Tuesday, 6 March 2012
Dimensioni accelerazione centripeta
[a c] = [v2/R] = [v2 R-1] = [L2 t-2 R-1] =
[L2 T-2 L-1] = [L T -2] = [accelerazione]
Thursday, 1 March 2012
Why is the ocean blue?
"Why is the ocean blue? Speculation about the blue color of the ocean, as seen from above, goes way back. Lord Rayleigh claimed it was simply reflection of the blue sky. The correct explanation required combining the 19th-century ideas of Robert Bunsen, who felt that the color depended on light absorption by water, and Jacques-Louis Soret, who felt that the color was entirely due to scattering. C. V. Raman pointed out the importance of molecular scattering, and in 1923 Vasily Shuleikin combined those ideas to develop a complete explanation of the color of the sea."
In Physics Today, Shedding new light on light in the ocean
Tommy D. Dickey, George W. Kattawar, and Kenneth J. Voss
April 2011, http://dx.doi.org/10.1063/1.3580492
Recent advances are making it possible for optical oceanographers to solve a host of pressing environmental problems.
Tommy D. Dickey, George W. Kattawar, and Kenneth J. Voss
April 2011, http://dx.doi.org/10.1063/1.3580492
Recent advances are making it possible for optical oceanographers to solve a host of pressing environmental problems.
More planets than stars
Microlensing suggests that our galaxy has more planets than stars, buBertram M. Schwarzschild
March 2012, http://dx.doi.org/10.1063/PT.3.1463
Gravitational bending of light reveals exoplanets with large orbital radii.
"Most of the more than 600 exoplanets discovered to date have been found through Doppler evidence of periodic host-star motion or photometric evidence of transits across a star’s face. Both methods are strongly biased in favor of planets with orbital radii much smaller than Earth’s, which defines 1 astronomical unit (AU). Gravitational microlensing is an alternative technique that’s most sensitive to planets a few AU from their stars. It favors very distant stars and it’s relatively unbiased as to stellar mass. Though microlensing’s discovery rate is still modest, it appeals to those who seek a representative galactic survey of planets with orbits like those of the solar system." http://www.physicstoday.org/resource/1/phtoad/v65/i3/p19_s1
March 2012, http://dx.doi.org/10.1063/PT.3.1463
Gravitational bending of light reveals exoplanets with large orbital radii.
"Most of the more than 600 exoplanets discovered to date have been found through Doppler evidence of periodic host-star motion or photometric evidence of transits across a star’s face. Both methods are strongly biased in favor of planets with orbital radii much smaller than Earth’s, which defines 1 astronomical unit (AU). Gravitational microlensing is an alternative technique that’s most sensitive to planets a few AU from their stars. It favors very distant stars and it’s relatively unbiased as to stellar mass. Though microlensing’s discovery rate is still modest, it appeals to those who seek a representative galactic survey of planets with orbits like those of the solar system." http://www.physicstoday.org/resource/1/phtoad/v65/i3/p19_s1
Master of the Mint
I read today that Sir Isaac Newton was a "Master of the Mint." It is quite interesting this activity of the great scientist. But, what is the Mint? It is the "place where money is coined." The term derived from a Latin moneta, that we have, as it is, in Italian.
http://www.etymonline.com/index.php?term=mint
The online etymology dictionary tells that the adjective meaning "perfect" (like a freshly minted coin) is from 1902; hence "mint condition". I like this coin as a fresh mint candy.
http://www.etymonline.com/index.php?term=mint
The online etymology dictionary tells that the adjective meaning "perfect" (like a freshly minted coin) is from 1902; hence "mint condition". I like this coin as a fresh mint candy.
Measuring latitude using a pendulum
Question: is it possible to determine the latitude using a pendulum?
Yes. Use the Foucault's pendulumForm http://en.wikipedia.org/wiki/Foucault_pendulum
At either the North Pole or South Pole, the plane of oscillation of a pendulum remains fixed relative to the distant masses of the universe while Earth rotates underneath it, taking one sidereal day to complete a rotation. So, relative to Earth, the plane of oscillation of a pendulum at the North Pole undergoes a full clockwise rotation during one day; a pendulum at the South Pole rotates counterclockwise.
When a Foucault pendulum is suspended at the equator, the plane of oscillation remains fixed relative to Earth. At other latitudes, the plane of oscillation precesses relative to Earth, but slower than at the pole; the angular speed, ω (measured in clockwise degrees per sidereal day), is proportional to the sine of the latitude, φ:
where latitudes north and south of the equator are defined as positive and negative, respectively. For example, a Foucault pendulum at 30° south latitude, viewed from above by an earthbound observer, rotates counterclockwise 360° in two days.
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