Find the greatest possible error for each measurement. 19.2 m

Mathematics

1 answer:

il63 [147K]3 years ago

5 0

19.25

Working;
The number is measured to the nearest 0.1
absolute error is 0.1/2=0.05
Add absolute error to 19.2

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The 6-lb particle is subjected to the action of its weight and forces F1 = 52i + 6j - 2tk6 lb, F2 = 5t 2 i - 4tj - 1k6 lb, and F

olga2289 [7]

Answer:

r=294.9m

Step-by-step explanation:

The forces on the particle are

$W=mg\hat{j}\\F_{1}=52\hat{i}+6\hat{j}-2t\hat{k}\\F_{2}=5t^{2}\hat{i}-4t\hat{j}-1\hat{k}\\F_{3}=(5-2t)\hat{i}$

Now , we sum all these forces to get the net force

$F_{T}=W+F_{1}+F_{2}+F_{3}\\F_{T}=(52+5t^{2}+5-2t)\hat{i}+((6+6-4t)\hat{j}+(-2t-1)\hat{k}\\F_{T}=(57-2t+5t^{2})\hat{i}+(12-4t)\hat{j}+(-2t-1)\hat{k}\\$

we can use the fact F=m*a and integrate the acceleration

$a(t)=\frac{1}{m}F(t)\\\\v(t)=\int a(t)dt=\frac{1}{m}\int{F_{T}}dt\\\\v(t)=\frac{1}{m}[(57t-t^{2}+\frac{5}{3}t^{3})\hat{i}+(12t-2t^{2})\hat{j}+(-t^{2}-t)\hat{k}]\\\\r(t)=\int v(t)dt=\frac{1}{m}[(\frac{57}{2}t^{2}-\frac{1}{3}t^{3}}+\frac{5}{4}t^{4})\hat{i}+(6t^{2}-\frac{2}{3}t^{3})\hat{j}+(-\frac{1}{3}t^{3}-\frac{1}{2}t^{2})]$

and we evaluate in r(2) an we take the norm to obtain the distance

$r(2)=\frac{1}{m}[\frac{394}{3}\hat{i}+\frac{56}{3}\hat{j}-\frac{14}{3}\hat{k}]\\|r(2)|=\frac{1}{m}\sqrt{[(\frac{394}{3})^{2}+(\frac{56}{3})^{2}+(\frac{14}{3})^{2}]}\\|r(2)|=\frac{132.73}{0.45}=294.9m$