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sleet_krkn [62]
3 years ago
7

Determine the radial acceleration of the ultracentrifuge using calculations

Physics
1 answer:
Dmitry [639]3 years ago
3 0

When something moves on a round track, the guidance of the something's velocity must continually switch. A switching velocity means that there must be an acceleration. This acceleration is horizontal to the guidance of the velocity. This is said as “the radial acceleration”, or “centripetal acceleration” ("centripetal" means "center searching"). The “radial acceleration” is equal to “the square of the velocity”, divided by “the radius of the circular path of the object”. The unit of the “centripetal acceleration” is m/s².

\text { Centripetal acceleration }=\frac{\text {velocity}^{2}}{\text {radius of motion}}

\mathrm{a}_{\mathrm{rad}}=\frac{V^{2}}{r}

where,

\text { and }=\text { radial, or centripetal, acceleration }(\mathrm{m} / \mathrm{s} ^2)

"v" = "velocity" (m/s) and "r" = "radius of motion of the object" (m)

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This equation is known as the ideal gas law, and it can be used to predict the behavior of many gases at relatively low pressure
Masja [62]
The correct answer is 
<span>C) either the pressure of the gas, the volume of the gas, or both, will increase.

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</span>pV=nRT
<span>where 
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6 0
3 years ago
Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm
Mariulka [41]

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.

Explanation:

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:

\eta\cdot \dot W = \dot K

\dot W = \frac{\dot K}{\eta} (Eq. 1)

Where:

\eta - Efficiency of fan, dimensionless.

\dot W - Electric power supplied fan, measured in watts.

\dot K - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2} (Eq. 2)

Where:

\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.

5 0
3 years ago
Freeeeeeeeeeeeeeeeeeeeeeeee points
andreyandreev [35.5K]

Answer:

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3 0
3 years ago
Read 2 more answers
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