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sweet [91]
3 years ago
7

Which of the following is an example of a dry lubricant? oil grease graphite petroleum

Physics
2 answers:
Naddika [18.5K]3 years ago
7 0

Answer:

Dry lubricants or solid lubricants are materials that, despite being in the solid phase, are able to reduce friction between two surfaces sliding against each other without the need for a liquid oil medium.

Explanation:

Naddik [55]3 years ago
7 0

Answer:

Your answer is graphite :) Have a nice day

Explanation:

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Neurons in our bodies carry weak currents that produce detectable magnetic fields. A technique called magnetoencephalography, or
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I = (1.80 × 10⁻¹⁰) A

Explanation:

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4 years ago
An obiect of mass weighing 5,24 k acceleration due to gravity is 9 8 meters/second2 is raised to a height of 1.63 meters. What i
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What happens if you move a bar magnet back and forth along the axis of the
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A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
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Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is v  = 0.0002808 \ m/s

Explanation:

From the question we are told that

    The current on the copper is  I  = 20 \ A

     The cross-sectional area is  A =  5.261 \ mm^2 =  5.261 *10^{-6} \ m^2

The number of copper atom in the wire is  mathematically evaluated

      n  =  \frac{\rho *  N_a}Z}

Where \rho is the density of copper with a value \rho =  8.93 \ g/m^3

          N_a is the Avogadro's number with a value N_a  = 6.02 *10^{23}\ atom/mol

         Z  is the molar mass of copper with a value  Z =  63.55 \ g/mol

So

     n  =  \frac{8.93 * 6.02 *10^{23}}{63.55}

     n  = 8.46 * 10^{28}  \  atoms /m^3

Given the 1 atom is equivalent to 1 free electron then the number of free electron is  

         N  = 8.46 * 10^{28}  \  electrons

The current through the wire is mathematically represented as

         I  =  N * e * v * A

substituting values

        20 =  8.46 *10^{28} * (1.60*10^{-19}) * v *  5.261 *10^{-6}

=>     v  = 0.0002808 \ m/s

       

8 0
3 years ago
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