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san4es73 [151]
3 years ago
8

A single brick falls with acceleration g. The reason a double brick falls with the same acceleration is

Physics
2 answers:
rewona [7]3 years ago
6 0

EVERYTHING falls with the same acceleration near the earth's surface.

I thought about this for many years.  I finally understood WHY, and here's the way I like to explain it:

-- It takes MORE FORCE to accelerate a larger mass, and LESS FORCE to accelerate a smaller mass.

-- But guess what !  Gravity exerts MORE FORCE on a larger mass, and LESS FORCE on a smaller mass.

So everything averages out, and the acceleration is always the same, no matter what's falling.

To say it just a little bit differently:

-- A larger mass NEEDS more force to accelerate, but it GETS more force from gravity.

-- A smaller mass NEEDS less force to accelerate, but it GETS less force from gravity.

So everything averages out, and the acceleration is always the same, no matter what's falling.

LekaFEV [45]3 years ago
5 0

The reason is because the force due to the acceleration from gravity is constant. It's the same as the typical "dropping a bowling ball and feather (with no air resistance) at the same time". Gravity acts on all object with the same acceleration regardless of physical properties.

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Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

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Complete statement is: <em>Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.</em>

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

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\rho_{a} - Density of air, measured in kilograms per cubic meter.

\dot V - Volume flow, measured in cubic meters per second.

A_{s} - Cross-sectional area of fan, measured in square meters.

If we know that \rho_{a} = 1.20\,\frac{kg}{m^{3}}, \dot V = 0.05\,\frac{m^{3}}{s}, \eta = 0.3 and A_{s} = 20\times 10^{-4}\,m^{2}, the power needed to be supplied by the fan is:

\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}

\dot K = 62.5\,W

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

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