Estimate the current density in copper wire at 20 °C if an electric field of 0.923 V/m exists within

Physics

1 answer:

hoa [83]3 years ago

4 0

We have that the estimated the current density in copper wire at 20 °C is

$J=38*10^6A/m^2$

From the question we are told

the current density in copper wire at 20 °C
if an electric field of 0.923 V/m exists within the wire

Generally the equation for the current density is mathematically given as

current density

$J=\alpha E$

Where

$\alpha=\frac{1}{e}\\\\\\alpha=\frac{1}{2.44*10^{-8}}\\\\\alpha=4.09*10^7$

Therefore

$J=4.09*10^7* 0.923$

$J=38*10^6A/m^2$

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Karen runs sets in basketball practice. She starts from a line runs 2.0 m, returns to the line, runs 4.0 m, to the line, runs 6.

Aleonysh [2.5K]

D. distance = 23 m, displacement = + 1 m

Explanation:

Let's remind the difference between distance and displacement:

- distance is a scalar, and is the total length covered by an object, counting all the movements in any direction

- displacement is a vector connecting the starting point and the final point of a motion, so its magnitude is given by the length of this vector, and its direction is given by the direction of this vector.

In this case, the distance covered by Karen is given by the sum of all its movements:

$distance = 2.0 m + 2.0 m+4.0 m+4.0 m +6.0 + 5.0 m=23.0 m$

The displacement instead is given by the difference between the final point (1.0 m in front of the starting line) and the starting point (the starting line, 0 m):

$displacement = +1.0 m-0 m=+1 m$