솔루션피아

$$ \begin{aligned} & 1\ut{fortnight} \\ =& 1\ut{fortnight} \cdot \frac{14\ut{days}}{1\ut{fortnight}}\cdot \frac{24\ut{h}}{1\ut{days}}\cdot\frac{3600\ut{s}}{1\ut{h}}\cdot\frac{10^6[\mu s]}{1\ut{s}} \\ =& 14 \cdot24\cdot3600\cdot10^6[\mu s] \\ =& 1209600000000[\mu s] \\ \approx& 1\times10^{12}[\mu s] \end{aligned} $$

$$ \%=100\times \frac{(\text{TrueValue}-\text{ApproxValue})}{\text{TrueValue}} $$ (a) $1\ut{\mu Centry} = ?$ $$ \begin{aligned} & 1\ut{\mu Centry} \\ &= 1\times 10^{-6}\ut{Centry} \cdot \frac{100\ut{years}}{1\ut{Centry}} \cdot \frac{365.2422\ut{days}}{1\ut{years}}\cdot \frac{24\ut{h}}{1\ut{days}}\cdot \frac{60\ut{min}}{1\ut{h}} \\ &= 1\times 10^{-6}\cdot100\cdot 365.2422\cdot 24\cdot 60\ut{min} ..

$$ \text{put} \begin{cases} t_A(t_B)&=a_A\cdot t_B +b_A \\ t_B&=ref \\ t_C(t_B)&=a_C\cdot t_B +b_C \end{cases} $$ $$ \text{picture 1-4}, \begin{cases} t_A(125)&=312&=125a_A +b_A \\ t_A(290)&=512&=290a_A +b_A \\ t_C(25.0)&=92.0&=25.0a_C +b_C \\ t_C(200)&=142&=200a_C +b_C \end{cases} $$ $$ \therefore \begin{cases} t_A(t_B)&=\frac{40}{33}t_B +\frac{5818}{33} \\ t_B&=ref \\ t_C(t_B)&=\frac{2}{7}t_B ..

$$ \begin{aligned} & 10^{10}\ut{years} : 10^6\ut{years} = 1.0\ut{days} : x\ut{s} \end{aligned} $$ $$ \begin{aligned} \\ x &= \frac{10^6\ut{years} \times 1.0\ut{days}}{10^{10}\ut{years}} \\ &= \frac{10^6\ut{years} \times 1.0\ut{days}}{10^{10}\ut{years}} \times \frac{24\ut{h}}{1\ut{days}} \times \frac{3600\ut{s}}{1\ut{h}} \\ &= \frac{10^6 \times 1\times 24\times 3600}{10^{10}\times 1\times 1}\ut{s..

$$ \begin{aligned} & 100.0\ut{years} : \pm0.010\ut{s} = 1.0\ut{s} : \pm x\ut{s} \end{aligned} $$ $$ \begin{aligned} \\ x &= \frac{0.010\ut{s} \cdot 1.0\ut{s}}{100.0\ut{years}} \\ &= \frac{0.010\ut{s} \cdot 1.0\ut{s}}{100.0\ut{years}}\cdot \frac{1\ut{years}}{365.2422\ut{days}} \cdot \frac{1\ut{days}}{24\ut{h}}\cdot \frac{1\ut{h}}{3600\ut{s}} \\ &= \frac{0.010 \cdot 1.0\cdot 1\cdot 1\cdot 1}{100.0..

$$ \begin{aligned} A_n= A_0 \times (1-0.43)^n

$$ \begin{aligned} & \Ans \\ =& 1.00\ut{cm} \cdot \sqrt[3]{1\ut{mol}} \\ =& 1.00\ut{cm}\cdot \frac{1\ut{m}}{100\ut{cm}} \cdot \sqrt[3]{1\ut{mol}\cdot \frac{6.02\times10^{23}\ut{EA}}{1\ut{mol}}} \\ =& \frac{1.00\cdot1\cdot\sqrt[3]{6.02\times10^{23}}}{100}\ut{m} \\ =& 100000\sqrt[3]{602}\ut{m} \\ \approx& 844368.773356757\ut{m} \\ \approx& 8.44\times 10^{5}\ut{m} = 844\ut{km} \end{aligned} $$

$$ \begin{aligned} & \Ans \\ =& 3.8\ut{mm} \cdot \frac{1}{100} \\ =& 3.8\ut{mm} \cdot \frac{1\ut{m}}{1000\ut{mm}} \cdot \frac{1}{100} \\ =& \frac{38\cdot1\cdot1}{10\cdot1000\cdot100}\ut{m} \\ =& \frac{19}{500000}\ut{m} \\ =& 0.000038\ut{m} \\ =& 3.8\times 10^{-5}\ut{m} = 38\ut{\mu m} \end{aligned} $$

$$ \begin{aligned} & \Ans \\=& 70.0\ut{mi/h} \cdot 1.00\ut{year} \\=& \frac{70\ut{mi}}{1\ut{h}} \cdot 1\ut{year}\cdot \frac{365.2422\ut{day}}{1\ut{year}}\cdot\frac{24\ut{h}}{1\ut{day}} \\=& \frac{70\cdot 1\cdot 365.2422\cdot 24}{1\cdot 1\cdot 1}\ut{mi} \\=& \frac{76700862}{125}\ut{mi} \\=& 613606.896\ut{mi} \\ \approx& 6.14\times10^{5}\ut{mi} \end{aligned} $$

(a) $$ \begin{aligned} &5.00\ut{almude} \\ &= 5.00\ut{almude} \cdot \frac{2\ut{medio}}{1\ut{almude}} \\ &= 10\ut{medio} \\ &= 10.0\ut{medio} \end{aligned} $$ (b) $$ \begin{aligned} &5.00\ut{almude} \\ &= 5.00\ut{almude} \cdot \frac{1\ut{cahiz}}{144\ut{almude}} \\ &= \frac{5}{144}\ut{cahiz} \\ &\approx 0.03472222222222222\ut{cahiz} \\ &\approx 3.47\times 10^{-2}\ut{cahiz} \end{aligned} $$ (c) $$ ..