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ziro4ka [17]
3 years ago
15

A silver bar 0.125 meter long is subjected to a temperature change from 200°C to 100°C. What will be the length of the bar after

the temperature change? A. 0.0000189 meter B. 0.00002363 meter C. 0.00023635 meter D. 0.124764 meter
Physics
2 answers:
Ad libitum [116K]3 years ago
6 0
ΔL=Lo*α*ΔT
We know that α=2,0*10⁻⁵
ΔL=0,125m*2,0*10⁻⁵*100°C
ΔL=0,00025m
Now the new lengh would be:
0,125m+0,00025=0,12525 ≈(d)
Delicious77 [7]3 years ago
5 0
\Delta L=  \alpha L_0 (T_f-T_i)

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248

D
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\begin{array}{ccc}&\text{Radius} & \text{Drift Speed}\\d) & 2 \; r &2.5 \; v\\a) & 3 \; r &1 \; v\\b) & 4 \; r &0.5 \; v\\c) & 1 \; r &5 \; v\\\end{array}

<h3>Explanation</h3>

I = v \cdot A \cdot n \cdot q,

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  • n is the number of charge carrier per unit volume, and
  • q is the charge on each charge carrier.

Area of a circular cross-section:

A = \pi \cdot r^{2},

where

  • r is the radius of the wire.

n and q are the same for all four samples, for they are made out of the same material.

As a result, I of each wire is directly proportional to v \cdot r^{2} where the value of \pi \cdot n \cdot q is constant.

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How do the four wires rank by their current?

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