G You are riding a roller-coaster going around a vertical loop, on the inside of the loop. If the loop has a radius of 47.0 m, how fast must the cart be moving in order for you to feel 2.5 times as heavy at the bottom of the loop

Step-by-step explanation:

Let the speed of the roller coaster at the bottom is v.

radius, r = 47 m

Let the mass is m and the force at the bottom is 2.5 times the weight.

F = 2.5 mg

So, the force is

v² = 1.5 gr

v² = 1.5 x 9.8 x 47

v = 26.3 m/s

∠acb = 8°

step-by-step explanation:

∠xba lies on the circle and is half the measure of the intercepted arc ax

hence

measure of arc ax = 2 × 42° = 84°

∠acb is outside the circle and is

∠acb = (arcab - arcax ), hence

∠acb = 0.5 (100 - 84) = 0.5 × 16° = 8°

geometry: triangles

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call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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