3.6 Answers and Solutions

PROBLEM SET 1

1(a) \(VT^{-1}\)
 (b) \(MVT^{-1}\)
 (c) \(MV^2\)
2(a) \(LT^{-2}\) (by substituting \(LT^{-1}\) for \(V\)).
 (b) \(MLT^{-2}\)
 (c) \(ML^2T^{-2}\)
3(a) \(FL\)
 (b) \(FL\)
 (c) \(FLT^{-1}\)
4(b) \(ML^2T^{-2}\)
 (c) \(ML^2T^{-3}\)
5. 2(a) 1 m\(\cdot\)sec^-2
     (b) 1 kg\(\cdot\)m\(\cdot\)sec^-2
     (c) 1 kg\(\cdot\)m²\(\cdot\)sec^-2
   3(a) 1 nt\(\cdot\)m, or 1 joule
     (b) 1 nt\(\cdot\)m, or 1 joule
     (c) 1 nt\(\cdot\)m\(\cdot\)sec^-1, or 1 joule\(\cdot\)sec^-1, or 1 watt
   4(b) 1 kg\(\cdot\)m²\(\cdot\)sec^-2, or 1 nt\(\cdot\)m, or 1 joule
     (c) 1 kg\(\cdot\)m²\(\cdot\)sec^-3, or 1 nt\(\cdot\)m\(\cdot\)sec^-1, or 1 joule\(\cdot\)sec^-1, or 1 watt

PROBLEM SET 2

1. Given \(1 sec=\frac{1}{31,556,925.9747}\)trop.yr
   31,556,925.9747 sec = 1 trop.yr
   and since 1 solar day (24 hrs) = 86400 sec, then
   \[(31556925.9747 sec)\frac{1\mbox{ solar day}}{86400 sec}=1\mbox{ trop.yr}\] 365.2422 solar days = 1 trop.yr

Note: The sidereal year is the time required for the sun to go once around the ecliptic and resume its (apparent) original position among the stars. The sidereal year is 365.2564 mean solar days. The tropical and sidereal years would be of equal durations were it not for precession of the equinoxes, an effect owed primarily to the moon.

1. \(\theta_k=\frac{5K}{9^{\circ}F}\theta_f+255.37K\)
   
2. \(\theta_c=\frac{5^{\circ}C}{9^{\circ}F}\theta_f-17.78^{\circ}C\)

3. By letting \(n\) be the number of molecules and a the area of each molecule, we want \(n\cdot a\) to be 1 square cm of area: \[n\cdot a=1cm^2\] We are given \(a=10\mathring{A}^2\), and knowing that \(1\mathring{A}=10^{-8}cm\), then \[a=10\mathring{A}^2=10(10^{-8}cm)^2=10\times10^{-16}cm^2=10^{-15}cm^2\] whence, by the first equation, \[n=\frac{1cm^2}{a}=\frac{1cm^2}{10^{-15}cm^2}=10^{15}\] a pure number since the units (cm²/cm²) cancel.

\[\begin{align*} K.E. &= \frac{1}{2}mv^2 \\ &=\frac{1}{2}(2.0\times10^{-22}g)(4.0\times10^4cm\cdot sec^{-1})^2 \\ &=(1.0\times10^{-22}g)(16.0\times10^8cm^2\cdot sec^{-2}) \\ &=1.0\times16.0\times10^{-22}\times10^8g\cdot cm^2\cdot sec^{-2} \\ &=1.6\times 10^{-13}erg \end{align*}\]

5. 1 micron = 10^-6m [but 1 m = 10² cm]
   = 10^-6(10² cm)
   = 10^-4 cm
   7.5 microns = 7.5 \(\times\) 10^-4 cm

1. \(3.5\times10^{-3}l = 0.0035l\)
   
2. \(3.5\times10^{-1} l = 0.35l\)
   
3. \(10^3yr=1000yr\)
   
4. \(0.04\times10^6W=4\times10^4W=40000W\)
   
5. \(2\times10^{-9}g=0.000000002g\)
   
6. \(0.002\times10^{-9}g=2\times10^{-12}g\)
   
7. \(200.5\times10^{-12}g=2.005\times10^{-10}g\)
   
8. \(1.099\times10^{-6}mol\)
   
