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1-30 할리데이 10판 솔루션 일반물리학

(a) t(mMAX) =? m˙= ⁣dm ⁣dt= ⁣d ⁣dt(5.00t0.83.00t+20.00)=0=4t53=0 \begin{aligned} \dot m &= \dxt{m} = \dt\(5.00t^{0.8}-3.00t + 20.00\) = 0 \\ &= \frac{4}{\sqrt[5]{t}}-3 = 0 \end{aligned} t=1024243[s]4.2139917695473251029[s]4.21[s] \begin{aligned} \\ t &= \frac{1024}{243}\ut{s} \\ &\approx 4.2139917695473251029\ut{s} \\ &\approx 4.21\ut{s} \end{aligned} (b) mMAX =? $$ \begin{aligned} m(t_{MAX}) &= 5.00(t_{MAX})^{0.8}-3.00(t_{MAX}) + 20.00\ut{g} \\ &= 5.00\(\frac{1024}{243}\)^{0..

10판/1. 측정 2019.07.22

1-28 할리데이 10판 솔루션 일반물리학

{c=299792458[m/s]1[MeV]=1.60218×1013[J]1[u]=1.66054×1027[kg] \begin{cases} c &= 299792458\ut{m/s} \\ 1\ut{MeV} &= 1.60218 \times 10^{-13}\ut{J} \\ 1\ut{u} &= 1.66054 \times 10^{-27}\ut{kg} \end{cases} $$ \begin{aligned} \\ \Ans &= E = ? \\&= E = mc^2 \\&= 1\ut{u} \cdot (299792458\ut{m/s})^2 \\&= 1\ut{u} \cdot (299792458\ut{m/s})^2 \cdot \frac{1.66054 \times 10^{-27}\ut{kg}}{1\ut{u}}\cdot \frac{1\ut{MeV}}{1.60218 \times 10^{-13}\ut{J}} \\&= \frac{186..

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1-27 할리데이 10판 솔루션 일반물리학

물체=상자 로 가정함. {mB=2.500[kg]mG1=21.15[g]mG2=21.16[g]Ans=mBΣmG=mB(mG1+mG2)=2.500[kg](21.15[g]+21.16[g])=2.500[kg](21.15[g]+21.16[g])1[kg]1000[g]=2.45769[kg]2.458[kg] \begin{aligned} & \begin{cases} m_B &= 2.500\ut{kg} \\ m_{G1} &= 21.15\ut{g} \\ m_{G2} &= 21.16\ut{g} \end{cases}\\ \\ \Ans &= m_B - \Sigma m_G \\ &= m_B - (m_{G1}+m_{G2}) \\ &= 2.500\ut{kg} - (21.15\ut{g}+21.16\ut{g}) \\ &= 2.500\ut{kg} - (21.15\ut{g}+21.16\ut{g})\cdot \frac{1\ut{kg}}{1000\ut{g}} \\ &= 2.45769\ut{kg} \\ &\approx 2.458\ut{kg} \end{aligned}

10판/1. 측정 2019.07.21

1-24 할리데이 10판 솔루션 일반물리학

{R=60[μm]ρSiO2=2600[kg/m3]L=1.00[m] \begin{cases} R &= 60\ut{\mu m} \\ \rho_{\text{SiO}_2} &= 2600\ut{kg/m^3} \\ L &= 1.00\ut{m} \end{cases} nSparticle=Scuben=ScubeSparticle \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}} \\ &=..

10판/1. 측정 2019.07.21

1-23 할리데이 10판 솔루션 일반물리학

{S=6.00[m2]m=324[kg]ρ=2.70×103[kg/m3] \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=? Ans=d=VolumeArea=mρS=324[kg]2.70×103[kg/m3]6.00[m2]=150[m]=0.02[m]=2.00[cm] \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}

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1-22 할리데이 10판 솔루션 일반물리학

{ρ=19.32[g/cm3]m=29.34[g] \begin{cases} \rho &= 19.32\ut{g/cm^3} \\ m &= 29.34\ut{g} \end{cases} (a) S=? h=1.000[μm] 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..

10판/1. 측정 2019.07.21