Answer on Question #85542 - Physics - Mechanics | Relativity
The density of a ball is ( 300 ± 8 ) k g ⋅ m − 3 (300 \pm 8) \, \mathrm{kg} \cdot \mathrm{m}^{-3} ( 300 ± 8 ) kg ⋅ m − 3 35 c m 35 \, \mathrm{cm} 35 cm 78 c m 78 \, \mathrm{cm} 78 cm ± 1 c m \pm 1 \, \mathrm{cm} ± 1 cm
Given:
ρ = ( 300 ± 8 ) k g / m 3 x 1 = ( 35 ± 1 ) c m x 2 = ( 78 ± 1 ) c m m − ? \begin{array}{l}
\rho = (300 \pm 8) \, \mathrm{kg} / \mathrm{m}^{3} \\
x_{1} = (35 \pm 1) \, \mathrm{cm} \\
x_{2} = (78 \pm 1) \, \mathrm{cm} \\
\mathrm{m} - ? \\
\end{array} ρ = ( 300 ± 8 ) kg / m 3 x 1 = ( 35 ± 1 ) cm x 2 = ( 78 ± 1 ) cm m − ? Solution.
Let's find the mass of the ball:
m = ρ V = 4 3 π ρ R 3 = 4 3 π ρ ( D 2 ) 3 = 1 6 π ρ ( x 2 − x 1 ) 3 m = 1 6 ⋅ 3.14 ⋅ 300 ⋅ ( 0.75 − 0.35 ) = 62.8 ( k g ) \begin{array}{l}
m = \rho V = \frac{4}{3} \pi \rho R^{3} = \frac{4}{3} \pi \rho \left(\frac{D}{2}\right)^{3} = \frac{1}{6} \pi \rho (x_{2} - x_{1})^{3} \\
m = \frac{1}{6} \cdot 3.14 \cdot 300 \cdot (0.75 - 0.35) = 62.8 \, (\mathrm{kg}) \\
\end{array} m = ρ V = 3 4 π ρ R 3 = 3 4 π ρ ( 2 D ) 3 = 6 1 π ρ ( x 2 − x 1 ) 3 m = 6 1 ⋅ 3.14 ⋅ 300 ⋅ ( 0.75 − 0.35 ) = 62.8 ( kg )
Let's calculate the absolute error:
Δ m ≈ d m = ∣ ∂ m ∂ ρ ∣ ∣ Δ ρ ∣ + ∣ ∂ m ∂ x 1 ∣ ∣ Δ x 1 ∣ + ∣ ∂ m ∂ x 2 ∣ ∣ Δ x 2 ∣ = = ∣ ∂ ∂ ρ ( 1 6 π ρ ( x 2 − x 1 ) 3 ) ∣ ∣ Δ ρ ∣ + ∣ ∂ ∂ x 1 ( 1 6 π ρ ( x 2 − x 1 ) 3 ) ∣ ∣ Δ x 1 ∣ + ∣ ∂ ∂ x 2 ( 1 6 π ρ ( x 2 − x 1 ) 3 ) ∣ ∣ Δ x 2 ∣ = = 1 6 π ( x 2 − x 1 ) 3 Δ ρ + 1 6 π ρ ⋅ 3 ( x 2 − x 1 ) 2 Δ x 1 + 1 6 π ρ ⋅ 3 ( x 2 − x 1 ) 2 Δ x 2 = = 1 6 π ( x 2 − x 1 ) 3 Δ ρ + 1 6 π ρ ⋅ 3 ( x 2 − x 1 ) 2 ( Δ x 1 + Δ x 2 ) \begin{array}{l}
\Delta m \approx d m = \left| \frac{\partial m}{\partial \rho} \right| |\Delta \rho| + \left| \frac{\partial m}{\partial x_{1}} \right| |\Delta x_{1}| + \left| \frac{\partial m}{\partial x_{2}} \right| |\Delta x_{2}| = \\
= \left| \frac{\partial}{\partial \rho} \left( \frac{1}{6} \pi \rho (x_{2} - x_{1})^{3} \right) \right| |\Delta \rho| + \left| \frac{\partial}{\partial x_{1}} \left( \frac{1}{6} \pi \rho (x_{2} - x_{1})^{3} \right) \right| |\Delta x_{1}| + \left| \frac{\partial}{\partial x_{2}} \left( \frac{1}{6} \pi \rho (x_{2} - x_{1})^{3} \right) \right| |\Delta x_{2}| = \\
