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It will take 300000 btus at 12000 a ton and a ton of air will be 400 sq feet so. 10000 sq ft divided by 400 is 25 tons of air then multiply 25 times 12000 btus and you get 300000
this distance is about 2.38901 miles.
0.275 acres
The jet can reach the landing strip if tan(11) ≥ 12000 ft/10 milesthat is, if tan(11) ≥ 12000 ft/52800 ft = 0.2273tan(11) = 0.1944 which is not large enough. Oh no! CRASH!!!!
It could be area = 12000 sq ft. "decimal" simply refers to a representation of numbers so that the place value for any digit is ten times the place value of the digit to its right.
There are no decimals in 12000 * * * * * 12000 sq ft = 27.55 decimals.
It would take about 2 minutes to fall 30,000 ft if skydiving in freefall. The actual time may vary depending on factors such as air resistance and body position.
1 ft = 0.3048 meters (exact conversion ratio, this is a good one to memorize).(12000 ft) * (0.3048 m/ft) = 3657.6 m
12000 btu = 1 ton 1 ton per 400 sq ft 1200 / 400 = 3 3 times 12000 = 36000 btu
It will take 300000 btus at 12000 a ton and a ton of air will be 400 sq feet so. 10000 sq ft divided by 400 is 25 tons of air then multiply 25 times 12000 btus and you get 300000
39370.07 ft
usually around 400sq. ft room give or take
One acre is 43,560 square feet. Therefore, 12000 sq. ft is only 0.275 of one acre.
this distance is about 2.38901 miles.
I went tandem and you freefall for about a minute (which feels like much longer) and you parachute down for a little over 10 minutes. We jumped from 14,000 ft. Depends on the type, static line you only freefall for about 5 seconds before the chute deploys. For deployments above 12,000 ft, freefall lasts between 45-60 seconds with a chute ride for 4-12 minutes. After 50 mph you do not feel the sensation that you are falling. Freefall meets maximum velocity at 120 mph. Capt. Joe W. Kittinger achieved the highest and longest (14 min) parachute jump in history in 1960 as part of a United States Air Force program testing. Freefall lasted 4 minutes and 36 seconds.
0.275 acres
The jet can reach the landing strip if tan(11) ≥ 12000 ft/10 milesthat is, if tan(11) ≥ 12000 ft/52800 ft = 0.2273tan(11) = 0.1944 which is not large enough. Oh no! CRASH!!!!