This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Page 1 Assignment 10 100 points + 20 bonus total Section 18.1 Problem 14 : = ? + ? and ? = ?? + ?? so ? = ??? 6 2 + ??? = ? 2 2 2 6 = 16 Problem 22 : The force has no -component. Therefore the (positive) work done in the first half of 1 will be exactly canceled by the (negative) work done in the second half, so the total work over the path 1 is zero. The same holds true for 2 , again by virtue of the vertical symmetry of the path and the fact that is constant and because the horizontal part of 2 contributes zero work. For 3 , the total work will be greater than zero, since the diagonal part of 3 is in the same general direction as and the horizontal part of 3 contributes zero work. Problem 44 : The below figure shows the wind velocity vectors on each side of the square, where the speed is ? meter/sec on the south side and ? 12 meter/sec on the north side. The circulation is the sum of the line integrals along the four sides of the square. The line integrals along the eastern and western edges are both zero, since the wind velocity is perpendicular to these edges. The integral to the right along the south side equals (1000 km)( ? meter/sec)= ? 10 6 meter 2 /sec, and the integral to the left along the north side equals (1000 km)( ? 12 meter/sec)= ? 12 10 6 meter 2 /sec. ?? ??? = ? 10 6 + ? 12 10 6 = 1.2 10 7 ? 2 / ....
View Full Document
This note was uploaded on 03/21/2010 for the course AMS 261 taught by Professor Fortmann during the Fall '08 term at SUNY Stony Brook.
- Fall '08