HW 3 solution_pdf

HW 3 solution_pdf - ghei (mg42382) HW 3 Opyrchal (11130)...

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ghei (mg42382) – HW 3 – Opyrchal – (11130) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A 4100 kg helicopter accelerates upward at 3 . 42 m / s 2 . The acceleration oF gravity is 9 . 8 m / s 2 . What liFt Force is exerted by the air on the propellers? Correct answer: 54202 N. Explanation: The propellers exert a Force ± upward and gravity acts downward. The net Force (di- rected upward) on the helicopter is F = ma = s F up - s F down = F - mg Thus F = ma + mg = m ( a + g ) Note: Here a and g are absolute values oF the helicopter’s acceleration and the acceleration oF gravity, respectively. 002 (part 1 oF 2) 10.0 points A 13200 kg sailboat experiences an eastward Force 19300 N due to the tide pushing its hull while the wind pushes the sails with a Force oF 71900 N directed toward the northwest (45 westward oF North or 45 northward oF West). What is the magnitude oF the resultant ac- celeration oF the sailboat? Correct answer: 4 . 53258 m / s 2 . Explanation: According to Newton’s Second Law, mva = v F net = v F wind + v F tide . To fnd the magnitude oF the net Force, we draw the parallelogram For vector addition v F wind v F tide v F net θ α NW W E Note that 135 is the angle between the tide and the wind Forces. Use the Law oF Cosines: F 2 net = F 2 wind + F 2 tide - 2 F wind F tide cos θ = (71900 N) 2 + (19300 N) 2 - 2 (71900 N)(19300 N) cos135 = 3 . 57964 × 10 9 N 2 and the boat’s acceleration is a = F net m = 3 . 57964 × 10 9 N 2 13200 kg = 4 . 53258 m / s 2 . 003 (part 2 oF 2) 10.0 points What is the direction oF the boat’s accelera- tion?

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This note was uploaded on 05/02/2011 for the course PHYSICS 111 taught by Professor Wang during the Spring '09 term at NJIT.

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HW 3 solution_pdf - ghei (mg42382) HW 3 Opyrchal (11130)...

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