Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 90
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Misha, I respect you.
+10
alsu:
It's fair to say that M is indeed the centre of mass of the cake)) but that's a different problem!
(4) At the initial moment, a large number of bodies are launched simultaneously from the same point along differently directed straight chutes. All the chutes are in the same vertical plane. The initial velocity of the bodies is zero. There is no friction. On what curve will these bodies be placed after 1 second of falling? Why?
Let the angle of inclination of the plane to the vertical be a, the angle of inclination of the gutter to the formative plane (a line on the given plane having the same inclination to the vertical as the whole plane) be b, and the angle of inclination of the gutter to the vertical be c. The ratio cos(c) = cos(a)*cos(b) is evident from the figure. The value of a for a system of troughs on a given plane is a constant, the value of b is a variable.
Consider the forces acting on the body. Since there is no friction, the only forces remaining are gravity and the support reaction force. The first force is strictly downward, the second is strictly perpendicular to the surface. Consider the projections of forces on coordinate axis, one of which (y) is normal to the surface and the second (x) coincides with the direction of body movement. There is no displacement along the y-axis, so the resultant force on this axis is 0. Only the projection of gravity mgx acts along the x-axis. From the similarity of the triangles it follows that mgx= mg*cos(c) = mg*cos(a)*cos(b).
Thus the motion of the body is equi-accelerated with acceleration equal to a = g*cos(a)*cos(b).
The path travelled during time t will be written as s(t) = a*(t^2)/2 = g*cos(a)*cos(b)*(t^2)/2, which at t=1 gives s(1) = K*cos(b), where K = g*cos(a)/2 = const. for a given plane.
I.e. obtained an equation in polar coordinates for the geometric location of body positions in 1 second. To understand what this curve is (for those who are not yet in the know), let us perform a transition to Cartesian coordinates:
x'(1) = s(1)*cos(b) = K*cos(b)*cos(b)
y'(1) = s(1)*sin(b) = K*cos(b)*sin(b)
x'^2+y'^2 = K^2*(cos(b)^2*(cos(b)^2+sin(b)^2)) = K^2*cos(b)^2 = K*x'
i.e.
x'^2 - Kx' + y'^2 = 0
or
(x'-K/2)^2 + y'^2 = (K/2)^2
We give the equation of a circle of radius K/2, touching point 0 and lying in the plane downwards from it.
Yeah, Alexei. Did you draw all this in Paint?
Eh... you're in no danger, you've been in this thread for a long time. All means are good.
And I've been running a branch of mine on the Fourth for a long time now (although it's all dried up on the Fourth - but whatever, it's almost all megamos here anyway).
But I'm not looking forward to the Annals...
Yeah, Alexei. Did you draw all this in Paint?
By the way, I've been thinking about how to draw a rotated ellipse in Paint. Preferably a precise one. No ideas so far.
A couple of thoughts about affine transformations did come up though.
By the way, I've been wondering how to draw a rotated ellipse in Paint. Preferably accurate. So far, no ideas.
Instead of Paint - Paint .NET link
in it - layers/rotation and zoom
in it - layers/turn and scale
That's great, thanks. I'll draw in it.
(3) In Brainiac, one in a thousand is born with superpowers. Every newborn is given a DNA test to detect them. There is a 1% chance of testing error. The son of Brainiac is recognised as superhuman. What is the probability that he really isn't?
(5) The invaders once again put Megamind to the test. They stuck 30 flags in the ground in a large field and drew a circle with a radius of 100 metres. All Megamozg can do is choose a point on the circle from which to start the occupant runner. The runner runs at a speed of 10 metres per second. He should run out of the starting point, run to a flag, bring it to the starting point, run to the next flag, bring it to the starting point, etc. (pull flags out, drop them and turn around runner momentarily). If he manages to bring all flags to the starting point in 10 minutes, Megamozg gets shot. Can Megamozg always escape by choosing the correct starting point? The flags are stuck at different points.
(4) There are 2 blue, 2 red and 2 green balloons. In each colour, one of the balloons is heavier than the other. All the lighter balls have the same weight, all the heavier ones have the same weight. There are also scales with two cups without weights. How many weighings are minimally necessary to guarantee the identification of the heavy balls?