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vivado [14]
3 years ago
7

HALP me!! This is a physics question.

Physics
2 answers:
zepelin [54]3 years ago
5 0
Distance 6.9 m

Work done

345 kJ = 345000

Force 5 x 10 4 = 50000 N
emmasim [6.3K]3 years ago
4 0
<h2>Answer:</h2>

<u>Distance covered is 6.9 meters</u>


<h2>Explanation:</h2>

Data given:

Work Done = 345 kJ = 345000 J

Force = 5 x 10 ^ 4 =  50000 N

Distance = ?


Solution:

As we know that

Work Done = Force applied x Distance covered

By arranging the equation we get

Work / Force = Distance covered

By putting the values

345000 / 50000 = 6.9

So distance covered is 6.9 meters

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The blue mass is currently 4 meters away from the red mass.
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Answer:

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Explanation:

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The visible light portion of the electromagnetic spectrum is often subdivided into the colors of red, orange, yellow, green, blu
Ugo [173]

Light at the red end of the visible portion has the least energy, lowest frequency, same speed, and longer wavelength compared to the violet end.

<h3><u>Explanation:</u></h3>

The range in which the light exists is described as the electromagnetic spectrum. The light waves, radio waves, gamma rays,etc that exist in the world is not visible to human eyes. A kind of wave that modifies  magnetic and electric fields is light. Spectroscopy makes use of all the frequencies and the wavelengths of the electromagnetic radiation.

The part of the electromagnetic spectrum that can be seen by the human eyes is the visible spectrum. The light waves with the wavelengths of  380 to 740 nm can be sen by the human eyes. Light at the red end of the visible portion has the least energy, lowest frequency, same speed, and longer wavelength compared to the violet end.

5 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
Question 3. AP PHYSICS. Currently learning about torque so I assume you apply torque. I don't get it at all
8090 [49]

wow that is confusing


8 0
2 years ago
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