# Tag Info

## New answers tagged measurements

4

Yes, the experiment was famously done by Michelson and Morley in 1887. They used the Earth itself as the moving object, and measured the speed of light in the east-west direction (in the direction the Earth is moving) and compared it with the speed of light in the north-south direction (in which the Earth is roughly stationary, at least relative to the Sun)....

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Dip the rock into motor oil. Take it out and and let it drip dry for about a hour. Then place the rock into a container filled with water. For the next few days occasionally move the rock around in the water with the objective of removing the oil from the rock. The area of oil slick on the water’s surface will match the surface area of the rock. If needed ...

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Why not try electrophoretic deposition? You would know the average thickness based on the statistics from engineering specs/data for whatever deposited material. You would also be able to calculate the added volume from Archimedes' principle. You also know the mass density so you could then calculate the surface area of the deposited film/material.

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Per: https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/02%3A_Measurement_and_Problem_Solving/2.02%3A_Significant_Figures_-_Writing_Numbers_to_Reflect_Precision it appears "certain digits" refers to a single measurement on a graduated analog device such as a gauge (such as a thermometer or manometer), a clock,...

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I wouldn't consider them that way, no. As John Rennie mentions in the comments, thermometers work for solids, liquids and gasses, whereas the temperature being proportional to average kinetic energy is determined from gasses specifically. Also, kinetic energy doesn't measure speed. Although it changes in proportion to the speed, it also depends directly ...

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You cannot actually determine the concrete velocity of each particle, but by means of the Maxwell-Boltzmann distribution for the velocity vector: $$f (v) = \left( \frac{m}{2 \pi kT} \right) ^{2/3} \exp \left(-m \frac{v^2}{2kT} \right)$$ It is possible to compute the mean velocity as: $$\langle v\rangle= \int \mathrm dv \cdot f(v) v$$ And you end up ...

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When we say that something is observable, it doesn't mean it is immediate. Even mass is not immediate, we can not know the mass of an object only by look at it or touch it. Linear momentum and angular momentum need some operations to determine its value, but the same happens with position through latitude, longitude and altitude.

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