1122ˆˆ(0.500)(10.0i 12.0j)5.00i 6.00jˆˆ(0.750)(14.0)(cos110 i sin110 j)3.59i 9.87 jmv=+=+=°+°=−+GG(in SI units) and 312(2.65 0.500 0.750)kg 1.40 kgmmmm=− − =−−=, we solve for v3→and obtain3( 1.01 m/s)i ( 11.3 m/s)jv=−+−G.(a) The magnitude of 3vGis 3||v=G11.4 m/s. (b) Its angle is 264.9
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.