It does seem that math requirements are being reduced, even from college level course work. Many engineering majors offer "methods" classes in place of taking calc 3, which teaches use of major specific software. I think its a bad idea to reward pople for not understanding math.
A small and heavy metal ball (negligible air resistance) is attached to the end of a rod (of negligible mass) which other end is attached to an axis (with negligible friction) so that the ball can swing freely. In other words, it's a pendulum.
The ball starts at zero velocity from the position depicted in the picture. There's a mechanism in the rod which releases the ball exactly at the point depicted in the picture.
The observer is located at the initial position of the ball. The observer has no visual contact with the system, but he can detect the sound of the ball hitting the ground, and measure exactly how much time passes between the initial position and the sound of the ball hitting the ground reaching the observer.
Assume that g = 9.8 m/s^2, and the speed of sound = 340 m/s. As depicted in the picture, the altitude of the ball (and thus also the observer) from the ground in the initial position is 50 m.
Exactly 5 seconds pass between the initial position and the sound reaching the observer. The question is: How long is the rod (measured from the center of the axis to the center of the ball)?
Er, I guess it's assumed that the radius of the ball is negligible, isn't it?
The pendulum should release the ball at the same time regardless of its length. Additionally, the ball launches with a velocity perpendicular to the pendulum (thus, 20° up from the horizontal). We can use the first fact to determine the amount of time the ball spends in flight, and the second to determine how far (horizontally) it flew during that time. That gives us the horizontal velocity, with which we can solve for the vertical velocity using basic trig; that in turn gives us the ball's kinetic energy, which was derived from its potential energy, which depends on the length of the rod. I don't feel like plugging in the numbers, but this should be the right approach.
When the ball is released, its speed is sqrt(2*r*g*cos(pi/9)).
I'd need to integrate sqrt(sin(x)) or sqrt(cos(x)), and I'm not sure how to do this.
Hi, I have a question. I was thinking about this last night, but I couldn't come up with any answer: Consider the function p(x) satisfying: p(x) = 1 if x=p/q where p and q are both prime, p>=2, q>=2, and q>p. p(x) = 0 else. What I am wondering is, what properties does this function have on the interval D=[0,1]? Can the limit x->a p(x) be calculated at any point a in D? Is p continous in any point in D? What can be said about the set A={x: p(x)=1}? Is that a closed or open set? Because there are infinitely many primes, the set A will have infinitely many points, but I haven't been able to figure out much more than that.
A is countable and dense in [0,1]. This means that p(x) is nowhere continuous and its limit exists at no point. Additionally, it is not closed, as its closure is [0,1] =/= A, and it is not open, as clearly it cannot be expressed as the union of open intervals. I now offer proof that A is dense in [0,1]. The fact that it is countable follows immediately from the fact that it is a subset of Q. Let 1 > r1 > r2 > 0. We wish to find primes p and q such that r1*q > p > r2*q. The gap between our two limits, q*(r1-r2), grows linearly with q. However, for all sufficiently large primes, the gap between successive primes grows with O(p^(3/4)). Thus, by choosing q large enough, we can make q*(r1-r2) large enough that there must be a prime p between r1*q and r2*q. Thus, there exist p and q with r1 > p/q > r2 for all r1, r2 in [0,1], and thus A is dense in [0,1].
Because the harmonic series is made of converging series (all terms where the denominator is even for example. add them up, you will get a finite number), it must be made on an infinite number of converging series in order to diverge. for every converging series, there is a starting number (1/p) where p is prime, and yeh, you get the drift.
Not quite correct. The sum of the reciprocals of the even numbers, or the sum of the reciprocals of any set of multiples of a number, is just as divergent as the harmonic series.
The reciprocals of prime numbers produce a divergent series. I wonder how hard it is to prove that... (I'm assuming it's probably very hard, as almost anything involving primes.)
Also, was my answer to your earlier problem (with the pendulum) correct?
(correct me if I made any mistakes here)
The sum in the brackets always converges, and is therefore finite, and there are a finite number of products (t terms). Therefore this series is finite, and our initial assumption must be untrue: that there are a finite number of primes. Therefore there must be an infinite number of primes.
EDIT: also, maybe for that physics question there is more than one solution. I wouldn't doubt it.
