ENGINEERING PHYSICS - EXAMPLE 30.1 : Two viruses a and b with masses of Ma and Mb are moving at velocities of Va and Vb respectively, facing towards each other and collide. After collision both masses of Ma and Mb disappear. (a) Find the total momentum available for both a and b. Hint : momentum = mass x velocity = M x V. (b) Guess the total energy E generated from the disappearance of a and b. Let c to be the velocity of light. Hint : E is equal to M c square.
ENGINEERING PHYSICS - ANSWER 30.1 : (a) Momentum of a = Ma x Va. Momentum of b = Mb x Vb. Total momentum of a and b = Ma x Va + Mb x Vb. (b) After mass disappearance : Energy of a = Ma x c x c. Energy of b = Mb x c x c. Then E = Ma x c x c + Mb x c x c = (Ma + Mb)(c x c). 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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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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