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Showing content with the highest reputation on 03/20/20 in all areas

  1. 2 points
    Actually my wife and I have discussed this. Fortunately for us we can add some distance if only one of us is infected. Our plan is: - The sick person stays in the lower level of our home which has its own bedroom, bathroom, kitchen sink, microwave, television, and door to the outside. - Well person stays upstairs and takes care of all animals and supplies cooked food. - Ill person does their own dishes. - Luckily for us I found a handful of N95 masks in my workshop. - Hope like hell that sick person doesn't need care, otherwise entire plan might go to hell in a handcart (which is an appropriate phrase as its origin comes from hauling the dead off the streets of London in handcarts to protect horses during the plague.)
  2. 1 point
    ! Moderator Note You’ve proposed this before, still have no science to discuss, and were told not to bring this up again
  3. 1 point
    Then let us say that the arc has length \(\pi - 2\alpha\) and is placed symmetrically around the \(y\)-axis as in this diagram. Then the arc itself is the segment of the circle between the points \((\cos \alpha,\sin \alpha)\) and \((-\cos \alpha,\sin \alpha)\) in the upper half of \(\mathbb{R}^2.\) And its chord is the horizontal line segment connecting its endpoints as shown. If we take a random point \(p=(\sin t,\cos t)\) in the upper right quadrant, i.e. the point for which its distance to the \(y\)-axis along the arc is \(t,\) with \(0 < t < \pi/2,\) then the line defined by this point is the line \(L\) through \(p\) and the point \(q=(s,\sin \alpha)\) on the chord, such that the ratio \(s/\cos \alpha\) of the distance from the \(y\)-axis to \(q\) equals the ratio \(t/(\pi/2-\alpha)\) of the distance along the arc between the \(y\)-axis and \(p\) itself to the entire piece of arc in the upper right quadrant. So \(s = \frac{\cos \alpha}{\pi/2-\alpha}t\) follows. The equation for \(L\) becomes \(L : y = \frac{\cos t - \sin \alpha}{\sin t -s}(x - s) + \sin \alpha,\) which is checked by setting \(x = s,\sin t.\) Clearly the \(y\)-intercept of \(L\) is the point \((0,-\frac{\cos t - \sin \alpha}{\sin t - s}s+\sin \alpha)\) by setting \(x=0\) in the equation for \(L.\) If this is correct, then the question seems to be whether there exists a value of \(\alpha\) with \(0 < \alpha < \pi/2\) for which this intersection point does not depend on \(t.\)
  4. -1 points
    Look, I know you want to quibble with semantics because you can't refute the substance. So you talk about rules of the forum because you obviously want the thread moved to speculation. Sadly, for you, you're question is answered in the substance of the post and I'll repeat it again. Easily, it's what Quantum Field Theory says. QFT says particles are excitation's of underlying quantum fields. QFT doesn't say there aren't any particles. Here's another paper by Physicist Art Hobson. There are no particles, there are only fields https://arxiv.org/pdf/1204.4616.pdf Again, I understand why you want to quibble with semantics instead of debating substance. It's because you don't like the argument but you can't refute it. The existence of particles as real is something that has been debated for years because of QFT.
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