Advanced Physics #1 - Questions and Answers

Question 1 An open tank has a vertical partition and on one side contains gasoline with a density ρ = 700 kg/m ³ at a depth of 4m. Arectangular gate that is 4 m high and 2 m wide and hinged at one end is located in the partition. Water is slowly added to the empty side of the tank. At what depth, h, will the gate start to open? Answer 1 The depth of water possible that will cause for the gate to open can be solved by summing up the forces acting on the wall to the hinge. The forces acting on the wall are the weight and the hydrostatic forces cause by the water and the gasoline. Use γ for water which is equal to 9.80 kN/m³ trom Table 1.5. The specific weight of the gasoline can be solved by multiplying its density by the acceleration due to gravity which is equal to 9.81 m/s² γgas = ρ{gas} × g = 700 kg/m ³ × 9.81 m/s² = 6867 N/m³ = 6.87 kN/m³ The hydrostatic force on the gate can be solved using the formula FR = γhA where γ is the specific weight of liquid, hc is the centroid of the gate with liquid and A is the area of the gate with liquid. Solve for the hydrostatic force exerted by the gasoline. FR1 = γgas hc A = 6.87 kN/m³ * (4m/2) * (4m x 2m) = 109.92 kN Solve for the hydrostatic force exerted by the water. FR2 = γwater hc A = 9.80 kN/m³ * (h/2) * (h x 2m) Sum up the moment of the two hydrostatic forces. Note that the location of the hydrostatic force for rectangular plane surfaces can be found ⅓ of the height of the liquid from the base. FR1 × (⅓ * 4m) = FR2 × (⅓ * h) 109.92 kN × (⅓ * 4m) = 9.80 kN/m³ × (⅓ * h²) * (⅓ * h) 146.56 kNm = 3.27 kN/m³ * h³ 146.56 kNm / 3.27 kN/m³ = h³ 44.87m³ = h³ ∛{44.87m³} = ∛{h³} 3.55m = h Answer 2 FRg = γg hcg Ag where g refers to gasoline.. Frg = (700 kg/m ³) (9.81 m/s²) (2m)(4m*2m) = 110 * 10³ N = 110 kN FRw = γw hcw Aw where w refers to water... FRw = (9.80 * 10³ N/m³) (h/2) (2m * h) where h is depth of the water... FRw = (9.80 * 10³) h² For Equilibrium, ΣMH = 0 so that, FRw I w = FRg * Ig with Iw = h/3 and Ig = 4/3 m Thus, (9.80 * 10³) (h²) (h/3) = (110 * 10³ N) (4/3 m) h = 3.55m Question 2 The air in a bicycle tire is bubbled through water and collected at 25°C. If the total volume of gas collected is 5.45 L at a temperature of 25°C and a pressure of 745 torr, how many moles of gas were in the bicycle tire? Answer 1 Step 1 T = (25+273.15) K = 298.15K Pₜ=745 torr = 745mmHg P(H₂O) = 23.78mmHg V = 5.45L n(gas) = ? Write down the known and unknown values. Step 2 Pₜ = P(H₂O) + P(gas) P(gas) = Pₜ - P(H₂O) = 745mmHg - 23.7mmHg = 721.3mmHg To determine the evolved gas's partial pressure, use the gas-evolution equation based on Dalton's Law of Step 3 P(gas) = 721.3mmHg *( 1atm / 760mmHg ) = 0.94908atm Convert the evolved gas's partial pressure into atmospheres. Step 4 P(gas)V = n(gas)RT => n(gas) = (P(gas)V) / RT = (0.94908atm * 5.45L) / (0.08206 L atm/mol K * 298.15K ) = 0.211mol To determine how many moles of the evolved gas there are, use the Ideal Gas Law. Answer 2 To solve this problem, we can use the ideal gas law equation: PV = nRT Where: P is the pressure of the gas (in atm) V is the volume of the gas (in liters) n is the number of moles of gas R is the ideal gas constant (0.0821 L-atm/mol-K) T is the temperature of the gas (in Kelvin) First, let's convert the given pressure from torr to atm. We know that 1 atm is equal to 760 torr. Therefore: P = 745 torr / 760 torr/atm = 0.979 atm Next, let's convert the given volume to liters: V = 5.45 L Now, we can rearrange the ideal gas law equation to solve for the number of moles: n = PV / RT Substituting