meters are based on something to do with light in relation to traveling around the earth. so what are ounces based upon and why?
209932
The ounce is the name for a number of different units of mass (oz), and also of two units of fluid volume (fl oz) and of one unit of force, the ounce-force (ozf). See Imperial system of units and U.S. customary units for more about the frameworks within which these units are defined.\n\navoirdupois ounce defined by the avoirdupois system of mass (see pound (mass)). There are 16 ounces in the avoirdupois pound. The avoirdupois ounce is approximately 28.35 gram. \ntroy ounce defined by the troy system of mass. In troy weight, there are 12 ounces in a pound, and a troy pound is 5760 grains (about 373.24 g), rather than 7000 (about 453.59 g). Note: at roughly 31.104 g, the troy ounce is about 10 per cent more than the more-common avoirdupois ounce. These troy ounces are now used only when weighing precious metals like gold, platinum and silver. One ounce of gold is always 31.1 g. "Maria Theresa Ounce" was once introduced in Ethiopia and some European countries, which was equal to 31.1025 g. \napothecaries' ounce defined by the now-obsolete apothecaries' system of mass. This ounce was the same size as the troy ounce, and its symbol was ℥ (U+2125). \nThe fluid ounce unit of volume also exists in the two versions of 28.41 ml (Imperial) and 29.57 ml (U.S.). \nThe ounce-force (see pound-force) is the force exerted by a mass of 1 oz avoirdupois at the standard acceleration of gravity, usually borrowing the value which is official for defining kilograms-force, 9.80665 m/s². Thus 1 ozf is approximately 278.0 millinewton. \nThe Dutch term ons is a unit of mass that is considerably more, 100 g, a redefinition adopted with the introduction of the metric system together with a redefinition of the Dutch pond as 1 kg. (Though the ons of 100 g remains in informal usage, this pond as a kilogram has fallen by the wayside, replaced by the metric pound of 500 g adopted in other countries.)
416036
I was curious about your question. The website below was an interesting source. Metric Time is decimal based instead of base 6. Standard time measurement is now called Anglo-Babylonian Time (ABT). The second is defined as 1/60 1/60 1/24 of the earth's rotation. 24 hours ABT = 86,400 seconds. 25 hours MT = 250,000 seconds according to the proposal. That would mean 25 hours in a metric day. Another proposal had 100 units per day and noon was called 50.\n\nMetric Time is a proposal, but proponents may still want to use the term second, but the interval could be different.
233114
1 picometer is 1 x 10^-12 meters.\n1 cm is 1 x 10^-2 meters.\n\n36 cm = .36 m\n145 pm = 1.45 x 10^-10 m\n\n.36 m / 1.45 E-10 = 2,482,758,621\nWhich would mean you would need almost 2.5 trillion gold atoms placed right next to eachother,
660935
I am not sure I understand your question in the way it is phrased.\n\nAll measurements need to be taken in a reference frame...it is impossible not to.\nSince there is no absolute coordinate system in the universe, every measurement is relative to the reference frame it is taken in.\n\nFor example,\nlets say that you (meaning in YOUR reference frame) measured a space ship's velocity to be some value, v.\nsomeone on the space ship on the other hand, who is in the Space Ships reference frame, would measure the speed of the ship to be zero and your speed to be v (in the opposite direction).\nIt does not make any sense to say that the space ship is traveling at a speed of v, since that leaves off the reference frame.\n\nOrdinarily (under non relativistic conditions), the reference frame is assumed to be with respect to something at "rest".\nFor example, if you are in a car and look at the speedometer, it might read 55 mph.  This speed obviously does not mean you are traveling at 55 mph with respect to the sun, mars, or alpha centauri, but rather with respect to the surface of the Earth.\n\nIt doesn't just apply for velocities either, it also would apply to distances, times, mass, momentum, energy, .....\nIn different reference frames, measurements can have all sorts of values, each just as "correct" as the next.\n\nSo for all measurements, a reference frame is required to make a measurement.
