Quick Answers to small questions
- Thread starter peter grimes
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This morning my wife and I got into a argument discussion. She administers the 'Dollar Tuesday' event where she works. Simple: Everybody who wishes to participate writes their name on a $1 [US] bill. All the bills go into a bag. She draws out a bill at the end of the day, and whoever's name is on the bill gets the entire bag. The prize is usually around $100. But this is the last day of work until the new year, so it's $5 Super-Tuesday!
The conflict arose when she mentioned that she was going to stop by the bank on the way to work to withdraw $200 worth of $5 bills, since most of the people would have $20 bills on hand, and she'd have to make change. I thought that was excessive. 40 bills?? Not necessary. But then she asked how many bills she should withdraw. I thought 40 was too many, but 10 too few. Is there a way to determine the optimal amount to have on hand?
For those who are unfamiliar with US currency, the denominations are 1, 5, 10, and 20.
The conflict arose when she mentioned that she was going to stop by the bank on the way to work to withdraw $200 worth of $5 bills, since most of the people would have $20 bills on hand, and she'd have to make change. I thought that was excessive. 40 bills?? Not necessary. But then she asked how many bills she should withdraw. I thought 40 was too many, but 10 too few. Is there a way to determine the optimal amount to have on hand?
For those who are unfamiliar with US currency, the denominations are 1, 5, 10, and 20.
This is complicated by the fact that if you change the 10's before the 20's, you can use the 10's you now have to give change for the 20s
But if you change the 20's first you have give all the change out in 5's
Is it easier to assume everyone has 20's ?
But if you change the 20's first you have give all the change out in 5's
Is it easier to assume everyone has 20's ?
Mise
isle of lucy
I think it's optimal to simply guess 
I'm sure she can find a use for the $5s some other time; she's a woman afterall.
I'm sure she can find a use for the $5s some other time; she's a woman afterall.
Only by making assumptions about how many will be playing, and what they'll bring.
Don't forget, for everyone who pays with something bigger than a $5 note, you need to supply 2 $5 notes from the bank. 1 for change, one to have a name written on and go in the bag. Because those who pay with $5 notes don't add to your change supply, since their note goes in the bag too.
Assuming there is nearly 100 participants, and assuming most people won't have a $5 of their own, 40 doesn't sound like enough, as that only covers 20% of participants not supplying their own.
Veritass
Emperor
I was musing on the way home that an object being "sucked" by a vacuum was really being "pushed" by the collision of the molecules on the other side. I assume that the maximum theoretical speed that such an object could attain by being pushed would be the velocity of the atoms bouncing around on the other side of it. For air at sea level, what is that speed? I realize that in a closed system, as the air pushed and expanded, it would lose its velocity, but assuming a long tunnel of vacuum on one side and infinite air intake on the other side, what would be the maximum velocity attainable?
The light speed is the limit. The velocity distribution of a gas is the Maxwell–Boltzmann distribution. (Actually, if you get close to light speed it's not applicable any more and you need a relativistic version, and that's where the light speed limit comes from). This distribution has a very low, but non-zero probability for the particles to be at any high speed. So in an infinitely long tunnel with perfect vacuum the light speed would be the only limit.
In reality however, there is no perfect vacuum and you would have a few particles in that vacuum creating a bit of resistance that would eventually balance out the acceleration from the very fast particles. How fast that would be would depend on the actual vacuum (and the resistance of the air flowing into the tunnel and so on)
What you can calculate is the mean velocity of particles in air. For nitrogen at 300K that would be around 470 m/s (ignore the 600m/s figure on that Wikipedia page, it's wrong). The realistic maximum speed would be on that order of magnitude.
In vaccum. The speed of light in a medium has no relativistic significance. That's why it is possible for particles to be faster than the speed of light in that medium and create Cherenkov light.
My wife [why do all my questions start that way??] has a cousin named Jason. She calls him her 'double cousin' because her mom & his mom are sisters, and her dad & his dad are brothers. Two siblings married a pair of siblings.
From the perspective of heredity, are not my wife and Jason the equivalent of siblings?
My reasoning for thinking of them as brother and sister is that she and he have the same grandparents - just as my sister and I do.
Opinions?
From the perspective of heredity, are not my wife and Jason the equivalent of siblings?
My reasoning for thinking of them as brother and sister is that she and he have the same grandparents - just as my sister and I do.
Opinions?
The child of your parent's sibling is a cousin. All cousins share grandparents. E.g., my father had five siblings. They all had children, who are my cousins. All of us cousins share the same grandparents - my father's parents.
In the specific case you mentioned, your mom and Jason might be closer, genetically, than is common with cousins, but not necessarily. It is a probability only, and given how distinctly different the DNA can be between any two given siblings, I doubt the probability has increased by any significant amount in this case.
In the specific case you mentioned, your mom and Jason might be closer, genetically, than is common with cousins, but not necessarily. It is a probability only, and given how distinctly different the DNA can be between any two given siblings, I doubt the probability has increased by any significant amount in this case.
well, on average siblings have 50% of genetic information in common same is true with parents and children. Offspring of a sibling of yours (niece/nephew) or a half-sibling of yours (one parent in common) has 25% in common and offspring of a sibling of your parent (cousin) has 12.5% in common. Given no relations between the original families.
