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g100num [7]
3 years ago
15

An object at rest increases its velocity to 10m/s in 3 seconds. What is its acceleration

Physics
1 answer:
BabaBlast [244]3 years ago
5 0

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) - (speed at the beginning).

Change in speed = (10 m/s) - (zero) = 10 m/s

Time for the change = 3 sec

Acceleration = (10 m/s) / (3 sec)

<em>Acceleration = (3 and 1/3) m/s²</em>  or 3.333 m/s²

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the electric field strength of this charge is two times the strength of the other charge

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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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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

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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

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

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         N  = 8.46 * 10^{28}  \  electrons

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         I  =  N * e * v * A

substituting values

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

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