Physics, 10.10.2019 21:30 Reebear8372

You apply a potential difference of 5.70 v between the ends of a wire that is 2.90 m in length and 0.654 mm in radius. the resulting current through the wire is 17.6
a. what is the resistivity of the wire?

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Answer from: HistoryLee

1) First of all, let's find the resistance of the wire by using Ohm's law:
V=IR

where V is the potential difference applied on the wire, I the current and R the resistance. For the resistor in the problem we have:
$R= \frac{V}{I}= \frac{5.70 V}{17.6 A}=0.32 \Omega$

2) Now that we have the value of the resistance, we can find the resistivity of the wire $\rho$ by using the following relationship:
$\rho = \frac{RA}{L}$
Where A is the cross-sectional area of the wire and L its length.
We already have its length L=2.90 m

, while we need to calculate the area A starting from the radius:
$A=\pi r^2 = \pi (0.654\cdot 10^{-3}m)^2=1.34 \cdot 10^{-6}m^2$

And now we can find the resistivity:
$\rho = \frac{RA}{L}= \frac{(0.32 \Omega)(1.34 \cdot 10^{-6}m^2)}{2.90m}= 1.48 \cdot 10^{-7}\Omega \cdot m$

Answer from: Quest

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod. this generates an electric current in the coil, which powers the bulb.

Answer from: Quest

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You apply a potential difference of 5.70 v between the ends of a wire that is 2.90 m in length and 0...

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