, but for a boiling fluid, this value can range between 2,500 and 10,000. Basically you'll have to estimate this value. I don't know if this is way too much, or if this is what you're looking for. If anybody can see a simpler way, be my guest. Peace.

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I have a physics question too actually.

A 200 gram piece of metal is 30 degrees Celsius and is put into a beaker of boiling water, how much heat energy does the metal absorb?

If the metal is left in there until equilibrium is reached, the metal and water will eventually be the same temperature. In other words, you set up two equations, one for the metal with the initial temperature at 30, and the final temperature being what you want to calculate, and the water with an initial temperature at 100, and the final temperature being the same value.

You can manipulate the equations so that you are trying to calculate the final temperature for each, then set them equal to each other. I'm guessing the only unknown in the equation will be the change in heat energy on each side. Since the energy will be lost in one equation and gained on the other, you probably will be able to move them both to the same side of the equation and end up with 2 times the unknown energy, divide both sides by two, and sort out what the heat transfer is from the masses of metal and water and the specific heats and initial temperatures of both substances.

I'm doing this without actually seeing the equations in front of me, but I'm betting that's it.

There could also potentially be a twist. If the water is boiling, some of the water in an enclosed beaker could potentially be steam. Steam at the same temperature as water has higher thermal energy, because energy is required to make a phase change from liquid to gas state. In that case you'd have to take into account the proportion of water that is steam and add that energy in to your equation.

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I have a physics question too actually.

A 200 gram piece of metal is 30 degrees Celsius and is put into a beaker of boiling water, how much heat energy does the metal absorb?

If the metal is left in there until equilibrium is reached, the metal and water will eventually be the same temperature. In other words, you set up two equations, one for the metal with the initial temperature at 30, and the final temperature being what you want to calculate, and the water with an initial temperature at 100, and the final temperature being the same value.

You can manipulate the equations so that you are trying to calculate the final temperature for each, then set them equal to each other. I'm guessing the only unknown in the equation will be the change in heat energy on each side. Since the energy will be lost in one equation and gained on the other, you probably will be able to move them both to the same side of the equation and end up with 2 times the unknown energy, divide both sides by two, and sort out what the heat transfer is from the masses of metal and water and the specific heats and initial temperatures of both substances.

I'm doing this without actually seeing the equations in front of me, but I'm betting that's it.

There could also potentially be a twist. If the water is boiling, some of the water in an enclosed beaker could potentially be steam. Steam at the same temperature as water has higher thermal energy, because energy is required to make a phase change from liquid to gas state. In that case you'd have to take into account the proportion of water that is steam and add that energy in to your equation.

If there's steam in the beaker, it can't transfer to the piece of metal. If the steam isn't in contact with the metal, there's no means for a heat transfer to occur. If the metal was half submerged, you would have to worry about it.

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My dad just had a sweet idea. I will post pics later.

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I have a physics question too actually.

A 200 gram piece of metal is 30 degrees Celsius and is put into a beaker of boiling water, how much heat energy does the metal absorb?

If the metal is left in there until equilibrium is reached, the metal and water will eventually be the same temperature. In other words, you set up two equations, one for the metal with the initial temperature at 30, and the final temperature being what you want to calculate, and the water with an initial temperature at 100, and the final temperature being the same value.

You can manipulate the equations so that you are trying to calculate the final temperature for each, then set them equal to each other. I'm guessing the only unknown in the equation will be the change in heat energy on each side. Since the energy will be lost in one equation and gained on the other, you probably will be able to move them both to the same side of the equation and end up with 2 times the unknown energy, divide both sides by two, and sort out what the heat transfer is from the masses of metal and water and the specific heats and initial temperatures of both substances.

I'm doing this without actually seeing the equations in front of me, but I'm betting that's it.

There could also potentially be a twist. If the water is boiling, some of the water in an enclosed beaker could potentially be steam. Steam at the same temperature as water has higher thermal energy, because energy is required to make a phase change from liquid to gas state. In that case you'd have to take into account the proportion of water that is steam and add that energy in to your equation.