562808
Well, this is an interesting lesson in geometry, eh? Since the earth is a sphere (more or less) that rotates on an axis - the N & S points were chosen as the intersecting points for longitudinal lines of measurement - which all go back to the length of a day and segments thereof - as well as 360 degrees which were derived roughly from the concept of a 360 day year (it is actually 365+ but back in time the 360 measurement worked wonders for calculationsand thus it fit into geometry of all kinds .. including the longitudinal slices of the earth in 1 degree increments). If you chose to - you could select another two points each an oposite 90 degrees from the N and S poles (which made sense becauseof the magnetic N and the rotational axis). The other two points would be arbitrary. Somebody would have to mark those spots. A secondary longitudinal series could then overlay the primary longitudinal lines and create a kind of variable grid based on an arbitrary selection of the two secondary points. Actually ONE of those lines would lie on top of one of the primary (current) longitudianal lines and it would be a double use line or a kind of disection of hemispheres.  \nBUT\nIt would not be measuring anything very well because the secondary points are arbitrary and not related to anything else.  \nSO\nThe reason the latitude disectors came into play with their resulting grids over the face of earth'ssphere, was becuase they help measure something.  Since stars, the sun and the moon are  always - based on measured timepieces - in the same place and the same time repeatedly, you could measure the distance above the horizon to a star and figure out how far you were N or S of the equator - a halfway point from N to S.  It became easy this way to use the North Star to detect your position N or S by the angle of THAT star above the horizon becuase it was always amost right on top of the N Pole.  So the slices of LATITUDE N and S were critical to navigation. Once the clock was perfected to hold accurate time within a few seconds a month, the accuracy of E and W postion on earth relative to the stars also became possible.  If people did not need to know were they were on earth (especially ships at sea) there would have been no need for a good grid system.
341381
I think instead of "weight" you mean "mass".  "Weight" refers only to the measurment of matter in a gravatational field.  But that's another issue.\n\nThe answer is "no" light (or photons) have no mass.  Or as close to none that it isn't worth mentioning. \n\nRemember, light is the fastest thing we know of.  The closer any bit of matter gets to moving at the speed of light, the more massive (if it helps, think "heavier") it becomes.  Eventually it becomes so massive that the engergy it would take to actually get it to move AT the speed of light is infinate.  So any object with mass cannot move at the speed of light.  Since light DOES move at the speed of light (duh) it can't have mass.\n\nBUT...light does have energy.  And Einstein says energy = mass.  And that's true.  If you took a great big sheet of aluminum foil a mile across and stuck in into space, all the photons (bits of light) coming from the sun would whack into the foil.  Eventually all that whacking would start to make the foil move. Just like the wind blows a sail.  In fact, this sort of aluminum foil gizmo is called a "solar sail".  So light DOES have a sort of mass when it is moving.\n\nBut for the most part when you're talking about an object having mass, you're talking about it's mass when it is sitting still.  This is called it's "rest mass".  And sitting still (if it could sit still) light has zero mass.
531674
Yeah, it's E=mc^2 and it relates energy and mass.  E is energy, m is mass, c is the speed of light (celeritas).  The equation could come useful in physics, like atomic physics.\n\nWhat did you wanted to know?
520636
A gravimeter is a device designed to measure the local gravitational field. Also sometimes called a gravitometer, a gravimeter is a type of accelerometer. According to the equivalence principle of Albert Einstein, the effects of a gravitational field are rigorously equivalent to acceleration. Therefore, gravimeters are susceptible to vibration which causes small, oscillatory accelerations. This is counteracted by integral vibration isolation and signal processing.
656280
It is a means of measurement of events that are taking place around us.   Think  of the days when people have relied on shadows produced by sun to be a measure of time.  They only understood that day and night repeat.  When the earth rotation of sun was understood, what we now term as 24 hrs a day came into being.  The past also was surmised to be at this time as Roman, mayan civilisation etc.,   hence wthout time the relative intervals of events cannot be placed and with new scientific inventions we are able to pin point happenings to part of seconds also.\nThe various calendars that have been made at different parts of the world also have taken events that have taken place as the basis or zero.   When scientific inventions result in exact no of rotations of earth that have taken place or earth got formed from the Sun, then that can be a basis and a new unit can be had for a fourth dimension.\nRamachandran V.
392776
Hi,\n\nFirst think about space as a big sheet of saran wrap. Then imagine that mass distorts the sheet, causing wells, or dents in it. now imagine that the photons need to travel along the saran wrap (i.e. traveling through space) and it makes sense that they would seem to change directions with respect to gravity. \n\nThat is a gross oversimplification, but I think that the analogy is roughly correct.\n\nHope that helps, \nMatt
381257
The size of an object you can resolve with an optical system has a theoretical limit given by the wavelenght of the light you're using divided by the diameter of the optics multiplied by 1.22. If you plug the numbers in you find for a 1 foot diameter optical system at visible light (~400 nm) at an altitude of 100km, the smallest thing you could see is on the order of a foot, give or take (rough calculation). You could definitely not read a postage stamp or a wrist watch. Maybe a license plate with some bigger optics, but this is ignoring all sorts of atmospheric scintillation which screws things up.
370201
Well we need two equatons here:\nF = ma : this will give us the mass\nF = uN : this is just needed\n\nSet these two equal to each other.\nu = .30\nN = 80N\nm = whats needed\na = acceleration due to gravity on earth 9.8m/s^2\n\nThus setting those two equations equal to each other results as follows:\nForce = Force\nuN = ma\n(.3)*(80N) = m * 9.8m/s^2\nm = (.3*80)/9.8\nm = 2.44kg\n\nI have no clue where that above guy created his equation from, but it's sooo wrong.  I mean who weighs items in Newtons.  Newtons is a measure of Force, not weight!
