# Esercizi risolti sugli errori di misura **Convenzioni** - Errore assoluto: $\Delta x$ - Errore relativo: $\varepsilon_x=\dfrac{\Delta x}{x}$ - Propagazione semplificata: - somme e differenze: $\Delta z=\Delta x+\Delta y$ - prodotti e quozienti: $\varepsilon_z=\varepsilon_x+\varepsilon_y$ - potenze: $\varepsilon_{x^n}=n\varepsilon_x$ --- ## Esercizio 1 – Somma Dati: - $a=(12.30\pm0.05)\,\text{cm}$ - $b=(8.70\pm0.02)\,\text{cm}$ Calcolare $L=a+b$. **Soluzione (F4)** $$ L=a+b=(12.30\,\text{cm})+(8.70\,\text{cm})=21.00\,\text{cm} $$ $$ \Delta L=\Delta a+\Delta b=(0.05\,\text{cm})+(0.02\,\text{cm})=0.07\,\text{cm} $$ $$ \boxed{L=(21.00\pm0.07)\,\text{cm}} $$ --- ## Esercizio 2 – Differenza Dati: - $x=(5.40\pm0.03)\,\text{m}$ - $y=(2.15\pm0.04)\,\text{m}$ Calcolare $d=x-y$. **Soluzione (F4)** $$ d=x-y=(5.40\,\text{m})-(2.15\,\text{m})=3.25\,\text{m} $$ $$ \Delta d=\Delta x+\Delta y=(0.03\,\text{m})+(0.04\,\text{m})=0.07\,\text{m} $$ $$ \boxed{d=(3.25\pm0.07)\,\text{m}} $$ --- ## Esercizio 3 – Prodotto (area) Dati: - $L=(2.50\pm0.02)\,\text{m}$ - $W=(1.20\pm0.01)\,\text{m}$ Calcolare $A=L\cdot W$. **Soluzione (F4)** $$ A=L\cdot W=(2.50\,\text{m})(1.20\,\text{m})=3.00\,\text{m}^2 $$ $$ \varepsilon_A=\frac{0.02\,\text{m}}{2.50\,\text{m}}+\frac{0.01\,\text{m}}{1.20\,\text{m}} =0.0163 $$ $$ \Delta A=\varepsilon_A\cdot A=(0.0163)(3.00\,\text{m}^2)=0.05\,\text{m}^2 $$ $$ \boxed{A=(3.00\pm0.05)\,\text{m}^2} $$ --- ## Esercizio 4 – Quoziente (velocità) Dati: - $s=(120.0\pm0.5)\,\text{m}$ - $t=(14.2\pm0.1)\,\text{s}$ Calcolare $v=s/t$. **Soluzione (F4)** $$ v=\frac{s}{t}=\frac{120.0\,\text{m}}{14.2\,\text{s}}=8.45\,\text{m s}^{-1} $$ $$ \varepsilon_v=\frac{0.5\,\text{m}}{120.0\,\text{m}}+\frac{0.1\,\text{s}}{14.2\,\text{s}} =0.0112 $$ $$ \Delta v=\varepsilon_v\cdot v=(0.0112)(8.45\,\text{m s}^{-1})=0.09\,\text{m s}^{-1} $$ $$ \boxed{v=(8.45\pm0.09)\,\text{m s}^{-1}} $$ --- ## Esercizio 5 – Potenza (volume di un cubo) Dati: - $a=(4.00\pm0.02)\,\text{cm}$ Calcolare $V=a^3$. **Soluzione (F4)** $$ V=a^3=(4.00\,\text{cm})^3=64.0\,\text{cm}^3 $$ $$ \varepsilon_V=3\frac{0.02\,\text{cm}}{4.00\,\text{cm}}=0.015 $$ $$ \Delta V=\varepsilon_V\cdot V=(0.015)(64.0\,\text{cm}^3)=1.0\,\text{cm}^3 $$ $$ \boxed{V=(64.0\pm1.0)\,\text{cm}^3} $$ --- ## Esercizio 6 – Densità Dati: - $m=(245.0\pm0.2)\,\text{g}$ - $V=(30.0\pm0.3)\,\text{cm}^3$ Calcolare $\rho=m/V$. **Soluzione (F4)** $$ \rho=\frac{m}{V}=\frac{245.0\,\text{g}}{30.0\,\text{cm}^3} =8.17\,\text{g cm}^{-3} $$ $$ \varepsilon_\rho=\frac{0.2\,\text{g}}{245.0\,\text{g}}+\frac{0.3\,\text{cm}^3}{30.0\,\text{cm}^3} =0.0108 $$ $$ \Delta\rho=\varepsilon_\rho\cdot\rho=(0.0108)(8.17\,\text{g cm}^{-3}) =0.09\,\text{g cm}^{-3} $$ $$ \boxed{\rho=(8.17\pm0.09)\,\text{g cm}^{-3}} $$ --- ## Esercizio 7 – Periodo di oscillazione Dato: - $t_{20}=(31.6\pm0.2)\,\text{s}$ Calcolare il periodo $T=t_{20}/20$. **Soluzione (F4)** $$ T=\frac{t_{20}}{20}=\frac{31.6\,\text{s}}{20}=1.58\,\text{s} $$ $$ \Delta T=\frac{0.2\,\text{s}}{20}=0.01\,\text{s} $$ $$ \boxed{T=(1.58\pm0.01)\,\text{s}} $$ --- ## Esercizio 8 – Accelerazione Dati: - $s=(1.20\pm0.01)\,\text{m}$ - $t=(0.62\pm0.02)\,\text{s}$ Calcolare $a=\dfrac{2s}{t^2}$. **Soluzione (F4)** $$ a=\frac{2s}{t^2} =\frac{2(1.20\,\text{m})}{(0.62\,\text{s})^2} =6.24\,\text{m s}^{-2} $$ $$ \varepsilon_a=\frac{0.01\,\text{m}}{1.20\,\text{m}} +2\frac{0.02\,\text{s}}{0.62\,\text{s}} =0.0729 $$ $$ \Delta a=\varepsilon_a\cdot a =(0.0729)(6.24\,\text{m s}^{-2}) =0.45\,\text{m s}^{-2} $$ $$ \boxed{a=(6.24\pm0.45)\,\text{m s}^{-2}} $$ --- ## Esercizio 9 – Energia cinetica Dati: - $m=(0.250\pm0.001)\,\text{kg}$ - $v=(3.40\pm0.05)\,\text{m s}^{-1}$ Calcolare $K=\frac12 mv^2$. **Soluzione (F4)** $$ K=\frac12 mv^2 =\frac12(0.250\,\text{kg})(3.40\,\text{m s}^{-1})^2 =1.45\,\text{J} $$ $$ \varepsilon_K=\frac{0.001\,\text{kg}}{0.250\,\text{kg}} +2\frac{0.05\,\text{m s}^{-1}}{3.40\,\text{m s}^{-1}} =0.0334 $$ $$ \Delta K=\varepsilon_K\cdot K =(0.0334)(1.45\,\text{J}) =0.05\,\text{J} $$ $$ \boxed{K=(1.45\pm0.05)\,\text{J}} $$ --- ## Esercizio 10 – Resistenza elettrica Dati: - $V=(12.0\pm0.1)\,\text{V}$ - $I=(1.50\pm0.03)\,\text{A}$ Calcolare $R=V/I$. **Soluzione (F4)** $$ R=\frac{V}{I} =\frac{12.0\,\text{V}}{1.50\,\text{A}} =8.00\,\Omega $$ $$ \varepsilon_R=\frac{0.1\,\text{V}}{12.0\,\text{V}} +\frac{0.03\,\text{A}}{1.50\,\text{A}} =0.0283 $$ $$ \Delta R=\varepsilon_R\cdot R =(0.0283)(8.00\,\Omega) =0.23\,\Omega $$ $$ \boxed{R=(8.00\pm0.23)\,\Omega} $$