\( B_{i}=1.10 \mathrm{~T} \) (upward)
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$\begin{array}{l}{\text { Determine the activities of }(a) 1.0 \mathrm{g} \text { of } \frac{131}{53} \mathrm{I}\left(T_{\frac{1}{2}}=8.02 \text { days) }\right.} \\ {\text { and }(b) 1.0 \mathrm{g} \text { of }_{92}^{238} \mathrm{U}\left(T_{1}=4.47 \times 10^{9} \mathrm{yr}\right)}\end{array} $
$$ \begin{aligned} &t=\frac{u \sin \theta}{g} \text { and } t^{\prime}=\frac{u \sin \theta}{g^{\prime}}=\frac{u \sin \theta}{\frac{11 g}{10}}=\frac{10}{11} t\\ &\text { \% decrease in } t=\frac{t-t^{\prime}}{t} \times 100=\left(1-\frac{10}{11}\right) \times 100=9 \%\\ &\text { Hence, the correct answer is option (b). } \end{aligned} $$
Compute the velocity and acceleration at time $t=t_{0} .$ Is the object going up or down? $$ h(t)=10 t^{2}-24 t,(a) t_{0}=2(b) t_{0}=1 $$
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