Physics, 10.12.2019 03:31 depaze6652

Anumber 10 awg copper wire has a resistance of 0.9989 ohms per 1,000 feet and a cross sectional area of 10,380 cm. if the resistance of copper wire is inversely proportional to its cross-sectional area, what would be the resistance of 1,000 feet of copper wire with a cross section of 5,178 cm? (round the final answer to four decimal places.)

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Answer from: katiekern5207

The resistance of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. In this case we are talking about copper, and its resistivity at $0\ºC$ is:

$\rho=1.72(10)^{-8}m\Omega$

is the length of the wire, which in this case is 1000ft , but we will make the conversion to meters, in order to work with the same units:

$l=1000ft\frac{0.3048m}{1ft}=304.8m$

is the transversal area of the wire. In this case is a circumference, so we will use the formula of the area of the circumference:

$s=\pi{r}^{2}$

Now, in this problem we have two transversal areas:

$s_{1}=10380cm=103.8{m}^{2}$

$s_{2}=5178cm=51.78{m}^{2}$

And two resistances:

$R_{1}=0.9989\Omega$ and $R_{2}$ which is the one we are asked to find.

Taking into account we are working with the same material (copper), the $\rho$ will be the same. In addition, the length of the wire is the same, the only thing that changes is the transversal area.

According to this given data, we have a system of two equations, as follows:

$R_{1}=\rho\frac{l}{s_{1}}$ (2)

$R_{2}=\rho\frac{l}{s_{2}}$ (3)

We know all the values, except $R_{2}$ .

At this point of the problem, we will approach it in the following way:

From (2) let’s isolate $l\rho$ :

$l\rho=R_{1}S_{1}$ (4)

From (3) let’s also isolate $l\rho$ :

$l\rho=R_{2}S_{2}$ (5)

If equation (4)=equation (5)

$R_{1}S_{1}=R_{2}S_{2}$ (6)

Now, we have to find the value of $R_{2}$ :

$R_{2}=\frac{R_{1}S_{1}}{S_{2}}$ (7)

Finally:

$R_{2}=2.002\Omega$

Rounding:

$R_{2}=2\Omega$ >>>>>This is the result

It is important to note that as the resistance of the wire is inversely proportional to its cross-sectional area, when the area decreases the resistance increases.

Answer from: Quest

This is the answer: body 1 + engine 2

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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Anumber 10 awg copper wire has a resistance of 0.9989 ohms per 1,000 feet and a cross sectional area...

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