9. \(10^2are=100are\) [1 are = 100 m², hence 1 hectare = 10000 m²]
   
10. \(10 (dekacm)^2=10(10^2cm^2)=10^3cm^2=1000cm^2\), or, in meters
    \(10(dekacm)^2= 10(10\times10^{-2}m)^2 =10(10^{-1}m)^2=10(10^{-2}m^2) =10^{-1}m^2=0.1m^2\)

1. 7.35 gigaliters
   
2. 7.35 nanoliters
   
3. 1 megawatt
   
4. 0.20 \(\times\) 10^-5 mol = 2.0 \(\times\) 10^-6 mol = 2.0 micromoles
   
5. 8.575 megamicrocuries = 8.575 (10⁶\(\cdot\) 10^-6 curies) = 8.575 curies
   
6. 2.10 kilocalories per gram = 2.10 Kcal\(\cdot\) g^-1
   
7. 1 milliphoton/cm² = 1 mphot\(\cdot\)cm^-2

PROBLEM SET 3

1. \([m][v(x)][dv]=MV^2\) (dimensions of kinetic energy), or \(MV^2=M(LT^{-1})^2=ML^2T^{-2}=FL\) (the dimensions of mechanical work).
   
2. \(\Big[\frac{\partial}{\partial x}\Big][m][v][dv]=\frac{MV^2}{L}=MLT^{-2}=F\)
   
3. \([R]=[\mu][f(x)][dx]\), whence
        \([f(x)]=\frac{[R]}{[\mu][dx]}\), and since frictional resistance has the dimensions of force, then
        \([f(x)]=\frac{F}{FL^{-2}TL}=LT^{-1}=V\) (hence, the function \(f(x)\) is fluid velocity).

2. Given \([f(x,y)]=\frac{FL^{-2}}{FL^{-1}}=L^{-1}\), then \(\Big[\frac{\partial^2 \phi}{\partial x^2}\Big]=L^{-1}\) or \(\Big[\frac{\partial^2\phi}{\partial y^2}\Big]=L^{-1}\) (since the terms of the Laplacian must have like dimensions). Therefore,
   \[\Big[\frac{\partial^2\phi}{\partial x^2}\Big]=L^{-1}\] \[\frac{[\phi]}{L^2}=L^{-1}\] \[[\phi]=L\]

3. \([\ddot{x}]=[\omega^2x]=[f(t)]\) But \(\ddot{x}\equiv\frac{d^2x}{dt^2}\); therefore
     \(\frac{[d^2x]}{dt^2}=\frac{L}{T^2}\) (the dimensions of acceleration). Consequently,
     \([f(t)]=LT^{-2}\), and \([\omega^2][x]=LT^{-2}\) \[[\omega^2]=T^{-2}\] \[[\omega]=T^{-1}\]

4. \[\rho\frac{\partial p}{\partial z}[\rho]\frac{[p]}{[z]}= ML^{-3}\frac{MLT^{-2}/L}{L}=M^2L^{-5}T^{-2}\] \[\mu\frac{\partial^2\omega}{\partial z^2}=[\mu]\frac{[\omega]}{[z^2]}= ML^{-1}T^{-2}\frac{LT^{-1}}{L^2}=ML^{-2}T^{-2}\]   and because the equation is not dimensionally homogeneous, we conclude that it is not correctly formulated.

1. The concept of specific heat of a substance under conditions of constant pressure (\(C_p\)) permits of volumetric change, while the concept of specific heat of a substance under conditions of constant volume (\(C_v\)) permits of pressure change. In either case, \[[C_p]=[C_v]=\frac{H}{M\Theta}\]
2. cal\(\cdot\)g^-1\(\cdot\)K^-1. (calories per gram per degree Kelvin).

1. \[q_x\hat{i}+q_y\hat{j}+q_z\hat{k}=-k\Big(\frac{\partial\theta}{\partial x}\hat{i}+\frac{\partial\theta}{\partial y}\hat{j}+ \frac{\partial\theta}{\partial z}\hat{k}\Big)\] Because the equation must be dimensionally homogeneous in its terms, we can resolve the dimensional problem from any component. By taking the component in the x-direction, \[[q_x]=[k]\Big[\frac{\partial\theta}{\partial x}\Big]\] (signs have no influence on dimensions of terms)
   \(\frac{H}{L^2T}=[k]\frac{\Theta}{L}\)
   \([k]=\frac{H}{L\Theta T}\) (the quantity of heat transferred per unit thickness of substance per unit temperature difference per unit time).

