[填空题]
A force of 10.0 ± 0.2 N is applied to p pc.gm.6/+ w3wncnq2 yxdgo )a mass off l/6nk*o mxfww2w-7o 2.0 ± 0.01 kg, causing it to accelerate. What is the percentagmok*-xnwo f/ 7lww2f6e 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 usf;z gw/sw,0 sn xpc4a92vh 1 hmj*fq+ling the 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 $7 \%$.
Post time 2024-8-21 21:38:19