솔루션피아

$$ \begin{aligned} \Ans &= \rho = ? \\ &= \frac{\mathrm{Mass}}{\mathrm{Volume}} \\ &= \frac{5.324\ut{g}}{2.5\ut{cm^3}} \\ &= \frac{5.324\ut{g}}{2.5\ut{cm^3}} \cdot \frac{1\ut{kg}}{1000\ut{g}}\cdot \left(\frac{100\ut{cm}}{1\ut{m}}\right)^3 \\ &= \frac{10648}{5}\ut{kg/m^3} \\ &= 2129.6\ut{kg/m^3} \\ &\approx 2.1\times 10^3\ut{kg/m^3} =2.1\ut{tonne/m^3} \\ &= 2.1\ut{kg/L}=2.1\ut{g/cm^3} \end{aligne..

(a) $$ \begin{cases} R &= 1.90\ut{\AAA} \\ N_0 &= 6.023 \times 10^{23}\ut{EA/mol} \\ m_{\mathrm{Na}} &= 22.9898\ut{g/mol} \end{cases} $$ $$ \begin{aligned} \Ans &= \rho_{\mathrm{Na}} = ? \\ &= \frac{\mathrm{Mass}}{\mathrm{Volume}} \\ &= \frac{m_{\mathrm{Na}}}{N_0} \cdot \frac{3}{4\pi R^3} \\ &= \frac{22.9898\ut{g/mol}}{6.023 \times 10^{23}\ut{EA/mol}} \cdot \frac{3}{4\pi (1.90\ut{\AAA})^3} \\ &=..

$$ \begin{cases} R &= 60\ut{\mu m} \\ \rho_{\text{SiO}_2} &= 2600\ut{kg/m^3} \\ L &= 1.00\ut{m} \end{cases} $$ $$ \begin{aligned} \\ n\cdot S_{\text{particle}} &= S_{\text{cube}} \\ n &= \frac{S_{\text{cube}}}{S_{\text{particle}}} \end{aligned} $$ $$ \begin{aligned} \\ \Ans &= \Sigma m = nm =n\rho V_{\text{particle}} \\ &= \frac{S_{\text{cube}}}{S_{\text{particle}}}\rho V_{\text{particle}} \\ &=..

$$ \begin{cases} S &= 6.00\ut{m^2} \\ m &= 324\ut{kg} \\ \rho &= 2.70 \times 10^3 \ut{kg/m^3} \end{cases} $$ d=? $$ \begin{aligned} \\ \Ans &= d = \frac{Volume}{Area} \\ &= \frac{m}{\rho \cdot S} \\ &= \frac{324\ut{kg}}{2.70 \times 10^3\ut{kg/m^3} \cdot 6.00\ut{m^2}} \\ &= \frac{1}{50}\ut{m} \\ &= 0.02\ut{m} = 2.00\ut{cm} \end{aligned} $$

$$ \begin{cases} \rho &= 19.32\ut{g/cm^3} \\ m &= 29.34\ut{g} \end{cases} $$ (a) S=? $$ h = 1.000\ut{\mu m} $$ $$\begin{aligned} \\ \Ans &= S = \frac{\text{Volume}}{\text{Hight}} \\ &= \frac{m}{\rho}\cdot\frac{1}{h} \\ &= \frac{29.34\ut{g}}{19.32\ut{g/cm^3}}\cdot \frac{1}{1.000\ut{\mu m}} \\ &= \frac{489}{322}\ut{cm^3}\cdot \frac{1}{1.000\ut{\mu m}}\cdot \frac{10^6\ut{\mu m}}{1\ut{m}}\cdot \left..

$$ \begin{aligned} \Ans &= \rho = \frac{\text{Mass}}{\text{Volume}} \\ &= \frac{9.05\ut{g}}{3.5\ut{cm^3}} \\ &= \frac{181}{70}\ut{g/cm^3} \\ &\approx 2.5857142857142857143\ut{g/cm^3} \\ &\approx 2.6\ut{g/cm^3} \end{aligned} $$

$$ \begin{cases} \rho_{WATER} &= 1000\ut{kg/m^3} \\ 1\ut{gal} &= 231\ut{inch^3} \\ 1\ut{inch} &= 2.540\ut{cm} \\ v &= 1.5\ut{g/min} \end{cases} $$ (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..