If there's steam in the beaker, it can't transfer to the piece of metal. If the steam isn't in contact with the metal, there's no means for a heat transfer to occur. If the metal was half submerged, you would have to worry about it.

If the system is closed, the steam-water mixture will have to turn fully to water before the heat transfer to the metal will allow the temperature to start decreasing. If the problem is one of equilibrium, you can't end up with non-boiling water and steam in the same closed system at equilibrium.

The original poster was nowhere near specific enough on the terms of the question to settle exactly what we are trying to solve and under what conditions, however. Details are important in physics. You can't half-tell someone about a question and don't tell them all your information or the actual wording of what is being asked and expect to get a single, straightforward answer. Maybe it's heat transferred after a period of time, in which case you were probably on the right track, maybe it's heat transfer after equilibrium has been reached, in which case I was probably on the right track. Original poster, are you following this, or what?

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I have a physics question too actually.

A 200 gram piece of metal is 30 degrees Celsius and is put into a beaker of boiling water, how much heat energy does the metal absorb?

If the metal is left in there until equilibrium is reached, the metal and water will eventually be the same temperature. In other words, you set up two equations, one for the metal with the initial temperature at 30, and the final temperature being what you want to calculate, and the water with an initial temperature at 100, and the final temperature being the same value.

You can manipulate the equations so that you are trying to calculate the final temperature for each, then set them equal to each other. I'm guessing the only unknown in the equation will be the change in heat energy on each side. Since the energy will be lost in one equation and gained on the other, you probably will be able to move them both to the same side of the equation and end up with 2 times the unknown energy, divide both sides by two, and sort out what the heat transfer is from the masses of metal and water and the specific heats and initial temperatures of both substances.

I'm doing this without actually seeing the equations in front of me, but I'm betting that's it.

There could also potentially be a twist. If the water is boiling, some of the water in an enclosed beaker could potentially be steam. Steam at the same temperature as water has higher thermal energy, because energy is required to make a phase change from liquid to gas state. In that case you'd have to take into account the proportion of water that is steam and add that energy in to your equation.

If there's steam in the beaker, it can't transfer to the piece of metal. If the steam isn't in contact with the metal, there's no means for a heat transfer to occur. If the metal was half submerged, you would have to worry about it.

If the system is closed, the steam-water mixture will have to turn fully to water before the heat transfer to the metal will allow the temperature to start decreasing. If the problem is one of equilibrium, you can't end up with non-boiling water and steam in the same closed system at equilibrium.

The original poster was nowhere near specific enough on the terms of the question to settle exactly what we are trying to solve and under what conditions, however. Details are important in physics. You can't half-tell someone about a question and don't tell them all your information or the actual wording of what is being asked and expect to get a single, straightforward answer. Maybe it's heat transferred after a period of time, in which case you were probably on the right track, maybe it's heat transfer after equilibrium has been reached, in which case I was probably on the right track. Original poster, are you following this, or what?

Agreed, more information is required, like we don't know if this is silver or a chunk of lead.

We did it.

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Sorry about not having enough info to solve the problem, I was posting by memory =/

So here is the EXACT problem, though it still doesn't help from my perspective.

A 250 gram glass marble is taken from a freezer at -23 degrees C and placed into a beaker of boiling water [just consider it as 100 C]. How much thermal energy is absorbed by the marble?

I don't have a clue how to solve this, we never did one in class or lab lack this. The ones we did gave us the mass of the water. Anyone have a clue? (I don't think I need the mass of the water though).

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is that q = m * c * (T2 - T1) ?

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is that q = m * c * (T2 - T1) ?

Problem is, you don't know the final temperature, and to solve it using that equation would require you to know the mass of the water.

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*woot first post *

The m needed would be the marble's weight, 0.250 kg. Also, the you can assume that if the water keeps boiling, the final temperature of the marble will be 100 degrees C.

Q = m[marble] . c[marble] . (100 - T[marble])

m[marble] = 0.250 kg

c[marble] would be = c[glass] (look that one up)

T[marble] = -23 C

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