Chapter 5121P5.4Fmgg==weight of ballvvrelease=and time to accelerate =t:ai=∆∆vtvtvt±(a)Distance xvt=:xvtvt=FHGIKJ=22(b)FjipggFFvgt−=±±FijpgggtF=+P5.5m=400. kg,vii=300.±ms,j8800100.±.±ejt=savij+∆t500.±.±.±ms2Faij+m250.±.±NF=( )+( )=559...NP5.6(a)Let the x-axis be in the original direction of the molecule’s motion.ataafi=+ −=+×=−×−:..6706703 00 10447 101315mss2(b)For the molecule,
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .