1. Estimate the computer power consumed in the building. Answer should be in Watts [W] and Horse Power [HP]
- Estimated power consumption for a computer 130 watts per hour. Source
- 28 computers per computer lab + teacher’s computers in classrooms. Seven computer classrooms + 80 computer lab = 276, rounded to 280 and multiplied by five (five floors) = 1400. This will account for there not being as many computer labs on all floors. Plus there is the gaming lab on the fifth floor. There are also offices and wasted floor space on some floors. 1400 times 130 watts = 182000, rounded to 185,000 Watts. This should give a rough estimate of power consumption per hour in Watts.
- Using Google’s power converter I input 18500 Watts which equals 248.09 British Horse Power, Rounded to 250 BHP.
2. Estimate the average building height on your campus. Answer should be in feet [ft] amd meter [m]
- A quick search provided that the average height of a story is 10 feet. I estimated and that is fairly accurate. But I will tack on two feet to account for drop ceilings and the fact that the ceilings are office building height. So 12 feet per story, 12 times 5 = 60 feet plus 10 feet for roof and other building features. 70-ish feet. I think that this building is taller than most of the others, but there are others about this height. I'd say that 65 feet is fairly reasonable average for all of the buildings on Central Campus
- There are 3.3 meters per foot and using Google’s feet to meters converter we get approximately 20 meters.
This next part was done by my classmate: Yang Xu.
3. Estimate the height of your building. The answers should be in feet [ft] and meters[m]
- 20 meters which is about 65.62 feet. The height of each floor is about 3 meters. There are 5 levels in this building so it’s about 15 meters. Including the top of the building and the error. I estimate the height of this building is about 20 meters (65.62 feet).
4. Estimate the weight of air in the building. The answers should be in Newtons [N] and pound force [lbf]
- The surface area of each flood is about 51 m * 30 m = 1530 m2. The height of each flood is about 3 m. So the volume of this building is about 51m*30m*3m*5= 22950 m3. The room temperature is about 25 oC so the density of air is about 1.184 kg/ m3
- So the weight of the air in kg is 27173.8 kg. Chang to Newtons is about 266293.44 N (59865.17 lbf)
The rest of this work is my own
- First I needed to find the square footage of the building. I went onto Google Maps and looked up Central Campus.
Source: Google Maps
- I found the building and using the measuring tool I found out the square footage. The length is ≈ 175 ft and the width is ≈ 112.5 ft. So the rough square footage is 19,687.5 ft2, ≈19,700 ft2. I multiplied the square footage by the height and got the cubic area. ≈ 1,379,000 ft3.
- A quick search for the weight of a cubic foot air turned up this, 0.0807 lbs. per cubic foot. I multiplied the cubic footage by 0.0807 lbs. ≈ 111,285 lbs. Another search tuned up that the Moon's gravity is 83.3% of Earth's. ≈ 92,700 lbs. on the Moon.
- Again, another search sows that the gravity on Mars is ≈ .38 of Earth's. ≈ 42288 lbs. on Mars.
- This is assuming that the building is completely hollow and is very thin walled.
6. Estimate the amount of grassy area in Charlotte in [km2] and [mi2]
- The Charlotte-Metro area covers 3,149 sq mi (8,160 km2), sourced from Wikipedia. I'm going to say that 1/3 of the Charlotte area is grassy. ≈ 1039 mi2 or 2693 km2
7. Estimate how long it would take an astronaut to travel to the moon using current technology.[s] and [hours]
- The distance to the moon at it's closest point is ≈ 225,623 miles. A rocket can go at 4.9 miles per second, or 17600 miles per hour. If you divide 225,623 miles by 4.9 miles per second you get ≈ 46,045 seconds.
- If you divide 225,623 miles by 17600 miles per hour you get ≈ 12.82 hours. to double check I multiplied 12.82 hours by 36000 seconds and I got 46150 seconds, which is only 105 seconds of difference.
- The rocket would not need to slow down as the question does not ask about or mention saftey.
