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seropon [69]
2 years ago
8

Definition of Xenophobia​

Physics
2 answers:
Ivanshal [37]2 years ago
8 0

Answer: dislike of or prejudice against people from other countries.

Explanation:

Delicious77 [7]2 years ago
4 0

Answer:

dislike of or prejudice against people from other countries.

Explanation:

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What is the current if 4C of charge passes in 2 s?
julia-pushkina [17]

Answer:

I hope 2 amperes of current passes

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2 years ago
Is a force that opposes the motion between two objects in contact with each other.
Sergeeva-Olga [200]

Answer: Friction  :)        

7 0
2 years ago
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Give 2 reasons for fitting heavy commercial vehicles with many tyres​
forsale [732]
Better weight distribution and more stability
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2 years ago
A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting
rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is \sqrt{725} ft/ sec

Explanation:

Given:

h(t) =  25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

D(t) = \sqrt{h^2 + x^2}

Lets find the derivative of distance with respect to time

\frac{dD}{dt} (t)  = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}

Substituting the values of h(t) and  x(t) and simplifying we get,

\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}

\frac{dh}{dt} = 25ft/sec

\frac{dx}{dt} = 10 ft/sec

\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}  = \frac{725}{\sqrt{725}}  = \sqrt{725} ft / sec

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
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