Problem-2.13: An object is placed in front of a convex mirror at a distance of 50 cm. A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and the plane mirror is 30 cm, it is found that there is no parallax between the images formed by the two mirrors. What is the radius of curvature of the convex mirror?

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Answer 1

The radius of curvature of the convex mirror is 1500 cm.

In this problem, the absence of parallax between the images formed by the convex mirror and the plane mirror indicates that the virtual image formed by the plane mirror coincides with the real image formed by the convex mirror. By applying the mirror formula and considering the object distance of 50 cm and the virtual object distance of 30 cm, we can derive an equation to solve for the focal length of the convex mirror. Solving this equation yields a focal length of 750 cm, which is twice the radius of curvature. Therefore, the radius of curvature of the convex mirror is 1500 cm.

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compare the gate/inverter/chip counts in the schematic diagrams prepared for eachsection of the procedure. if a minimum gate/inverter/chip count is desirable, whatwould you recommend?

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In comparing the gate/inverter/chip counts in the schematic diagrams prepared for each section of the procedure, if a minimum gate/inverter/chip count is desirable, I would recommend optimizing the design by minimizing the number of gates and inverters used.

Reducing the gate/inverter/chip count in a schematic diagram can have several advantages, including improved efficiency, reduced power consumption, and cost savings. To achieve this, one can employ various design techniques such as logic simplification, gate sharing, and reusing common subcircuits. By carefully analyzing the circuit and identifying opportunities for simplification, it is possible to eliminate unnecessary gates and inverters. Additionally, utilizing more advanced integrated circuits with higher gate density can help reduce the chip count further. However, it's important to strike a balance between gate/inverter/chip count reduction and maintaining circuit performance and reliability. Thorough simulation and testing should be performed to ensure that the optimized design meets the desired specifications.

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Given Two parts of a machine are held together by a bolt. The clamped member stiffness is 24 Lb/in while that of the bolt is 14 (one-fourth) of the stiffness of the clamped member. The bolt is preloaded to an initial tension of 1,200 Lb. The external force acting to separate the joint fluctuates between 0 and 6,000 Lb. Find: a) The total bolt load b) The load on the clamped member when an external load is applied c) The load in which the joint would become loose.

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a) The total bolt loadThe bolt load is the tension applied to the bolt to hold the two parts of the machine together. When the external load acts on the joint, the bolt load increases. The total bolt load is equal to the sum of the initial bolt load and the increase in bolt load due to the external force. Therefore, the total bolt load can be calculated using the equation:

Total bolt load = Initial bolt load + Increase in bolt load due to external forceThe initial bolt load is given as 1,200 lb. To calculate the increase in bolt load, we need to first calculate the deflection in the clamped member due to the external force. The deflection can be calculated using the stiffness of the clamped member as follows:Deflection = External force / Clamped member stiffness= 6,000 / 24= 250 inThe increase in bolt load can be calculated using the deflection and the stiffness of the bolt as follows:Increase in bolt load = Deflection x Bolt stiffness= 250 x (1/4) x 14= 875 lbTherefore, the total bolt load is:

Total bolt load = Initial bolt load + Increase in bolt load due to external force= 1,200 + 875= 2,075 lbTherefore, the total bolt load is 2,075 lb.b) The load on the clamped member when an external load is appliedWhen an external load is applied, the clamped member experiences a force due to the deflection caused by the external load. The force on the is equal to the total bolt load. Using the value of total bolt load calculated above, we can find the load in which the joint would become loose as follows:External force = Total bolt load= 2,075 lbTherefore, the load in which the joint would become loose is 2,075 lb.

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5- The following elementary liquid phase reversible reaction: A↔2B is taking place in an isothermal CSTR reactor at T=450 K. The inlet volumetric flow is 10 m^3/h. The inlet molar flow is 20 mol/h. The equilibrium constant Kc=6. What is the equilibrium conversion of the reaction: a) 0.79 b) 0.90 c) 0.86 d) None of the above. 6- For the same reaction mentioned in question 5 , what is the equilibrium concentration of A : a) 0.42 mol/m^3
b) 0.20 mol/m^3
c) 0.28 mol/m^3
d) None of the above

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5- The equilibrium conversion of the reaction A↔2B in the isothermal CSTR reactor at T=450 K is 0.79. 6- The equilibrium concentration of A in the isothermal CSTR reactor for the reaction A↔2B is 0.20 mol/m^3

The equilibrium conversion (X) of a reversible reaction is determined by the equilibrium constant (Kc). For the given reaction, A↔2B, the equilibrium constant is Kc=6. The equilibrium conversion can be calculated using the equation:

X = (1 - sqrt(1 + 4Kc)) / 2

Plugging in the value of Kc=6:

X = (1 - sqrt(1 + 4*6)) / 2 ≈ 0.79

To determine the equilibrium concentration of A, we can use the equilibrium conversion (X) and the inlet molar flow of A. The equilibrium concentration of A can be calculated using the equation:

Ca = Fa / (V0 * (1 + X))

Where Ca is the equilibrium concentration of A, Fa is the inlet molar flow of A (20 mol/h), V0 is the inlet volumetric flow (10 m^3/h), and X is the equilibrium conversion.

Plugging in the given values and X=0.79:

Ca = 20 mol/h / (10 m^3/h * (1 + 0.79)) ≈ 0.20 mol/m^3

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