= \frac{1}{6} \pi (x_{2} - x_{1})^{3} \Delta \rho + \frac{1}{6} \pi \rho \cdot 3 (x_{2} - x_{1})^{2} \Delta x_{1} + \frac{1}{6} \pi \rho \cdot 3 (x_{2} - x_{1})^{2} \Delta x_{2} = \\
= \frac{1}{6} \pi (x_{2} - x_{1})^{3} \Delta \rho + \frac{1}{6} \pi \rho \cdot 3 (x_{2} - x_{1})^{2} (\Delta x_{1} + \Delta x_{2}) \\
\end{array} Δ m ≈ d m = ∣ ∣ ∂ ρ ∂ m ∣ ∣ ∣Δ ρ ∣ + ∣ ∣ ∂ x 1 ∂ m ∣ ∣ ∣Δ x 1 ∣ + ∣ ∣ ∂ x 2 ∂ m ∣ ∣ ∣Δ x 2 ∣ = = ∣ ∣ ∂ ρ ∂ ( 6 1 π ρ ( x 2 − x 1 ) 3 ) ∣ ∣ ∣Δ ρ ∣ + ∣ ∣ ∂ x 1 ∂ ( 6 1 π ρ ( x 2 − x 1 ) 3 ) ∣ ∣ ∣Δ x 1 ∣ + ∣ ∣ ∂ x 2 ∂ ( 6 1 π ρ ( x 2 − x 1 ) 3 ) ∣ ∣ ∣Δ x 2 ∣ = = 6 1 π ( x 2 − x 1 ) 3 Δ ρ + 6 1 π ρ ⋅ 3 ( x 2 − x 1 ) 2 Δ x 1 + 6 1 π ρ ⋅ 3 ( x 2 − x 1 ) 2 Δ x 2 = = 6 1 π ( x 2 − x 1 ) 3 Δ ρ + 6 1 π ρ ⋅ 3 ( x 2 − x 1 ) 2 ( Δ x 1 + Δ x 2 ) ε = Δ m m = 1 6 π ( x 2 − x 1 ) 3 Δ ρ + 1 6 π ρ ⋅ 3 ( x 2 − x 1 ) 2 ( Δ x 1 + Δ x 2 ) 1 6 π ρ ( x 2 − x 1 ) 3 = Δ ρ ρ + 3 Δ x 1 + Δ x 2 x 2 − x 1 \varepsilon = \frac{\Delta m}{m} = \frac{ \frac{1}{6} \pi (x_{2} - x_{1})^{3} \Delta \rho + \frac{1}{6} \pi \rho \cdot 3 (x_{2} - x_{1})^{2} (\Delta x_{1} + \Delta x_{2}) }{ \frac{1}{6} \pi \rho (x_{2} - x_{1})^{3} } = \frac{\Delta \rho}{\rho} + 3 \frac{\Delta x_{1} + \Delta x_{2}}{x_{2} - x_{1}} ε = m Δ m = 6 1 π ρ ( x 2 − x 1 ) 3 6 1 π ( x 2 − x 1 ) 3 Δ ρ + 6 1 π ρ ⋅ 3 ( x 2 − x 1 ) 2 ( Δ x 1 + Δ x 2 ) = ρ Δ ρ + 3 x 2 − x 1 Δ x 1 + Δ x 2
Relative error:
ε = 8 300 + 3 0.01 + 0.01 0.75 − 0.35 ≈ 0.18 \varepsilon = \frac{8}{300} + 3 \frac{0.01 + 0.01}{0.75 - 0.35} \approx 0.18 ε = 300 8 + 3 0.75 − 0.35 0.01 + 0.01 ≈ 0.18
Than absolute error:
Δ m = m ε = 62.8 ⋅ 0.18 = 11.3 ≈ 12 ( k g ) \Delta m = m \varepsilon = 62.8 \cdot 0.18 = 11.3 \approx 12 \, (\mathrm{kg}) Δ m = m ε = 62.8 ⋅ 0.18 = 11.3 ≈ 12 ( kg )
Finally the mass of the ball:
m = ( 62.8 ± 12 ) k g ≈ ( 63 ± 12 ) k g m = (62.8 \pm 12) \, \mathrm{kg} \approx (63 \pm 12) \, \mathrm{kg} m = ( 62.8 ± 12 ) kg ≈ ( 63 ± 12 ) kg
Answer: m = ( 63 ± 12 ) k g m = (63 \pm 12) \, \mathrm{kg} m = ( 63 ± 12 ) kg
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#339914 on Dec 2023