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Shalnov [3]
3 years ago
15

To construct a non-mechanical water meter, a 0.500-T magnetic field is placed across the supply water pipe to a home and the Hal

l voltage is recorded.
Required:
a. Find the flow rate through a 3.00-cm-diameter pipe if the Hall voltage is 60.0 mV.
b. What would the Hall voltage be for the same flow rate through a 10.0-cm-diameter pipe with the same field applied?
Physics
1 answer:
Alex73 [517]3 years ago
6 0

Answer:

a.  4 m/s b. 0.2 V

Explanation:

a. Find the flow rate through a 3.00-cm-diameter pipe if the Hall voltage is 60.0 mV.

The hall voltage V = vBd where v = flow-rate, B = magnetic field strength = 0.500 T and d = diameter of pipe = 3.00 cm = 0.03 m

Since V = vBd

v = V/Bd    given that V = 60.0 mV = 0.060 V, substituting the values of the other variables, we have

v = 0.060 V/(0.500 T × 0.03 m)

v = 0.060 V/(0.015 Tm)

v = 4 m/s

b. What would the Hall voltage be for the same flow rate through a 10.0-cm-diameter pipe with the same field applied?

Since the hall voltage, V = vBd and v = flow-rate = 4 m/s, B = magnetic field strength = 0.500 T and d' = diameter of pipe = 10.0 cm = 0.10 m

Substituting the variables into the equation, we have

V = vBd

V = 4 m/s × 0.500 T × 0.10 m

V = 0.2 V

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The speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

The angular momentum(L) of an electron moving in a circular path is given by the formula,

L = mvr ........(i)

We know that the radius of the path of an electron in a magnetic field is

r = mv/qB

Putting this value in equation (i),

L = mv x mv/qB

or L = (mv)^2/qB

Putting the given values in the above equation,

4 x 10^-25 = (9.1x10^-31)^2 x v^2/ 1.6 x 10^-19 x 1 x 10^-3

v comes out to be 8.88 x 10^7 m/s.

Hence, the speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

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brainly.com/question/15104254?referrer=searchResults

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Mature salmon swim upstream, returning to spawn at their birthplace. During the arduous trip they leap vertically upward over wa
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The minimum speed is 7.5 m/s.

Explanation:

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Height = 2.87 m

We need to calculate the minimum speed

Using equation of motion

v^2=u^2-2gh

u=\sqrt{2gh}

Where, u = minimum velocity

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h = height

Put the value into the formula

u=\sqrt{2\times9.8\times2.87}

u=7.5\ m/s

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Charge q1 = +2.00 μC is at -0.500 m along the x axis. Charge q2 = -2.00 μC is at 0.500 m along the x axis. Charge q3 = 2.00 μC i
Kobotan [32]

The magnitude of <em>electrical</em> 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 <em>position</em> 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 <em>same</em> sign attract each other and particles with charges of <em>opposite</em> 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 <em>electrical</em> force on charge q_{3} (R), in newtons, due to the others is found by Pythagorean theorem:

R = 0.102\,N

The magnitude of <em>electrical</em> 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
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