[填空题]
A force of 10.0 ± 0.2 N is applied to a mass of hr6 e tfe(z 55y7tfdo7fpq(tgo3y d/ 12.0 ± 0.0*ref u, szvxjnm+ 2,/r8az qn01 kg, causing it to accelerat vz0 xrae8uj*, /q+s mn2,fnrze. What is the percentage uncertainty of its acceleration? % (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 the formula:v45r7xlyn4 e3is;j6lo4wm gg $$ a=\frac{F}{m} $$ 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 $7 \%$.
Post time 2024-8-21 21:38:19