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Solutions to Midterm Exam, Spring 2008

# Solutions to Midterm Exam, Spring 2008 - Therefore the...

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Solutions to Midterm, Spring 2008 1. Dimensional analysis suggests that the wave speed is proportional to √(gλ). a) V=6 √2 = 8.49 knots b) V=6 √15 = 23.2 knots 2. a) The voltage divider relationship given by Ohm’s law is: V/V 0 = R/(R+R 1 ), or R= VR 1 /(V 0 – V) b) Again, using Ohm’s law, P = VI = V 2 /R V 0 = 0.1 volt R 1 = 20 ohms V = (10/2 8 ) = 0.04 volts d) ( R) 2 = ( R/ V) 2 ( V) 2 + ( R/ V 0 ) 2 ( V 0 ) 2 + ( R/ R 1 ) 2 ( R 1 ) 2 R/ V = R 1 /(V 0 – V) + VR 1 /(V 0 – V) 2 = 1381 R/ V 0 = VR 1 /(V 0 – V) 2 = 866 R/ R 1 = V/(V 0 – V) = 1.68 R=108 ohms R= VR 1 /(V 0 – V) = 3.26x1000/(5.2-3.26) = 1680 ohms e) P = V 2 /R ( P) 2 = ( P/ V) 2 ( V) 2 + ( P/ R) 2 ( R) 2 P/ V = 2V/R P/ R = - V 2 /R 2 ( P) 2 = 4 (V/R) 2 ( V) 2 + (V/R) 4 ( R) 2

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or, ( P/P) 2 = 4 ( V/V) 2 + ( R/R) 2 P/P = 0.0688; P=(3.26) 2 /1680 = 6.33 mW; P = 0.44 mW 3. a) The uncertainty in the mean is proportional to 1/√N.
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Unformatted text preview: Therefore the uncertainty (noise) in the mean decreases by a factor of 10 when the number of averaged cycles is increased to 10,000 from 100. The signal-to-noise ratio increases by a factor of 10. b) See the graph of C D vs Re d for spheres given in lecture 4. The transition from laminar to turbulent occurs at about 2-3x10 5 . 4. mdv/dt=mg-F d-F b = mg - 3πdμv - (4π/3)(d/2) 2 ρg dv/dt + (3πdμ/m)v + πd 3 ρg/6m – g = 0 λ = 3πdμ/m α = g - πd 3 ρg/6m v = (α/λ) [1 – exp(-λt)] V t = α/λ = [g - πd 3 ρg/6m]m/3πdμ Re d =ρV t d/μ ≤ 1 for Stokes flow; V t ≤ μ/ρd...
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