solution-hw6

solution-hw6 - lai (yl8859) HW06 Ross (15200) 1 This...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: lai (yl8859) HW06 Ross (15200) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points A boat moves through the water with two forces acting on it. One is a 2140 N forward push by the motor on the propeller, and the other is an 1879 N resistive force due to the water around the bow. What is the acceleration of the 1111 kg boat? Correct answer: 0 . 234923 m / s 2 . Explanation: Given : F 1 = 2140 N , F 2 = 1879 N , and m = 1111 kg . 2140 N 1879 N The forces are unbalanced, so the net force is F net = m a = F 1 F 2 and a = F 1 F 2 m = 2140 N 1879 N 1111 kg = . 234923 m / s 2 forward. 002 (part 2 of 3) 10.0 points If it starts from rest, how far will it move in 18 s? Correct answer: 38 . 0576 m. Explanation: Given : t = 18 s Under acceleration, x = v i t + 1 2 a ( t ) 2 = 1 2 a ( t ) 2 since v i = 0 m/s. Thus x = 1 2 a ( t ) 2 = 1 2 ( . 234923 m / s 2 ) (18 s) 2 = 38 . 0576 m . 003 (part 3 of 3) 10.0 points What will its velocity be at the end of this time interval?...
View Full Document

This note was uploaded on 02/24/2011 for the course PHYS 152 taught by Professor Button during the Winter '08 term at IUPUI.

Page1 / 3

solution-hw6 - lai (yl8859) HW06 Ross (15200) 1 This...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online