PM2017 wrote:
How many medals are won in Science Olympiad each season?
Assuming all 50 states participate in scioly, medals are on average given out to top 4 placings (often varies from 3 to 6 but 4 seems pretty average to me). 2 divisions, 23 events each, 184 medals per tournament. Let's assume 10 invitationals, 6 regionals, and obviously 1 state tournament. This gives us 3 medals won each season per state, so 5 total.
Not exactly sure where to find this information, so I'll let PM2017 give us the answer.
Since someone else just posted, check the post above for the newest question (Don't want to create 2 strands of questions lol)
Time to waltz into the realm of craziness too late at night: Amount of individual pasta noodles created each year divided by the amount of energy in joules that the entire universe consumes each year, added on by the total area of the LOL map in DPI units.
Sleep is for the week; one only needs it once a week
Estimating E13 pasta noodles per year. Energy the universe consumes doesn't seem to make sense to me as a concept, but I'll go with E70 Joules. The map area is probably about E8; being a MOBA I assume the map isn't that huge. Ultimately the latter is the only thing that's relevant due to it be added - hence Fermi Answer: 8
The question doesn't actually make sense, since you can't exactly render a digital game map in DPI. The answer is probably in the range of E6-E8 though, since the energy in the universe is very high.
How many M&Ms are consumed each year by the number of people that could fill the stands of Wright State University's Nutter Center?
Unome wrote:
How many M&Ms are consumed each year by the number of people that could fill the stands of Wright State University's Nutter Center?
Not exactly sure but I'll just assume around 20,000 for the capacity of the Nutter Center. I don't think that many M&M's are eaten by a single person, so I'll assume around 5 packets a year? Around 20 pieces of M&M's per packet would gives us gives us 100 M&M's per person, so E6 total.
The Nutter Center can hold 10,400 people. Can't find how many are eaten by one person a year, but there are 400,000,000 produced each day, so 1.46E9 M&M's produced each year. Divide that by the world population gives us 19 M&M's a year per person. 10,400*19 gives us a final answer of E5.
What is the number of people over the age of 90 in the world divided by the number of children under the age of 15 in Australia?
7E9 people worldwide, about 1.5E-3 proportion are over 90 years old - therefore, E7 people worldwide over age 90. The population of Australia is in the range of 8E7, approximately 1/6 are under age 15 so around E7. Fermi Answer: 0
It's hard to find good data, but according to population pyramids approximately 20 million people worldwide are over the age of 90 - 2E7. 18% of Australia is under age 15, but I overestimated; the total population of Australia is around 25 million - hence, 4.4E6 people.
Dividing yields Fermi Answer: 1
How many times have AK-47s been fired in the entirety of human history through the end of 2016? (i.e. total number of shots)
7E9 people worldwide, about 1.5E-3 proportion are over 90 years old - therefore, E7 people worldwide over age 90. The population of Australia is in the range of 8E7, approximately 1/6 are under age 15 so around E7. Fermi Answer: 0
It's hard to find good data, but according to population pyramids approximately 20 million people worldwide are over the age of 90 - 2E7. 18% of Australia is under age 15, but I overestimated; the total population of Australia is around 25 million - hence, 4.4E6 people.
Dividing yields Fermi Answer: 1
How many times have AK-47s been fired in the entirety of human history through the end of 2016? (i.e. total number of shots)
oh dear
let's say that there exist E6 somewhat actively used AK46s. a magazine prob has abt E2 shots, and each actively used one probably fires on average E2 magazines a year. that leaves us a total of E10 shots in history
i am currently on a school computer so i'd rather not do the detailed research required to find a solution, and i do doubt that there is a readily available solution anyway, so i hope no one minds that i'm skipping over that
New question: how many layers of mylar would it take to stop a bullet?
i can't feel my arms wtf i think i'm turning into a lamp
voted least likely to sleep 2018, most likely to sleep in class 2017+2018, biggest procrastinator 2018
Mylar is about E-4 meters thick, and fairly strong. This is essentially a guess, but E3 layers should be sufficient - 40 cm should definitely do it, but I'm not quite sure about 4 cm.
Can't find good sources on this. Mylar seems to vary in thickness by a factor of about 20-30, and likewise with bullet strength.
How strong is a 5 mm wire of aluminum under tension in terms of a strand of human hair? (specifically terminal hair)
Mylar is about E-4 meters thick, and fairly strong. This is essentially a guess, but E3 layers should be sufficient - 40 cm should definitely do it, but I'm not quite sure about 4 cm.
Can't find good sources on this. Mylar seems to vary in thickness by a factor of about 20-30, and likewise with bullet strength.
How strong is a 5 mm wire of aluminum under tension in terms of a strand of human hair? (specifically terminal hair)
Mylar is about E-4 meters thick, and fairly strong. This is essentially a guess, but E3 layers should be sufficient - 40 cm should definitely do it, but I'm not quite sure about 4 cm.
Can't find good sources on this. Mylar seems to vary in thickness by a factor of about 20-30, and likewise with bullet strength.
How strong is a 5 mm wire of aluminum under tension in terms of a strand of human hair? (specifically terminal hair)
By 5mm, do you mean diameter or length?
Diameter, sorry. I was intending for a comparison of tensile strengths.
Mylar is about E-4 meters thick, and fairly strong. This is essentially a guess, but E3 layers should be sufficient - 40 cm should definitely do it, but I'm not quite sure about 4 cm.
Can't find good sources on this. Mylar seems to vary in thickness by a factor of about 20-30, and likewise with bullet strength.
How strong is a 5 mm wire of aluminum under tension in terms of a strand of human hair? (specifically terminal hair)
By 5mm, do you mean diameter or length?
Diameter, sorry. I was intending for a comparison of tensile strengths.
Ok. Good. That's what I thought
so i read somewhere that hair is comparable in strength to an aluminum wire of the same dimension. So essentially, this is asking to divide 5mm by the diameter of a human hair. I'm going to estimate this to be 0.01mm (probably too thin tbh, but since it will be rounded up overall at the end, since the five will cause a round-up, I'll go with it.). Since this is an easy calculation, I'll just do 5mm/0.01mm, which is 500, or Fermi answer 3. But this needs to be squared, so the fermi answer is 6.
This source (http://www.spearfish.k12.sd.us/~sgabriel/physics/chap7/Tensile%20Strength%20of%20Metals.pdf) says aluminum wire has a tensile strength of 2.4e8 N/m^2 (or 240 megapascals) A human hair a tensile strength of approx 200-250 mPa according to this wikipedia (https://en.wikipedia.org/wiki/Ultimate_tensile_strength) so, I'll just say that I was right in my assumption that they are equal in tensile strength, as one range includes the other value. So then, its just a question of division of the cross-sectional areas. A a circle with diameter 5mm has area 19.63 mm^2. A terminal hair is, well, I couldn't find a source, so I just went with this site, whose credibilty I am not sure of... (http://www.schwarzkopf.com/en/hair-care/split-ends/hair-dictionary.html) who say that a normal hair is between 0.06 mm and 0.08 mm, so I'll go with 0.07mm, for an area of ~0.0038 mm^2. Dividing 19.63mm^2 by 0.0038mm^2 yields 5165, which just barely takes it to 4. I must have done something seriously wrong in my calculations, but I can't seem to figure out what...
What is the mass of all the rivets in the wings of all the Boeing 747s ever made in terms of moles of botulinum toxin?