2. \(cal\cdot cm^{-1}\cdot ^{\circ}C^{-1}\cdot sec^{-1}\)
   

3. \(J\cdot m^{-1}\cdot K^{-1}\cdot sec^{-1}\)
   

4. Since heat is energy, then \([k]=\frac{ML^2T^{-2}}{L\Theta T}=ML\Theta^{-1}T^{-3}\)
   

5. \(g\cdot cm\cdot K^{-1}\cdot sec^{-3}\)
   

6. \([k]=\frac{FL}{L\Theta T}=F\Theta^{-1}T^{-1}\)
   

7. \(nt\cdot K^{-1}\cdot sec^{-1}\)
   

8. \(W\cdot m^{-1}\cdot K^{-1}\)
   

9. Units are \(cal\cdot cm^{-1}\cdot ^{\circ}C^{-1}\cdot sec^{-1}\), \(J\cdot m^{-1}\cdot K^{-1}\cdot sec^{-1}\)
   Conversion chains are
      \(H\): calorie \(\rightarrow\) joule
      \(L\): centimeter \(\rightarrow\) meter
      \(\Theta\): degree Celsius \(\rightarrow\) degree Kelvin
   The identities are
      1 cal = 4.1840 J
      1 m = 100 cm
      1\(^{\circ}C\) temp. diff. = 1K temp. diff.
   Make the obvious statement

\[\begin{align*} \frac{cal}{cm\cdot ^{\circ}C\cdot sec}&=\frac{cal}{cm\cdot ^{\circ}C\cdot sec} \\ &=\frac{\rlap{-----}cal}{cm\cdot ^{\circ}C\cdot sec}\Big(\frac{4.1840 J}{\rlap{-----}cal}\Big) \\ &=\frac{1}{\rlap{-----}cm\cdot ^{\circ}C\cdot sec}\Big(\frac{4.1840J}{1}\Big)\Big(\frac{100\rlap{-----}cm}{m}\Big) \\ 1\frac{cal}{cm\cdot ^{\circ}C\cdot sec}&=418.40\frac{J}{m\cdot K\cdot sec} \end{align*}\]

10. Units are \(J\cdot m^{-1}\cdot K^{-1}\cdot sec^{-1}\), \(g\cdot cm\cdot K^{-1} \cdot sec^{-3}\).
    Conversion chains are
       \(H\): joule \(\rightarrow\) erg \(\rightarrow\) g\(\cdot\)cm²\(\cdot\)sec^-2
       \(L\): meter \(\rightarrow\) centimeter
    Identities are
       1 J = 10⁷ erg
       1 erg = 1 g\(\cdot\)cm²\(\cdot\)sec^-2
       1 m = 100 cm

\[\begin{align*} \frac{J}{m\cdot K\cdot sec}&=\frac{J}{m\cdot K \cdot sec} \\ &=\frac{J}{\rlap{---}m\cdot K\cdot sec}\Big(\frac{10^7 \rlap{-----}erg}{J}\Big)\Big(\frac{g\cdot cm^2 sec^{-2}}{\rlap{-----}erg}\Big)\Big(\frac{\rlap{---}m}{100cm}\Big) \\ 1\;J\cdot m^{-1}\cdot K^{-1}\cdot sec^{-1}&=10^5g\cdot cm\cdot K^{-1}\cdot sec^{-3} \end{align*}\]

\[\begin{align*} 1 ft&=\frac{1}{3}yd \\ &=\frac{1}{3}(0.9144 m) \\ &=\frac{1}{3}(0.9144)(100cm) \\ r&=1.5ft=1.5\Big(\frac{1}{3}(0.9144)(100cm)\Big) \\ vol&=\frac{4}{3}\pi r^3 \\ &= \frac{4\pi}{3}\Big((1.5)\frac{1}{3}(0.9144)(100cm)\Big)^3 \\ &=400320cm^3 \end{align*}\]

