Physics, 25.03.2020 23:42 larahanna

A cylindrical glass rod is heated with a torch until it conducts enough current to cause a light bulb to glow. The rod has a length, a diameter, and its ends, plated with material of infinite conductivity, are connected to the rest of the circuit. When red hot, the rod's conductivity varies with position measured from the center of the rod as, with go = 4 x 10-2 (22 m).
1) What is the resistance of the glass rod?
2) When a voltage AV is applied between the two ends, what is the current density j?
3) What is the steady-state magnitude of the electric field E (1)?

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Answer from: Worksmarter72641

1) The resistance of the glass rod is 2122.1 Ω

2) The current density is $\frac{12\sigma_0 \Delta V }{L}$

3) the steady-state magnitude of the electric field E is $= \frac{\Delta V\cdot12}{L^3} x^2i$ .

Explanation:

Here we note that

the conductivity varies with position x measured from the center of the rod as

σ(x) = σ₀L²/x² with σ₀ = 4 × 10⁻² (Ω·m⁻¹)

the resistance of a small length x is given by

dR = $\frac{\rho dx}{A} = \frac{dx}{\sigma _c A} = \frac{4x^{2}dx }{\sigma _0 L^2\pi d^2 }$

Where:

ρ = Resistivity

From the center, we integrate over the entire length of the cable as follows

$R = \int\limits^{\frac{L}{2}}_{-\frac{L}{2}} {} \, dR = \int\limits^{\frac{L}{2}}_{-\frac{L}{2}} {} \, \frac{4x^{2}dx }{\sigma _0 L^2\pi d^2 } = \frac{4x^{3}dx }{3\sigma _0 L^2\pi d^2}|_{-L/2}^{L/2} =\frac{L}{3\sigma_0 \pi d^2 }$

Which gives

R = $\frac{0.02 }{3\times 0.04\times \pi \times0.05^2}$ = 2122.066 Ω

2) The current density $|\overrightarrow{\rm J } (x)|$ = $\frac{I}{A}$

Where:

I = Current = $\frac{Pd}{R}$ and pd = ΔV

A = Area = πd²/4

Therefore current desity = $\frac{4\Delta V}{R\pi d^2}$

Substituting = $R =\frac{L}{3\sigma_0 \pi d^2 }$

Then we have

Current density = $\frac{12\sigma_0 \Delta V }{L}$

3) The steady-state magnitude of the electric field is given by

Magnitude of the Electric field

$\overrightarrow{\rm E}(x) = \frac{\overrightarrow{\rm J } (x)}{\sigma} = \frac{12\sigma_0 \Delta V /L}{\sigma_0 L^2/x^2}i = \frac{\Delta V\cdot12}{L^3} x^2i$

Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

Answer from: Quest

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

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