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Citrus2011 [14]
3 years ago
13

A student determines the density ρ of steel by taking measurements from a steel wire

Physics
1 answer:
Marina CMI [18]3 years ago
6 0

Answer:

The percentage uncertainty in his calculated value of density is \pm 0.713\,\%.

Explanation:

We can estimate the absolute uncertainty by the definition of total differential. That is:

\Delta \rho \approx \frac{\partial \rho}{\partial m}\cdot \Delta m + \frac{\partial \rho}{\partial d}\cdot \Delta d + \frac{\partial \rho}{\partial l}\cdot \Delta l (1)

Where:

\frac{\partial \rho}{\partial m} - Partial derivative of the density with respect to mass, measured in \frac{1}{mm^{3}}.

\frac{\partial \rho}{\partial d} - Partial derivative of the density with respect to diameter, measured in grams per cubic milimeter.

\frac{\partial \rho}{\partial l} - Partial derivative of the density with respect to length, measured in grams per cubic milimeter.

\Delta m - Mass uncertainty, measured in grams.

\Delta d - Diameter uncertainty, measured in milimeters.

\Delta l - Length uncertainty, measured in milimeters.

\Delta \rho - Density uncertainty, measured in grams per cubic milimeters.

Partial derivatives are, respectively:

\frac{\partial \rho}{\partial m} = \frac{4}{\pi\cdot d^{2}\cdot l} (2)

\frac{\partial \rho}{\partial d} = -\frac{8\cdot m}{\pi\cdot d^{3}\cdot l} (3)

\frac{\partial \rho}{\partial l} = - \frac{4\cdot m}{\pi\cdot d^{2}\cdot l^{2}} (4)

And we expand (1) as follows:

\Delta \rho \approx \frac{4\cdot \Delta m}{\pi\cdot d^{2}\cdot l} - \frac{8\cdot m\cdot \Delta d}{\pi\cdot d^{3}\cdot l}-\frac{4\cdot m\cdot \Delta l}{\pi\cdot d^{2}\cdot l^{2}}

\Delta \rho \approx \left(\frac{4}{\pi\cdot d^{2}\cdot l}\right)\cdot \left(\Delta m -\frac{m\cdot \Delta d}{d}-\frac{m \cdot \Delta l}{l}  \right) (5)

If we know that d = 2\,mm, l = 25\,mm, m = 6.2\,g, \Delta m = \pm 0.1\,g, \Delta d = \pm 0.01\,mm and \Delta l = \pm 0.1\,mm, then the absolute uncertainty is:

\Delta \rho \approx \pm\left[\frac{4}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)} \right]\cdot \left[(0.1\,g)-\frac{(6.2\,g)\cdot (0.01\,mm)}{2\,mm} -\frac{(6.2\,g)\cdot (0.1\,mm)}{25\,mm} \right]

\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}

And the expected density is:

\rho = \frac{4\cdot m}{\pi\cdot d^{2}\cdot l} (6)

\rho = \frac{4\cdot (6.2\,g)}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)}

\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}

The percentage uncertainty in his calculated value of density is:

\%e = \frac{\Delta \rho}{\rho}\times 100\,\% (7)

If we know that \Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}} and \rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}, then the percentage uncertainty is:

\%e = \frac{\pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}} }{78.941\times 10^{-3}\,\frac{g}{mm^{3}} }\times 100\,\%

\%e = \pm 0.713\,\%

The percentage uncertainty in his calculated value of density is \pm 0.713\,\%.

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Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

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