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183_notes:model_of_a_wire [2015/09/19 11:20] – [Modeling the solid wire] caballero | 183_notes:model_of_a_wire [2015/09/28 20:06] – caballero | ||
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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}\: | ||
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=== Two springs connected end-to-end (series) === | === Two springs connected end-to-end (series) === | ||
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+ | [{{183_notes: | ||
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Let's consider attaching a 100N ball to a single 100N/m spring. If we let the weight just hang motionless (no change in momentum), we know from the [[183_notes: | Let's consider attaching a 100N ball to a single 100N/m spring. If we let the weight just hang motionless (no change in momentum), we know from the [[183_notes: | ||
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Let's consider attaching a 100N ball to two 100N/m springs where each spring is connected to the ball and not to each other. In this case, both springs must stretch by the same amount. If the ball hangs motionless (no change in momentum), we can use the momentum principle to determine how much these springs stretch. | Let's consider attaching a 100N ball to two 100N/m springs where each spring is connected to the ball and not to each other. In this case, both springs must stretch by the same amount. If the ball hangs motionless (no change in momentum), we can use the momentum principle to determine how much these springs stretch. | ||
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+ | [{{183_notes: | ||
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$$\Delta \vec{p} = 0\: | $$\Delta \vec{p} = 0\: | ||
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$$s = \dfrac{mg}{2k} = \dfrac{100N}{200N/ | $$s = \dfrac{mg}{2k} = \dfrac{100N}{200N/ | ||
- | When we attach a second 100N/m spring to the ball, the springs both stretch 0.5m. That is, the overall stretch of the spring-mass system is half of what it is with one spring. //In parallel, each spring stretches the same amount//. | + | When we attach a second 100N/m spring to the ball, the springs both stretch 0.5m. That is, the overall stretch of the spring-mass system is half of what it is with one spring. |
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+ | //In parallel, each spring stretches the same amount//. | ||
== Modeling two side-by-side springs as one spring (effective spring constant) == | == Modeling two side-by-side springs as one spring (effective spring constant) == | ||
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$${k_{s, | $${k_{s, | ||
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+ | This way of modeling end-to-end and side-by-side springs will be very useful for modeling [[183_notes: |