Physics, 06.05.2020 05:41 JarretteWhitaker

A 1.50 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 15.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 C) the ammeter reads 18.8 A, while at 92.0 C it reads 17.4 A. You can ignore any thermal expansion of the rod.
1.Find the resistivity and for the material of the rod at 20 C. (rho= ? Ω*m)2.Find the temperature coefficient of resistivity at 20 C for the material of the rod.α= ? (C)^-1)

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Answer from: nacmp890

The answer to the questions are:

1. The resistivity for the material of the rod at 20 °C (ρ) is

$1.26378\times 10^{-5} \Omega m$ .

2. The temperature coefficient of resistivity at 20 °C for the material of the

rod(α) is $1.1169\times10^{-3}\ ^oC^{-1}\end{aligned}$ .

Given to us:

Voltage across the rod, V = 15.0 V

Length of rod, L = 1.5 m = 150 cm,

Diameter, d = 0.55 cm,

$Radius, r= \frac{d}{2}=\frac{0.550}{2}=0.275\ cm$

$\begin{aligned}Area, A&= \pi r^2\\&=\pi\times (0.275)^2\\&=0.075625\pi\ cm^2\\\end{aligned}$

Initial temperature, $T_o=20.0\ ^oC$

current at T_o , $I_o= 18.8\ A$

Final temperature, $T_1=92.0\ ^oC$

current at T_1 , $I_1=17.4\ A$

1.) To find out the resistivity of the rod(ρ),

Resistant of the rod(R),

$\begin{aligned}\\R_o&=\frac{Voltage}{Current(I_o)}\\&=\frac{15.0}{18.8} \\&=0.7979\ \Omega \\\end{aligned}$

Resistivity of the rod at $20^o\ C$ (ρ),

$\begin{aligned}\\\rho&=\frac{RA}{L}\\&=\frac{0.7979\times 0.075625\pi}{150}\\&=0.00126378\ \Omega cm\\&=1.26378\times 10^{-3} \Omega cm\\&=1.26378\times 10^{-5} \Omega m\\\end{aligned}$

Hence, the resistivity for the material of the rod at 20 °C (ρ) is $1.26378\times 10^{-5} \Omega m$ .

2.) To find out the temperature coefficient of resistivity at 20°C for the material of the rod (α) can be gotten from the equation,

${R_1} ={R_o}[ 1+\alpha(T_1-T_o)]$

we need resistant of the rod( R_1 ),

$\begin{aligned}\\R_1&=\frac{Voltage}{Current(I_o)}\\&=\frac{15.0}{17.4} \\&=0.862\ \Omega \\\end{aligned}$

Now, solving to get the value of α

$\begin{aligned}{R_1} &={R_o}[ 1+\alpha(T_1-T_o)]\\0.862&=0.7979[1+\alpha(92-20)]\\\frac{0.862}{0.7979}&= [1+\alpha(72)]\\1.0804&=[1+\alpha(72)]\\0.0804&=\alpha(72)\\\alpha&=0.0011169\ ^oC^{-1}\\\alpha&=1.1169\times10^{-3}\ ^oC^{-1}\end{aligned}$

Hence, the temperature coefficient of resistivity at 20 °C for the material of the rod(α) is $1.1169\times10^{-3}\ ^oC^{-1}\end{aligned}$ .

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Answer from: tess92

α = 1.114 × 10⁻³ (°C)⁻¹

Explanation:

Given that:

Length of rod (L) = 1.5 m,

Diameter (d) = 0.55 cm,

Area (A) = $\pi r^2$

Radius (r) = d / 2 = 0.275 cm,

Voltage across the rod (V) = 15.0 V.

At initial temperature (T₀) = 20°C, the current (I₀) = 18.8 A while at a temperature (T) = 92⁰C, the current (I) = 17.4 A

a) The resistance of the rod (R) is given as:

$R=\frac{Voltage(V)}{I_0} \\R=\frac{15}{18.8}=0.798\Omega$

Therefore the resistivity and for the material of the rod at 20 °C (ρ) is:

$\rho=\frac{RA}{L}=\frac{0.798*\pi *0.275^2}{1.5}=0.126\Omega m$

b) The temperature coefficient of resistivity at 20°C for the material of the rod (α) can be gotten from the equation:

$R_T=R_0[1-\alpha (T-T_0)]\\but,R_T=\frac{V}{I}=\frac{15}{17.4}=0.862\\$

Rearranging to make α the subject of formula:

$\frac{R_T}{R_0} =1+\alpha (T-T_0)\\\alpha (T-T_0)=\frac{R_T}{R_0}-1\\\alpha =\frac{\frac{R_T}{R_0}-1}{(T-T_0)} \\Substituting:\\\alpha =\frac{\frac{0.862}{0.798}-1 }{92-20} \\\alpha =\frac{0.0802}{72} =1.114*10^-3(^0C)^{-1$

Answer from: Quest

Answer sample (c); // would be best because the percentage of the energy in an incident wave that remains in a reflected wave from this material is the smallest;

Answer from: Quest

moisture is lackin, and organic acids are scarce.

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