Respuesta :
Answer:
a
The diameter is [tex]D =1.319*10^{-7} m[/tex]
b
The length to choose is [tex]L= 3.57m[/tex]
Explanation:
From the question we are told that
The mass of aluminum is [tex]m_A = 1.0g = 1.0*10^{-3}kg[/tex]
The expected power to dissipate is [tex]P_e = 6.5W[/tex]
The potential difference is [tex]E = 2.5V[/tex]
First we need to obtain the resistance R which is mathematically represented as
[tex]R = \frac{E^2}{P_e}[/tex]
Substituting values
[tex]R = \frac{2.5^2 }{6.5}[/tex]
[tex]= 0.962 \ \Omega[/tex]
Next we obtain the volume of the aluminium which is mathematically represented as
[tex]v =\frac{m_A}{\rho_A}[/tex]
The value of the density of aluminium is [tex]\rho_A=2700 kg/m^3[/tex]
Now substituting the value
[tex]v = \frac{1.0*10^{-3}}{2700}[/tex]
[tex]= 0.37 *10^{-6}m^3[/tex]
The resistance can also be mathematically represented as
[tex]R = \frac{\rho L}{A}[/tex]
Where L is the length , A is the area and [tex]\rho[/tex] is resistivity
and for aluminum the resistivity is [tex]\rho = 2.8*10^{-8} \Omega \ \cdot m[/tex]
Now multiplying the denominator and the numerator by L i.e [tex]\frac{L}{L} = 1[/tex]
[tex]R = \frac{\rho L}{A} [\frac{L}{L} ][/tex]
[tex]= \frac{\rho L ^2}{AL}[/tex]
But [tex]AL = v[/tex] i.e area × length = volume
[tex]R = \frac{\rho L ^2}{v}[/tex]
Now making L the subject of the formula in the question above
[tex]L = \sqrt{\frac{Rv}{\rho} }[/tex]
[tex]= \sqrt{\frac{(0.962)(0.37*10^{-6})}{2.8*10{-8} } }[/tex]
[tex]=3.57m[/tex]
Volume can also be mathematically represented as
[tex]v =AL[/tex]
Now making A the subject
[tex]A = \frac{v}{L}[/tex]
[tex]= \frac{0.37 *10 ^{-6}}{3.57}[/tex]
[tex]=1.036*10^{-7}m^2[/tex]
The area A is mathematically represented as
[tex]A = \pi r^2[/tex]
And [tex]r = \frac{D}{2}[/tex]
Now substituting this into the formula for area
[tex]A= \pi \frac{D^2}{4}[/tex]
Making D(diameter) the subject of the formula
[tex]D = \sqrt{\frac{4A}{\pi} }[/tex]
[tex]=\sqrt{\frac{4 * 1.036 *10^{-7}}{3.142} }[/tex]
[tex]=1.319*10^{-7}m[/tex]
This question involves the concepts of density, volume, dissipated power, Ohm's Law, and resistance.
a) The diameter of the wire will be "3.6 cm".
b) The length of the wire will be "3.66 m".
First, we will find out the volume of aluminum, using the formula of density:
[tex]Density=\frac{mass}{Volume}\\\\V =\frac{mass}{Density}\\\\V = \frac{0.001\ kg}{2710\ kg/m^3}\\\\V = 3.7\ x\ 10^{-7}\ m^3[/tex]
Now, we will use the formula for dissipated power to find out the resistance:
[tex]P=IE[/tex]
but according to Ohm's Law:
[tex]E=IR\\I=\frac{E}{R}[/tex]
Therefore,
[tex]P=\frac{E^2}{R}\\\\R=\frac{E^2}{P}[/tex]
where,
R = Resistance = ?
E = Voltage = 2.5 v
P = dissipated power = 6.5 W
Therefore,
[tex]R=\frac{(2.5\ v)^2}{6.5\ W}[/tex]
R = 0.96 Ω
b)
Now, we will use another formula for resistance of the wire:
[tex]R =\frac{\rho L}{A}\\\\R =\frac{\rho L}{A}\frac{L}{L}\\\\R =\frac{\rho L^2}{V}\\\\L=\sqrt{\frac{RV}{\rho}}[/tex]
where,
L = length of wire = ?
[tex]\rho[/tex] = resistivity of aluminum = 2.65 x 10⁻⁸ Ω.m
Therefore,
[tex]L=\sqrt{\frac{(0.96\ \Omega)(3.7\ x\ 10^{-7}\ m^3)}{2.65\ x\ 10^{-8}\ \Omega.m}}[/tex]
L = 3.66 m
a)
Now the diameter can be found by using the formula of volume:
[tex]V = A*L\\V = \frac{\pi d^2}{4}L\\\\d=\sqrt{\frac{4V}{\pi L}}\\\\d=\sqrt{\frac{4(3.7\ x\ 10^{-7}\ m^3)}{\pi (3.66\ m)}}[/tex]
d = 0.36 x 10⁻³ m = 3.6 cm
Learn more about Ohm's Law here:
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