솔루션피아

$$ \begin{cases} \rho_{WATER} &= 1000\ut{kg/m^3} \\ 1\ut{gal} &= 231\ut{inch^3} \\ 1\ut{inch} &= 2.540\ut{cm} \\ v &= 1.5\ut{g/min} \end{cases} $$ (a) $$ \begin{aligned} \Ans &= 1000000\ut{cm^3} - 193\ut{gal} \\ &= 1000000\ut{cm^3} - 193\ut{gal}\cdot\frac{231\ut{inch^3}}{1\ut{gal}}\cdot\left(\frac{2.54\ut{cm}}{1\ut{inch}}\right)^3 \\ &= 1000000\ut{cm^3} - \frac{91323059289}{125000}\ut{cm^3} \\ &..

$$ \begin{aligned} \omega &= \frac{2\pi\ut{rad}}{1\ut{days}} \\&= \frac{2\pi\ut{rad}}{1\ut{days}}\cdot \frac{1\ut{days}}{24\ut{h}}\cdot \frac{1\ut{h}}{3600\ut{s}} \\&=\frac{\pi }{43200}\ut{rad/s} \end{aligned} $$ $$ \begin{cases} H &= 1.70\ut{m} \\ t &= 11.1\ut{s} \\ \omega &= \frac{\pi }{43200}\ut{rad/s} \end{cases} $$ $$ \begin{cases} \theta &= \omega t \\ \cos \theta &= \frac{R}{R+H} \end{cas..

(현재가 21c 첫날이며, 하루는 정확히 24h 이라고 가정한다.) (문제에 필요한 자료가 부족하여 임의로 둠) $$ \begin{aligned} \Ans =& 24\ut{h} + \ut{time}\times\ut{AddDaysTime/Time} \\ =& 24\ut{h} + \(30\ut{c}-21\ut{c}\)\times1\ut{ms/c} \\ =& 24\ut{h} + 9\ut{ms} \end{aligned} $$

17. 시계별 1주일의 자신의 평균시각과의 차이는 아래표와 같다. (단위:초) 일 월 화 수 목 금 토 A -47 -31 -15 0 17 32 47 B -3 0 -5 5 5 -6 1 C -166 -108 -50 8 8 124 182 D 191 124 57 -10 -10 -144 -211 E 110 40 -15 -17 -17 -47 -57 이를 토대로 개별 시계의 표준편차를 구해보면 다음과 같다. 표준편차 A 33.9502437 B 4.466808811 C 122.7106546 D 141.7519631 E 57.67685019 따라서 표준편차가 가장 작은 순으로 나열하면 B, A, E, C, D 순으로 정확한 시계이다. 단순히 자신의 평균시각이 맞지 않는것은 시계를 맞추면 해결되나, 자신과의 편차가 크다면..

$$ T =1.55780644887275±3\ut{ms} $$ $$ \begin{cases} 1.55780644887278\ut{ms} &= T_{Max} &=T_M \\ 1.55780644887272\ut{ms} &= T_{min} &= T_m \end{cases} $$ (a) $x\ut{cycle/8days}=?$ (i) $T_M$ $$ \begin{aligned} & \frac{1\ut{cycle}}{1.55780644887278\ut{ms}} \cdot 8.00\ut{days} \\ &= \frac{1\ut{cycle}}{1.55780644887278\ut{ms}} \cdot 8.00\ut{days} \cdot \frac{24\ut{h}}{1\ut{days}}\cdot \frac{3600\ut{s..

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

$$ \%=100\times \frac{(\text{TrueValue}-\text{ApproxValue})}{\text{TrueValue}} $$ (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..