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3 years ago

An electric pump has 2kW power . How much water will it lift every minute if the height is 10 m ?

Physics

1 answer:

valkas [14]3 years ago

3 0

Answer:

Given that,

Power = 2000 W
time = 60 seconds
distance= 10m

Power = work done ÷ time

Here, since the movement is vertical, w = mgh

So,

Power = mgh÷t

2000 = (m × 9.8 ×10) ÷ 60

m = (2000 ×60) ÷98

m = 1224.5kg

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The magnitude of electrical force on charge $q_{3}$ due to the others is 0.102 newtons.

<h3>How to calculate the electrical force experimented on a particle</h3>

The vector position of each particle respect to origin are described below:

$\vec r_{1} = (-0.500, 0)\,[m]$

$\vec r_{2} = (+0.500, 0)\,[m]$

$\vec r_{3} = (0, +0.500)\,[m]$

Then, distances of the former two particles particles respect to the latter one are found now:

$\vec r_{13} = (+0.500, +0.500)\,[m]$

$r_{13} = \sqrt{\vec r_{13}\,\bullet\,\vec r_{13}} = \sqrt{(0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{13} =\frac{\sqrt{2}}{2}\,m$

$\vec r_{23} = (-0.500, +0.500)\,[m]$

$r_{23} = \sqrt{\vec r_{23}\,\bullet \,\vec r_{23}} = \sqrt{(-0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{23} =\frac{\sqrt{2}}{2}\,m$

The resultant force is found by Coulomb's law and principle of superposition:

$\vec R = \vec F_{13}+\vec F_{23}$ (1)

Please notice that particles with charges of same sign attract each other and particles with charges of opposite sign repeal each other.

$\vec R = \frac{k\cdot q_{1}\cdot q_{3}}{r_{13}^{2}}\cdot \vec u_{13} +\frac{k\cdot q_{2}\cdot q_{3}}{r_{23}^{2}}\cdot \vec u_{23}$ (2)

Where:

$k$ - Electrostatic constant, in newton-square meters per square Coulomb.
$q_{1}$ , $q_{2}$ , $q_{3}$ - Electric charges, in Coulombs.
$r_{13}$ , $r_{23}$ - Distances between particles, in meters.
$\vec u_{13}$ , $\vec u_{23}$ - Unit vectors, no unit.

If we know that $k = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q_{1} = 2\times 10^{-6}\,C$ , $q_{2} = 2\times 10^{-6}\,C$ , $q_{3} = 2\times 10^{-6}\,C$ , $r_{13} =\frac{\sqrt{2}}{2}\,m$ , $r_{23} =\frac{\sqrt{2}}{2}\,m$ , $\vec u_{13} = \left(-\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} \right)$ and $\vec u_{23} = \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$ , then the vector force on charge $q_{3}$ is:

$\vec R = \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$

$\vec R = 0.072\cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + 0.072\cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)\,[N]$

$\vec R = 0.072\cdot \left(0, -\sqrt{2}\right)\,[N]$

And the magnitude of the electrical force on charge $q_{3}$ ( $R$ ), in newtons, due to the others is found by Pythagorean theorem:

$R = 0.102\,N$

The magnitude of electrical force on charge $q_{3}$ due to the others is 0.102 newtons. $\blacksquare$

To learn more on Coulomb's law, we kindly invite to check this verified question: brainly.com/question/506926

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2 years ago