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Tanya [424]
2 years ago
5

The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho=rho0[1+α(T−T0)], where T0

is a reference temperature, usually 20∘C, and α is the temperature coefficient of resistivity. Part A First find an expression for the current I through a wire of length L, cross-section area A, and temperature T when connected across the terminals of an ideal battery with terminal voltage ΔV. Then, because the change in resistance is small, use the binomial approximation to simplify your expression. Your final expression should have the temperature coefficient α in the numerator. Express your answer in terms of L, A, T, T0, ΔV, rho0, and α.
Physics
1 answer:
Readme [11.4K]2 years ago
3 0

Answer:

I = ΔVA[1 - α (T₀ - T)]/Lρ₀

Explanation:

We have the following data:

ΔV = Battery Terminal Voltage

I = Current through wire

L = Length of wire

A = Cross-sectional area of wire

T = Temperature of wire, when connected across battery

T₀ = Reference temperature

ρ = Resistivity of wire at temperature T

ρ₀ = Resistivity of wire at reference temperature

α = Temperature Coefficient of Resistance

From OHM'S LAW we know that;

ΔV = IR

I = ΔV/R

but,  R = ρL/A   (For Wire)

Therefore,

I = ΔV/(ρL/A)

I = ΔVA/ρL

but,   ρ = ρ₀[1 + α (T₀ - T)]

Therefore,

I = ΔVA/Lρ₀[1 + α (T₀ - T)]

I = [ΔVA/Lρ₀] [1 + α (T₀ - T)]⁻¹

using Binomial Theorem:

(1 +x)⁻¹ = 1 - x + x² - x³ + ...

In case of [1 + α (T₀ - T)]⁻¹, x = α (T₀ - T).

Since, α generally has very low value. Thus, its higher powers can easily be neglected.

Therefore, using this Binomial Approximation, we can write:

[1 + α (T₀ - T)]⁻¹ = [1 - α (T₀ - T)]

Thus, the equation becomes:

<u>I = ΔVA[1 - α (T₀ - T)]/Lρ₀ </u>

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A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3
4vir4ik [10]

Answer:

d) 1.2 mT

Explanation:

Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.

First of all, we observe that:

- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is

I = 15 A

- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).

Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I = 15 A is the current in the conductor

r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field

Substituting, we find:

B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT

8 0
2 years ago
What two factors determine the energy production of the Hoover Dam
swat32
Mar 28, 2011 · The Hoover Dam generates electricity using hydropower. Hydropower is the power generated by moving water. Dams are often built on a large river, so the water falls down from the top of the dam. At the bottom of the dam, the water rotates a large turbine

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6 0
2 years ago
Read 2 more answers
A muon formed high in the Earth's atmosphere is measured by an observer on the Earth's surface to travel at speed V - 0.983c for
Alex_Xolod [135]

Answer:

The moun lives 2.198*10^-6 s as measured by its own frame of reference

The Earth moved 648 m as measured by the moun's frame of reference

Explanation:

From the point of view of the observer on Earth the muon traveled 3.53 km at 0.983c

0.983 * 3*10^8 = 2.949*10^8 m/s

Δt = d/v = 3530 / 2.949*10^8 = 1.197*10^-5 s

The muon lived 1.197*10^-5 s from the point of view of the observer.

The equation for time dilation is:

\Delta t' = \Delta t * \sqrt{1 - \frac{v^2}{c^2}}

Then:

\Delta t' = 1.197*10^-5 * \sqrt{1 - \frac{(0.983c)^2}{c^2}} = 2.198*10^-6 s

From the point of view of the moun Earth moved at 0.983c (2.949*10^8 m/s) during a time of 2.198*10^-6, so it moved

d = v*t = 2.949*10^8 * 2.198*10^-6 = 648 m

7 0
2 years ago
An adult elephant has a mass of about 5.0 tons. An adult elephant shrew has a mass of about 50 grams. How far r from the center
Diano4ka-milaya [45]

Answer:

2023857702.507m

Explanation:

f=\frac{GMm}{r^{2} }

recall from newton's law of gravitation

G=gravitational constant

mshew=50g

melephant=5*10^3kg

rearth=radius of the earth 6400km or 6400000m

mearth= masss of the earth

Gm(shrew)m(earth)/r(earth)^2 = Gm(elephant)m(earth)/r^2

strike out the left hand side and right hand side variables

m(shrew)/r(earth)^2 = m(elephant)/r^2

r^2 = m(elephant).r(earth)^2 / m(shrew)       .........make r^2 the subject of the equation

r^2=(5*10^{3} *(6400000)^{2} )/.05

r^2=40960000000000

r=2023857702.507m

4 0
2 years ago
An electric resistance heater is embedded in a long cylinder of diameter 30 mm. when water with a temperature of 25°c and veloc
victus00 [196]
KNOWN: Long, 30mm-diameter cylinder with embedded electrical heater; power required to maintain a specified surface temperature for water and air flows.
 FIND: Convection coefficients for the water and air flow convection processes, hw and ha, respectively.
 ASSUMPTIONS: Flow is cross-wise over cylinder which is very long in the direction normal to flow.
 The convection heat rate from the cylinder per unit length of the cylinder has the form
 q' = h*(pi*D)*(Ts-Tinf)
 and solving for the heat transfer convection coefficient, find
 Water
 hw = q'/((pi*D)*(Ts-Tinf))
 hw = (38*10^3 W/m) / ((pi*(0.030m))*(80-25)C)= 7330.77314  W/m^2K
 Air
 ha = (400W/m) / ((pi*(0.030m))*(80-25)C)=<span> 77.166033 </span> W/m^2K
 COMMENTS: Note that the air velocity is 10 times that of the water flow, yet
hw ≈ 95 × ha.
 These values for the convection coefficient are typical for forced convection heat transfer with liquids and gases
 Watter is a better convective heat transfer media than air 

3 0
2 years ago
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