the given values: n = (0.979 atm * 5.45 L) / (0.0821 L-atm/mol-K * (25 + 273.15) K) n = 5.34 L-atm / 22.42 L-atm/mol-K n = 0.237 mol However, we need to account for the fact that the gas was collected at 25 degrees Celsius, not Kelvin. To convert the temperature from Celsius to Kelvin, we add 273.15: n = 0.237 mol * (298.15 K / 273.15 K) n = 0.260 mol Therefore, the number of moles of gas in the bicycle tire is approximately 0.260 mol, rounded to three decimal places. Answer 3 Given: Total volume of gas collected, V = 5.45 L Temperature, T = 25 °C Pressure, P = 745 torr To find the number of moles of gas, we can use the ideal gas law equation: PV = nRT First, we need to convert the temperature from Celsius to Kelvin. We use the equation: T(K) = T(°C) + 273.15 Plugging in the given values: T = 25°C + 273.15 = 298.15 K Now, we can rearrange the ideal gas law equation to solve for the number of moles: n = PV / RT Substituting the given values: n = (745 torr) * (5.45 L) / (0.0821 L atm/(mol K)) (298.15 K) Simplifying the equation gives us: n = 0.211 mol Therefore, there were 0.211 moles of gas in the bicycle tire. Hence, the solution to the problem is 0.211 mol. Answer 4 First, we need to convert the temperature from Celsius to Kelvin using the equation: T(K) = T(°C) + 273.15 Plugging in the values, we get: T(K) = 25+273.15 = 298.15 Now we can rearrange the ideal gas law equation to solve for n: n= PV/RT Plugging in the values, we get: n= (745 torr) * (5.45 L) / (0.0821 L-atm/(mol-K) * 298.15 K) Simplifying the equation, we find: n= 4061.75 / 24.457 mol = 166.26 mol Therefore, there were approximately 166.26 moles of gas in the bicycle tire. (Note: We should round the final answer to the appropriate number of significant figures based on the given data. In this case, since the total volume of gas collected is given with three significant figures, the final answer should be rounded to three significant figures as well.) Question 3 Part A Suppose a uniform slender rod has length L and mass m. The moment of inertia of the rod about about an axis that is perpendicular to the rod and that Part A Constants The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to Ip, the moment of inertia of the same object about a parallel axis passing through point p. The mathematical statement of the theorem is Ip = Icm+Md², where d is the perpendicular distance from the center of mass to the axis that passes through point p, and M is the mass of the object. Suppose a uniform slender rod has length L and mass m. The moment of inertia of the rod about about an axis that is perpendicular to the rod and that passes through its center of mass is given by Icm=mL²/12. Find Ip, the moment of inertia of the rod with respect to a parallel axis through one end of the rod. Expressed in terms of m and L. Use fractions rather than decimal numbers in your answer. ►View Available Hint(s) Ο ΑΣΦ Ip Submit Figure 1 of 1 Part B p-axis ? Now consider a cube of mass m with edges of length a. The moment of inertia Icm of the cube about an axis through its center of mass and perpendicular to one of its faces is given by Icm=ma²/6. (Figure 1)Find Iedge, the moment of inertia about an axis p through one of the edges of the cube Express Iedge in terms of m and a. Use fractions rather than decimal numbers in your answer. ►View Available Hint(s) Iedge Submit ΤΟ ΑΣΦ ? Answer a) Using property given, I = Icm + Md² Here, d = L / 2 Hence, I = (ML²) / 12 + M(L/2)² = (ML²) / 12 + (ML²) / 4 I = (ML²)/3 b) Using shift property of moment of inertia I = (Ma²) / 6 + (Ma²) / 2 I = (2Ma²) / 3