Question

2 dans chaque cas, calculez le résultat de lopération en indiquant lincertitude correspondante.
a) (25,5 ± 0,5)g - (12,1 ± 0,1)g
b) (0,18 ± 0,01)kg ÷ (15,9 ± 0,1)cm³

Explanation:

Step1: Calculate value for a)

First compute each term: $25.5 \div 0.5 = 51$, $12.1 \div 0.1 = 121$. Then subtract: $51 - 121 = -70$.
For uncertainty: Use relative uncertainty for division: $\frac{\Delta A}{A} = \frac{\Delta x}{x} + \frac{\Delta y}{y}$.
For $25.5 \pm 0.5$: $\frac{0.5}{25.5} \approx 0.0196$. For $0.5$ (exact? No, wait—wait, the operation is $(25.5 \pm 0.5)g - (12.1 \pm 0.1)g$? Wait, original: $(25.5 ±0.5)g - (12.1 ±0.1)g$. So absolute uncertainty for subtraction: $\Delta = 0.5 + 0.1 = 0.6$.

Step2: Calculate value for b)

First compute each term: $0.18 \div 0.01 = 18$, $15.9 \div 0.1 = 159$. Then divide: $18 \div 159 \approx 0.1132$.
For uncertainty: Relative uncertainty for division: $\frac{\Delta (A/B)}{A/B} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$.
$\frac{\Delta (0.18)}{0.18} = \frac{0.01}{0.18} ≈0.0556$, $\frac{\Delta (15.9)}{15.9} = \frac{0.1}{15.9}≈0.0063$. Total relative uncertainty: $0.0556 +0.0063≈0.0619$. Absolute uncertainty: $0.1132×0.0619≈0.0070$.

Answer:

a) $-70 ±0.6$ g
b) $0.11 ±0.01$ kg/cm³ (rounded to appropriate sig figs)