305379
Well, that depends upon how one looks at MOVEMENT.  This is to say that though we tend to understand movement in relation to distance, to other objects, to speed, and so forth, when comparing movement in relation to the universe, the one all unifying factor of that which exists in a defined reality, actual movement becomes sort of relative, taking direction with it.\n\nBefore I pick out inherent directions of movement, let me explain a little of this subjective relativity.  What if I were to instead ask, "How much does the planet Earth weigh?"  This question similarly depends upon how one looks at the variable of weight.  See, if I ask how much a frying pan weighs or how much a barbell weighs, we can automatically presume that I am talking about its weight on Earth, how it is affected by the pull of the Earth's gravity.  So, how do we judge the pull of the Earth's gravity against the Earth itself?  Some would say you can't.  Others would maintain that the Earth itself weighs nothing, just floating in space.  Still others would determine the mass of the Earth against the pull of the sun's gravity.  Still others could match the whole solar system's weight against the pull of other systems or the pull or push from the center of the galaxy, thereafter attributing the very tiny percentage of mass for which the Earth is responsible in the equation.  Other folks would simply conceptually add up the mass of the planet and imagine what that mass would weigh if it were a great big ball sitting on our planet.  A few creative folks might pit our planet's mass against a gas giant like Jupiter, determining it's weight against that pull of gravity instead of the sun's, as if we were a theoretical super-satellite (moon) of another planet.  \n\nAll these subjective, objective, and conceptual differences come into play when we are talking about things greater than the finite.  Similar questions plague your query about movement.  So, there are not so many answers as there are different things to consider and different ways to look at it.  Here are a few options...\n\n1)  Relative to the sun, the Earth revolves.\n\n2)  Relative to its axis, the Earth rotates.\n\n3)  Relative to the plane of the solar system, the Earth wobbles on its axis.\n\n4)  Relative to a line between it’s own equator and a fixed point on the sun nearest us, the Earth tilts.\n\n5)  Relative in comparison to a planet that always has the same side facing the sun, like Mercury, or in other words, relative points on the surface of the Earth and the surface of the sun, the Earth is in an exaggerated or accelerated spin or spiral away from this presumed connector.\n\n6)  Relative to the galaxy, our solar system moves out from center in the Milky Way, the Earth moving with it.\n\n7)  Relative to other galaxies, our galaxy's mass gets caught in the collective gravitational pull of the mass of others, adjusting the direction and speed in which we travel as a galaxy and thereby affecting our solar system with Earth in it.\n\n8)  Relative to other stars in our galaxy, the Earth's proximity to the sun and small size seems more like a shake or a flicker than a revolution.\n\n9)  Relative to a center of the universe in a single big bang theory, everything, including us, moves on a vector out from center.\n\n10)  Relative to Einstein's use of LAMDA in an equation meant to unify his two theories of relativity; everything also collapses in toward center slightly in a very precise balance against the movement outward.\n\n11)  Relative to a multiple big bang theory, everything moves out from the last center of the last big bang, which in turn was already in a generalized outbound pattern from a previous big bang.\n\n12)  Relative to our branch in the implied spiral of the Milky Way Galaxy, a number of systems are moving in a sort of half-parabola ever slowing.\n\n13)  Relative to the idea of a unified greatness, like a universe, a single,
375884
light has no mass, therefore no weight.\n\nHowever, if all the energy of the light were somehow converted to mass, its mass would be m=E/c^2.
602102
It is an energy unit.\nJust like kilojoules.\nIt's not just based on fat but also sugar and stuff.\nCalorie is just older and a more crude way of measuring than kilojoules.\nTechnology advances and so do our ways to measure.\nFirst we used the sun and moon to define time but golden medal winners know it's about nano seconds these days.
206777
Pressure is force over area and would be per m^2.  They use the center of the earth because the math works out such that all the mass of the earth could be treated as being at that point.  It doesn't have as much physical meaning as you might think.
373589
You need to know that 16 ounces = 1 pound\n\nfour gallons = 16 ounces X 25 (pounds)\n                   =  400 ounces\n\n1 gallon = 100 ounces\n\n\nfifteen ounces of gold cost $6058.50\n\nyou need to find out how much one ounce of gold costs. \n\n$6058.50/15 = $403.90\n\nfor the third question about giant bamboo \n\n11.9/3 = 3.9666667= feet in one day\n3.9666667/24 = 0.16527777778= feet per hour\n0.16527777778 X 12 = 1.9833333333 inches per hour\n\nto the nearest hundredth = 1.98 inches per hour\n\nto show that it is reasonable,\ninches per hour X number of hours in day X 3days X 12 (inches to equal one foot) = 142.8 \n1.9833333333 X 24                                   X 3     X12     = 142.8