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AfilCa [17]
3 years ago
11

A car and a truck are moving next to each other on the highway. The truck has a mass of 910 kg and the car’s mass is 227 kg. IF

their momentums are equal and the truck is moving at 18 m/s, what is the speed of the car?
Physics
1 answer:
eimsori [14]3 years ago
3 0

Answer:

<h2>the car must move with the speed</h2><h2>v = 72.16 m/s</h2>

Explanation:

As we know that the momentum of the car and truck is same

so as per the formula of momentum we have

P = mv

so we have

m_1v_1 = m_2v_2

so we will have

910 \times 18 = 227 v

v = 72.16 m/s

So the car must move with the speed

v = 72.16 m/s

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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.

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:

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\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)

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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}

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