8. Identities are \(1L=10^3cm^3\), \(1mL=10^{-3}L\).
   The dimensional concept of cubical density is \[\mbox{density}=\frac{\mbox{mass}}{\mbox{volume}}\] and for the problem \[0.79g\cdot cm^{-3} = \frac{\mbox{mass (of 30 m alc.)}}{30 mL}\] Therefore, \[\begin{align*} \mbox{mass (of 30 mL of alc)}&=(0.79g\cdot cm^{-3})\cdot(30mL) \\ &=\Big(\frac{0.79g}{\rlap{-----}cm^3}\Big)\frac{30\rlap{-----}mL}{1}\Big(\frac{10^{-3}L}{\rlap{-----}mL}\Big)\Big(\frac{10^3\rlap{-----}cm^3}{\rlap{---}L}\Big) \\ &= 23.7g \end{align*}\]

9. Pressure is force per unit area.
   The area (\(L^2\)) is 15 cm \(\times\) 2.8 mm.
   The force (\(M\cdot LT^{-2}\)) is 150 lb \(\times\) 980 cm\(\cdot\)sec^-2.
   If we employ the cgs system of units, the conversion identities are

• 1 mm = 10^-3 m
  
• 1 m = 10² cm
  
• 1 lb = 0.4536 kg
  
• 1 kg=10³ g

\[\begin{align*} p&=F\cdot A^{-1} \\ &=\Big(\frac{150lb}{1}\cdot\frac{980cm}{sec^2}\Big)\Big(\frac{1}{15\rlap{-----}cm}\Big) \\ &=\frac{150\rlap{---}lb}{1}\Big(\frac{0.4536\rlap{---}kg}{\rlap{---}lb}\Big)\Big(\frac{10^3g}{\rlap{---}kg}\Big)\frac{980}{sec^2}\Big(\frac{1}{15}\Big)\frac{1}{2.8\rlap{-----}mm}\Big(\frac{\rlap{-----}mm}{10^{-3}\rlap{---}m}\Big)\Big(\frac{\rlap{---}m}{10^2cm}\Big) \\ &=15,876,000 g\cdot cm^{-1}\cdot sec^{-2} \end{align*}\]

In keeping with the concept of pressure as force per unit area, we can also write the pressure unit \(g\cdot cm^{-l}\cdot sec^{-2}\) as \[\frac{g}{cm\cdot sec^2}\cdot\frac{cm}{cm}=\frac{g\cdot cm\cdot sec^{-2}}{cm^2}=\frac{dyne}{cm^2}\] Hence the pressure on the ice can also be written \[p=15,876,000 \;dyn\cdot cm^{-2}\] On the other hand, the mass concentration (mass per unit area) is simply
\[\begin{align*} \frac{\mbox{mass}}{\mbox{area}}&=\frac{150lb}{15cm\times 2.8mm} \\ &=230.4\;lb/in^2 \\ &=16200\;g/cm^2 \end{align*}\]

10. For pressure in cgs units, the needed conversion identities are

• 1 \(\mu\)m = 10^-6 m
• 1 \(\mu\)m² = 10^-12 m²
• 1 m = 10² cm
• 1 m² = 10⁴ cm²
• 16 oz = 1 lb
• 1 lb = 0.4536 kg
• 1 kg = 10³ g

\[\begin{align*} A&=\pi r^2 = \pi(40\mu m)^2 = \pi\cdot 1600\mu m^2 \\ p&=F\cdot A^{-1} = (2oz)(980cm\cdot sec^{-2})\Big(\frac{1}{\pi\cdot 1600\mu m^2}\Big) \\ &=\frac{2\rlap{---}oz}{1}\Big(\frac{\rlap{---}lb}{16\rlap{---}oz}\Big)\Big(\frac{0.4536\rlap{---}kg}{\rlap{---}lb}\Big)\Big(\frac{10^3g}{\rlap{---}kg}\Big)\Big(\frac{980cm}{sec^2}\Big)\frac{1}{\pi\cdot 1600\rlap{---}\mu m^2}\Big(\frac{\rlap{----}\mu m^2}{10^{-12}\rlap{---}m^2}\Big)\Big(\frac{\rlap{---}m^2}{10^4cm^2}\Big) \\ &=1.105\times 10^9\;dyn\cdot cm^{-2} \end{align*}\]