Now make this case:
Brother A and Sister C have child E
Brother B and Sister D have child F
E has 50% in common with C and A
F has 50% in common with D and B
F has 25% in common with A of which E has half in common
F has 25% in common with C of which E has half in common
F has 25% in common with E
making the genetic relation similar to a Aunt/niece relation or a half-sibling relation rather than a sibling relation or a cousin relation.
Now make this case:
Brother A and Sister C have child E
Brother B and Sister D have child F
E has 50% in common with C and A
F has 50% in common with D and B
F has 25% in common with A of which E has half in common
F has 25% in common with C of which E has half in common
F has 25% in common with E
making the genetic relation similar to a Aunt/niece relation or a half-sibling relation rather than a sibling relation or a cousin relation.
I'm a woodworker, and I save off-cuts to burn in the firepit in our back yard. But getting them home presents a problem. I have a car, but I don't like to use it unless absolutely necessary (I live in New York City). So I take several small loads each week. Sometimes I ride my bike, other times I use the subway. It occurs to me that one method may be easier on me than the other. I wonder if we can figure this out?
So far, here's what I think we need to know:
-Mass of the load
-mass of bike
-mass of Peter
-distance from shop to apartment by bike
-change in elevation along bike route
-average speed of bike trip
-walking distance
-change in elevation along walking route (stairs involved)
-average speed of walking route
Am I missing anything? I'd love to be able to figure this out
So far, here's what I think we need to know:
-Mass of the load
-mass of bike
-mass of Peter
-distance from shop to apartment by bike
-change in elevation along bike route
-average speed of bike trip
-walking distance
-change in elevation along walking route (stairs involved)
-average speed of walking route
Am I missing anything? I'd love to be able to figure this out
That's not an easy calculation.
When walking, you are constantly using energy to hold the wood. It should be possible to compute the force needed to hold the load, for the time of the trip. Presumably the energy needed to pull the load as you walk is negligibly similar to just holding it. And similarly you should be able to compute the energy needed to lift the load by the change in elevation. However, there is an issue here. Walking is not like a frictionless roller coaster experiment, where the effort is proportional to the change in elevation. When you walk down a hill, you are spending significant energy to keep yourself from speeding up. So I'd estimate that the effort to walk down hill is near the energy to walk on flat ground. But the energy to walk up hill should be the same as roller-coaster physics. So you need to calculate the energy based on the total virtical distance climbed, not the change in elevation.
With biking, your load is on the bike rack, which means you are not holding it all the time. In fact part of the time you are not pedaling and therefore not spending any effort. The amount of energy used is proportional to resistance of your pedals, which is not constant. If you are constantly stopping at lights and such, then topological awareness will not give you much of an idea of how much energy you're using are doing. So you are left with estimating the typical resistance, and the fraction of time pedaling (or distance, it should give the same result). If you can replicate the pedaling you do on the trip on an exercise bikes, then it should tell you the calories burned.
When walking, you are constantly using energy to hold the wood. It should be possible to compute the force needed to hold the load, for the time of the trip. Presumably the energy needed to pull the load as you walk is negligibly similar to just holding it. And similarly you should be able to compute the energy needed to lift the load by the change in elevation. However, there is an issue here. Walking is not like a frictionless roller coaster experiment, where the effort is proportional to the change in elevation. When you walk down a hill, you are spending significant energy to keep yourself from speeding up. So I'd estimate that the effort to walk down hill is near the energy to walk on flat ground. But the energy to walk up hill should be the same as roller-coaster physics. So you need to calculate the energy based on the total virtical distance climbed, not the change in elevation.
With biking, your load is on the bike rack, which means you are not holding it all the time. In fact part of the time you are not pedaling and therefore not spending any effort. The amount of energy used is proportional to resistance of your pedals, which is not constant. If you are constantly stopping at lights and such, then topological awareness will not give you much of an idea of how much energy you're using are doing. So you are left with estimating the typical resistance, and the fraction of time pedaling (or distance, it should give the same result). If you can replicate the pedaling you do on the trip on an exercise bikes, then it should tell you the calories burned.
Since there are several loads, you could just find out by experiment. Take the first load one way and the second load the other way and compare how exhausted you are after each of them. This isn't going to be a very accurate measurement, but it's way more accurate than a theoretic calculation would be.
Describing the effort of carrying something is quite difficult, because from the physics point of view, you're hardly doing any work at all, but due to the way muscles work, they will get exhausted nevertheless.
In the end the experiment is the deciding thing in science. It doesn't help, if one way should be easier on you, when you feel the other way afterwards.
Describing the effort of carrying something is quite difficult, because from the physics point of view, you're hardly doing any work at all, but due to the way muscles work, they will get exhausted nevertheless.
In the end the experiment is the deciding thing in science. It doesn't help, if one way should be easier on you, when you feel the other way afterwards.
Atrebates
Prince
basic log rules: (Log A)(LogB) = Log AB
(Log A)/(Log B) = Log (A/B)
*edit* Log A^2 = 2Log A
Question becomes trivial with these rules.
(2 Log (x))/(Log ab) = cd
*corrected*
See below for actual. This is proof that even finalist physicists can be outright ******** and trivial errors are actually really really important. Sorry about that...
Hope this helps!
(Log A)/(Log B) = Log (A/B)
*edit* Log A^2 = 2Log A
Question becomes trivial with these rules.
(2 Log (x))/(Log ab) = cd
*corrected*
Spoiler :
See below for actual. This is proof that even finalist physicists can be outright ******** and trivial errors are actually really really important. Sorry about that...
Hope this helps!
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