QUANTUM COMPUTING - EXAMPLE 32.8 : In quantum computing, a quantum state is given by S = a | 00 > + b | 01 > + g | 10 > + d | 11 >. (a) Find S in term of | 0 > and | 1 > etc. (b) The probability of getting x is P(x). For S = 0.5 | 00 > + 0.5 | 01 > + 0.5 | 10 > + 0.5 | 11 >, find P(0) and P(1). Hint : P(00) + P(01) = P(0) = a x a + b x b, P(10) + P(11) = P(1) = g x g + d x d.
QUANTUM COMPUTING - ANSWER 32.8 : (a) S = a | 00 > + b | 01 > + g | 10 > + d | 11 > = | 0 > ( a | 0 > + b | 1 > ) + | 1 > ( g | 0 > + d | 1 > ). (b) By comparison of S, a = b = g = d = 0.5. P(0) = a x a + b x b = 0.5 x 0.5 + 0.5 x 0.5 = 0.5, P(1) = g x g + d x d = 0.5 x 0.5 + 0.5 x 0.5 = 0.5. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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NATURAL GAS ENGINEERING - QUESTION 26.1 : (a) In natural gas pipe sizing, the length of the pipe from the gas source metre to the farthest appliances is 60 feet. The maximum capacities for typical metallic pipes of 60 feet in length are : 66 cubic feet per hour for pipe size of 0.5 inches; 138 cubic feet per hour for pipe size of 0.75 inches; 260 cubic feet per hour for pipe size of 1 inch. By using the longest run method : (i) Find the best pipe size needed for the capacity of 75 cubic feet per hour. (ii) Estimate the suitable range of capacities for the pipe size of 1 inch. (b) The maximum capacities for typical metallic pipes of 50 feet in length are : 73 cubic feet per hour for pipe size of 0.5 inches; 151 cubic feet per hour for pipe size of 0.75 inches; 285 cubic feet per hour for pipe size of 1 inch. By using the branch method find the best pipe size needed for the capacity of 75 cubic feet per hour when the length of the pipe from the gas source metre to the appliance is 52 feet.
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PETROLEUM ENGINEERING - QUESTION 25.1 : Fact 1 : Dry air contains 20.95 % oxygen, 78.09 % nitrogen, 0.93 % argon, 0.039 % carbon dioxide, and small amounts of other gases by volume. Fact 2 : Volume occupied is directly proportional to the number of moles for ideal gases at constant temperature and pressure. Fact 3 : 12.5 moles of pure oxygen is required to completely burn 1 mole of pure octane. Fact 4 : Air-fuel ratio (AFR) is the mass ratio of dry air to fuel present in a combustion process such as in an internal combustion engine or industrial furnace. Fact 5 : Molecular weight of oxygen gas is 31.998 g / mole and molecular weight of nitrogen gas is 28.014 g / mole. (a) Find the molar ratio of nitrogen and oxygen, or (moles of nitrogen) / (moles of oxygen) in dry air, by assuming ideal features of nitrogen and oxygen gases. (b) How many moles of nitrogen are available if dry air is used to completely burn the 1 mole pure octane? (c) Find the mass of fuel of 1 mole of octane with molecular weight of 114.232 g / mole. (d) Find the mass of dry air with 12.5 moles of pure oxygen by assuming only oxygen and nitrogen gases exist in the air. (e) Find the air-fuel ratio (AFR) when octane is used as fuel. (f) Find the fuel-air ratio (FAR) when octane is used as fuel.
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PROCESS DESIGN - EXAMPLE 21.1 : According to rules of thumb in chemical process design, consider the use of an expander for reducing the pressure of a gas when more than 20 horsepowers can be recovered. The theoretical adiabatic horsepower (THp) for expanding a gas could be estimated from the equation : THp = Q [ Ti / (8130a) ] [ 1 - (Po / Pi) ^ a ] where 3 ^ 3 is 3 power 3 or 27, Q is volumetric flowrate in standard cubic feet per minute, Ti is inlet temperature in degree Rankine, a = (k - 1) / k where k = Cp / Cv, Po and Pi are reference and systemic pressures respectively. (a) Assume Cp / Cv = 1.4, Po = 14.7 psia, (temperature in degree Rankine) = [ (temperature in degree Celsius) + 273.15 ] (9 / 5), nitrogen gas at Pi = 90 psia and 25 degree Celsius flowing at Q = 230 standard cubic feet per minute is to be vented to the atmosphere. According to rules of thumb, should an expander or a valve be used? (b) Find the outlet temperature To by using the equation To = Ti (Po / Pi) ^ a.
Question 68 – Biochemical Oxygen Demand (BOD) could be calculated using the formula BOD = (DOi - DOf) (Vb / Vs) where Vb = Volume of bottle in ml, Vs = Volume of sample in ml, DOi = Initial dissolved oxygen in mg / L, DOf = Final dissolved oxygen in mg / L. (a) By using a bottle of Vb = 300 ml with sample Vs = 30 ml, find the BOD if DOi = 8.8 mg / L and DOf = 5.9 mg / L. (b) By using a bottle Vb = 600 mL with sample Vs = 100 mL, find the BOD if DOi = 8.8 mg / L and DOf = 4.2 mg / L. (c) Find the average BOD = [ Answer of (a) Answer of (b) ] / 2. (d) If the BOD-5 test for (a) - (c) is run on a secondary effluent using a nitrification inhibitor, find the nitrogenous BOD (NBOD) = TBOD - CBOD. Let TBOD = 45 mg / L and CBOD = Answer of (c).
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