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2a) b) Draw in perpendicular to tangent at A to a point E, through origin O. This is the diameter of the circle. OAP = x OPA = x (isos triangle) POA = 180 - 2x ROA = 2x OAR = 90 - 2x .: PRA = 90, so the lines are perpendicular 2b)a) RQB = o = 90 - x 2x + 90 + 90 + 180 - (90 - x) = 360 3x =...

how are you supposed to choose the kids when it already gives you the order that they're in? can you explain this further? :/

say what? the acceleration is the acceleration from impact>stop, ie final velocity divided by time of impact also, you might want to add a mass equivalent to the can to the no-can one, if you think this is appropriate for your experiment

how much working would you need? first one: keep one, say, man stationary. order the remaining 5 women and 4 men = 5!4! second one: keep the man stationary. order the other 4 men. order the 2 women that can't sit next to the first man in the other 3 spots. order the remaining 3 women = 4!3!3!

dividing in ratio m:n from point A to B; you're adding a ratio of m/AB to A, where AB is the distance from A to B represented either by m+n or n-m. draw a diagram to make this bit more clear dividing internally A + m/(m+n) * (B - A) divide externally A + m/(n-m) * (B - A) just do it by...

it didn't seem like that long working to me edit: did you use.. [s = sin@, c = cos@] b = v^2 s^2 / 2g a = 2 v^2 sc / g x = v c t y = v s t - 1/2 * g t^2 then just substitute into 4bx(a-x)/a^2. ?

just substitute the usual expressions for max height, max distance, and distance into the expression and verify that it is equivalent to the expression for height. or do that and then work backwards. I'm not sure which side you work with in show questions

I think it would just mean that the stone is thrown up a 30* hill, so you find the angle it would need to be thrown at to hit the hill at an angle of 0* and at a perpendicular angle to the hill; however my answers were a bit off. I got a) as 45* and b) as arctan 3 = 71*34*.

he made -23 into -115 and + 92 and made the -4.4^k into -5.4^k + 4^k

crumple zone increases stopping time, therefore decreasing force and power. It doesn't have any effect on impulse/momentum, except that the loss of momentum will occur over a longer time. Impulse is constant - it is change in momentum, ie I = mv - mu, and since this does not consider what...

(1/2)^n = 1600n n = 5/16 amount left: (1/2)^(5/16) = 0.805 amount decayed = 0.195

http://community.boredofstudies.org/showthread.php?t=167824&highlight=that+it+will+hit+a+bird+at+the+top+of+pole.+However%2C+instant+stone+is+thrown%2C

change sina to cos(90-a) or cosb to sin(90-b)

the discriminant is p^2 - 8p + 16 = (p -4)^2 putting this into the quadratic formula you get (p +- sqrt(pi4)^2)/4 = (p +- (p - 4))/4 (2p - 4)/4 and 4/4 = 1, so 1 is always a root. (check by subbing in x=1) so you can see that the other root will be rational if p is rational (unequal if p=/=4)...

Wow, no school. That would be pretty damn cool. What would the requirements be to do this? Would you have to be at a certain schooling level already, or could, say, an able 10 year old kid do this if they wanted? (just curious)

wait... what's the acceleration got to do with the crumple zone?

if you know that acceleration is moreorless constant

yeah.. that's because they're equivalent equations. If you're going to be pushing the car you won't be getting a constant acceleration, so why not let the car roll down a slope? (one with minimal friction)

acceleration is the change in velocity over time, not the change in displacement over time. 1.48m/s is the final velocity.

Re: Physics Question if you're assuming the acceleration is constant, then s/t = (v+u)/2 so if s = 1.7, t = 5, u = 0, then final velocity is 0.68m/s, and the acceleration is this divided by time (5 seconds) so 0.136ms^-2 I'm not too sure that I get exactly what you have to do for the...