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Ball Rack

How many triangles can you make out of a rack of snooker balls?
What is the formula for working out how many triangular arrangements (of more than one ball) can you make out of a triangular rack of 15 snooker balls.
we require three balls to make a triangle so that is the problem of selecting 3 balls out of 15 balls now we have to find out that how many different such collections of three balls can be so we use combination as order is unimportant 15C3=15!/3!(15-3)! = 15!/3!x12! = 455 so 455 different triangles can be formed out of 15 balls if no three balls lie on the same line
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After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear sp
After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.90 m/s, (An example with speed 2.85 m/s is shown in the figure). To reach the rack, the ball rolls up a ramp that rises through a vertical distance of 0.53 m.
(a) What is the linear speed of the ball when it reaches the top of the ramp? m/s
Well you didn't post any figure of the ramp.
Speed before hitting ramp: u = 2.90 m/s
Speed after reaching top of ramp: v = ? m/s
Vertical distance h= 0.53 m
The correct answer would incorporate that some momentum and kinetic energy *will* be lost when the ball hits ramp, assuming the ramp is a straight line. So this reduces starting speed to u' < u = 2.90 m/s.
(I always thought they would make the ramps curved).
But here is the "simple" (wrong) answer where we just do a simple energy balance and assume:
KE_after + PE_after = KE_before
½mv² + mgh = ½mu²
v² + 2gh = u²
v = √(u²-2gh)
So in this case:
v = √(2.90²-2(9.81)(0.53))
v = √(8.41-10.40)
so that seems to suggest the ramp is too high to climb(?).
The minimum critical initial speed needed so the ball could climb would be the one that gives v=0, i.e.
u²=2gh, u = √(2gh) = √10.40 = 3.22m/s
So ramp can't be climbed unless u >=3.22 m/s























































