181hw7s - Power density PD = 2450 W/11.0447 2 = 222 W/mm 2...

This preview shows page 1. Sign up to view the full content.

Full name:________________________________ Lab group: L 5 _ _ (-5 pt for wrong info) HW7, ET181, Spring 10. Due by 6 pm on Mon, Apr 5. 1) Multiple choice questions Page 705-706 Page 744-745 Question Answer Question Answer 29.1 B 30.2 A, B, E 29.3 D 30.4 A, B, D 29.6 B 30.6 C 29.8 B 30.7 B 29.10 B 30.8 D 1) Problem 29.1, page 706 Area A = π (3.75) 2 /4 = 11.045 mm 2 Power P = 0.70(3500) = 2450 J/s = 2450 W.
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Power density PD = 2450 W/11.0447 2 = 222 W/mm 2 . This power density is sufficient for welding since the minimum requirement is 10 W/mm 2 (see page 701). 2) Problem 29.4, page 706. Reference: Table 29.2 If using SI units, U m (aluminum) = 2.88 J/mm 3 U m (steel) = 10.32 J/mm 3 3) Problem 29.10, page 707 4) Problem 29.14, page 707 p1/1...
View Full Document

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern