Recent content by PeteSF

8 Tones in an Octave! (Not the best answer, though.)

Remaining solutions in Last Group: 26) 9 LIves of a Cat 23) 23 Pairs of Chromosomes in the Human Body 31) 6 Balls to an Over in Cricket[/color]

Not that I doubt you :wink:, but would you have any supporting references?

A 40mph-40mph head-on collision does the same damage as a 40mph collision with an immovable object. An 80mph collision with an object at rest of equal mass and strength object not attached to the ground (eg a car oil an oil slick) would also do the same damage. If the car at rest is in...

Sure we can. All you need is some kind of synchronization at different places. Radio telescope arrays and Very-Long-Baseline-Interferometry relies directly on the one-way speed of light. For that matter, the standard two-slit interference pattern relies directly on the one-way speed of...

I suspect that the pictures could be real, but taken with a lengthy exposure (perhaps several seconds). According to Wikipedia, high intensity lasers can be visible in clean air by Raman scattering or Rayleigh scattering. WiseGeek.com also suggests that green-beam lasers are visible in...

Note that high viscosity doesn't necessarily mean high density. Oil, for example, is more viscous and less dense than H2O.

We're already doing it - this thread is case study in BA (biological intelligence), and why it isn't. If I say "mouses", "gooses", "sheeps", "persons", "octopods", and "radiuses", instead of "mice", "geese", "sheep", "people", "octopodes", and "radii", why do grammarians get their knickers in...

Fantastic! How's the landing accuracy? Can you nail a 40cmx40cm square?

--> Homework forum??

Well, the article also says (correctly) that the atmosphere freezes regularly. What it fails to clarify is that if Pluto and Charon do share an atmosphere, then it would only be shared when the atmosphere is not frozen.

It's more likely that I've got mistakes in there (you picked one up already - angle b is indeed CHJ, as on the diagram). Yes, the idea was to add all the bits of the AC segment. See if you can go through and add them all yourself, without worrying about how I tried to do it. Start by drawing...

The original question mentioned a gyroscope with axis parallel to the ground - at 90° to the gravity field.

After a touch more work, it looks like you need basic calculus to find the minimum in that equation... Have you done derivatives using the quotient rule?

Consider the attached diagram (which I made too small. Sorry!) H is the midpoint of CE I is the center of the circle (where the perpendicular bisector of CE meets AB). J is where a perpendicular dropped from H meets CF. Now: AC = 28cm = HJ.tan(b) + HJ/tan(b) + x.tan(b) + x.tan(60°)...