50.24 sq ft
50.24 sq. ft.
16 sq. ft.
The answer depends on the shape: it can be any distance at all. Consider 10,000 square feet in the shape of a 10,000 ft x 1 ft rectangle. The distance from the centre to the edge can be as small as 0.5 ft or as large as 50,000 ft (approx 9.5 miles). Or consider a rectangle that is 1,000,000 ft x 0.01 ft. The distance can now be as small as 0.005 ft (0.06 inches) or as large as 500,000 ft (95 miles).
the awnser is 314 sq ft.
There are 192 inches around an 8 foot table.
The table could fit 150 knights, so if you allot 3 ft. per knight, you get a circumference of approximately 450 ft. and a diameter of around 143 ft.
If the 15 ft is the distance around the edge, then:Distance_round_edge = 15 ft If the 15 ft is the diameter of the round pool, then:Distance_round_edge = π x diameter = π x 15 ft= 15π ft≈ 47.12 ftIf the 15 ft is the radius of the round pool, then:Distance_round_edge = 2 x π x radius = 2 x π x 15 ft= 30π ft≈ 94.25 ft
The overall stopping distance would be around 122m (400ft) This is made up of a thinking distance of 24m (79ft) and an actual stopping distance of 98m (321ft). The thinking distance is around 3m for every 10mph of speed and the overall stopping distance is calculated as follows: 2x20 ft at 20mph 2.5x30 ft at 30mph 3x40 ft at 40mph 3.5x50 ft at 50mph 4x60 ft at 60mph 4.5x70 at 70mph 5x80 at 80mph = 400 ft james s
Ignoring air resistance, any falling object falls 4.23 feet in the first 0.513 second.The horizontal distance doesn't matter, and we don't need to know it. The time it takes to hitthe floor after it rolls off the edge has nothing to do with how fast it was rolling before it fell.
Somewhere around 375 ft.
Well if the table is 8ft around then all you have to do is devide by pi then you get the diamiter which is 2.54
Distance around circle = 2*pi*r = 157.1 so r = 157.1/(2*pi) = 25.00 ft so, to the nearest tenth of a foot, 25.0 ft