[填空题]
A force of 10.0 ± 0.2 N is appliedr ,-:7vqfa 3v: cf- lsdstrsn/m5 a-hv to a mass of 2.0 ± 0.01 kg, casf w9fcbwr 8g) 7r:2piusing it to accelerate. What is the percentage uncertainty of its accelerationgr)cf wbswr pi82f9:7 ? % (do not include the percent sign in your result)
参考答案: 2.50
本题详细解析: The acceleration is calculated using t- vvdo+bf9lvt;7p 7e*uig hjl y 2:xvw+q v)r+he formula: $$ a=\frac{F}{m} $$ so the percentage uncertainty on the acceleration is the sum of the percentage error on the force and on the mass. The percentage uncertainty on the force is: $$ \frac{0.2}{10.0} \times 100 \%=2 \% $$ The percentage uncertainty on the mass is:
$$ \frac{0.1}{2.0} \times 100 \%=5 \% . $$ Hence the percentage error on the acceleration will be $7 \%$.
