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| Both sides previous revision Previous revision Next revision | Previous revision Next revisionBoth sides next revision | ||
| 183_notes:model_of_a_wire [2015/09/20 12:17] – [Modeling the solid wire] caballero | 183_notes:model_of_a_wire [2018/05/29 20:54] – hallstein | ||
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| + | Section 4.4 and 4.5 in Matter and Interactions (4th edition) | ||
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| ===== Modeling a Solid Wire with Springs ====== | ===== Modeling a Solid Wire with Springs ====== | ||
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| $$\rho = 21.45 \dfrac{g}{cm^3} \left(\dfrac{1kg}{10^3g}\right)\left(\dfrac{100 cm}{1m}\right)^3 = 21.45 \times 10^3 kg/m^3$$ | $$\rho = 21.45 \dfrac{g}{cm^3} \left(\dfrac{1kg}{10^3g}\right)\left(\dfrac{100 cm}{1m}\right)^3 = 21.45 \times 10^3 kg/m^3$$ | ||
| - | Consider a cubic meter of Pt, which is a cube that is 1m long on each side((This amount of Pt would costs nearly $1, | + | Consider a cubic meter of Pt, which is a cube that is 1m long on each side((This amount of Pt would costs nearly $700, |
| $$21.45\times10^3 \dfrac{kg}{m^3} \left(\dfrac{1 mol}{0.195 kg}\right) \left(\dfrac{6.02\times10^{23} atoms}{1 mol}\right) = 6.62\times10^{28}\: | $$21.45\times10^3 \dfrac{kg}{m^3} \left(\dfrac{1 mol}{0.195 kg}\right) \left(\dfrac{6.02\times10^{23} atoms}{1 mol}\right) = 6.62\times10^{28}\: | ||