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A STUDENT’S SOLUTIONS MANUAL TO ACCOMPANY MECHANICS of FLUIDS
) m = F/a or kg = N/m/s 2 = N·s 2 /m 1.2 (B) [¿] = [Ä/(du/dy)] = (F/L L T L F T 2)/( / )/ = . /L 2 1.3 (A) 8 9 2.36 10 Pa 23.6 10 Pa 23.6 nPa − − × = × = 1.4 (C) The mass is the same on earth and the moon, so we calculate the mass using the weight given on earth as: m = W/g = 250 N/9.81 m/s 2 = 25.484 kg Hence, the weight on the moon is: W = mg = 25.484 × 1.6 = 40.77 N 1.5 (C) The shear stress is due to the component of the force acting tangential to the area: shear ...
- Exam (elaborations)
- • 196 pages •
) m = F/a or kg = N/m/s 2 = N·s 2 /m 1.2 (B) [¿] = [Ä/(du/dy)] = (F/L L T L F T 2)/( / )/ = . /L 2 1.3 (A) 8 9 2.36 10 Pa 23.6 10 Pa 23.6 nPa − − × = × = 1.4 (C) The mass is the same on earth and the moon, so we calculate the mass using the weight given on earth as: m = W/g = 250 N/9.81 m/s 2 = 25.484 kg Hence, the weight on the moon is: W = mg = 25.484 × 1.6 = 40.77 N 1.5 (C) The shear stress is due to the component of the force acting tangential to the area: shear ...