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3 years ago

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A manufacturer claims that a carpet will not generate more than 5.8 kV of static electricity What magnitude of charge would have

to be transferred between a carpet and a shoe for there to be a 5.8 kV potential distance d = 2.8 mm ? Approximate the area of a shoe as 30 cm x 8 cm. Express your answer using two significant figures.

1 answer:

DiKsa [7]3 years ago

6 0

Answer:

4.4×10⁻⁷ Coulomb

Explanation:

V = Voltage = 5.8 kV

d = Potential distance = 2.8 mm = 0.0028 m

A = Area = 0.3×0.08 = 0.024 m²

ε₀ = permittivity constant in a Vacuum= 8.85×10⁻¹² F/m

$\frac{Q}{V}=\frac{A\epsilon_0}{d}\\\Rightarrow \Q=V\frac{A\epsilon_0}{d}\\\Rightarrow Q=5.8\times 10^3\frac{0.024\times 8.85\times 10^{-12}}{0.0028}\\\Rightarrow Q=4.4\times 10^{-7}\ C$

Magnitude of charge transferred between a carpet and a shoe is 4.4×10⁻⁷ Coulomb.

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A circular copper wire is put in tension under a weight of 7000N. What is the ratio of its diameter after and before the load is

zaharov [31]

Answer:

$\frac{d_f}{d} =0.9999983$

Explanation:

Given:

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cross sectional area of the wire, $a=0.01\ m^2$

Poisson's ratio, $\mu=0.3$

we have, Young's modulus, $E=128\times 10^3\ MPa$

<u>Stress induced due to the applied force:</u>

$\sigma=\frac{F}{a}$

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$\sigma=700000\ Pa=0.7\ MPa$

<u>Now the longitudinal strain:</u>

<u /> $\epsilon=\frac{\sigma}{E}$ <u />

<u /> $\epsilon=\frac{0.7}{128\times 10^3}$ <u />

<u /> $\epsilon=5.468\times 10^{-6}$ <u />

Now from the relation of Poisson's ratio:

$\mu=\frac{\nu}{\epsilon}$

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$0.3=\frac{\nu}{5.468\times 10^{-6}}$

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<u>When the tensile load is applied its diameter decreases:</u>

The lateral strain is also given as,

$\nu=\frac{\Delta d}{d}$

$1.6406\times 10^{-6}=\frac{\Delta d}{112.8}$

$\Delta d=0.000185\ mm$

Now the final diameter will be:

$d_f=d-\Delta d$

$d_f=112.8-0.000185$

$d_f=112.799815\ mm$

Now the ratio:

$\frac{d_f}{d} =\frac{112.799815}{112.8}$

$\frac{d_f}{d} =0.9999983$

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