"thence along the arc of the curve 45.03 feet the radius point of which bears north 50°10'24" East 666.00"
I have another one I need to find the radius too as well
"thence northwesterly along the arc of a curve to the left 292.25 feet; the radius point of which bears south 76°33'00" west 3,044.00 feet"
2006-08-02 03:59:11 · 2 answers · asked by nightrider 1 in Education & Reference ➔ Homework Help
Isn't that last number the radius?
2006-08-03 05:41:14 · answer #1 · answered by Anonymous · 0⤊ 0⤋
9.8 m/s² = v²/r r = v² / 9.8 m/s² = 681.21 m²/s² / 9.8 m/s² = sixty 9.51 m the ideas-set from the outdoors of the tires on the outdoors of the ensue to the middle of gravity is ?. If ? < 40 5°, then skidding will ensue whilst r < sixty 9.51 m. If ? is larger, the truck will roll over even because of the fact the centripetal r < sixty 9.51 m * cos? i don't propose to point that those are safeguard turning radii at that velocity. it quite is continuously volatile to push the theoretical limits. while you're on the component to the shrink and your shipment shifts or you hit a patch of sand, the truck might roll or skid. If a canines runs interior the front of you, think of ofyou've have been given no determination yet to run over it, and that would desire to reason you to crash, too.
2016-12-11 05:05:32 · answer #2 · answered by ? 3 · 0⤊ 0⤋