Equations, Conversion Factors, and Constants - Exam 2 | PHYS 0002A, Exams of Physics

Download Equations, Conversion Factors, and Constants - Exam 2 | PHYS 0002A and more Exams Physics in PDF only on Docsity! Physics 2A EXAM 2 Name:___________________ Instructions: 1. Show your work on numerical problems. Do not expect any credit for a problem if the instructor cannot logically follow what you have put on your paper. 2. Make sure you include units with each answer. One point will be deducted for each infraction. 3. Round your answers to no more than 2 decimal places. 4. Circle your final answers. ________________________________________________________ Equations, Conversion Factors and Constants: 1 mile= 5280 feet (exact) 1 day= 24 hours (exact) 1 km =0.6214 miles 1 year=365.25 days 1 inch= 2.54 cm (exact) 1lb=0.454kg 1 hour=60 minutes (exact)=3600 sec 1 km=1000 meters (exact) 1 slug=14.59 kg 1ft=30.48cm 1 foot= 12 inches 1 mile= 1609.3 m 1 gallon =3.785 liters=231 in3 1 meter=39.37 inches 1 cm=0.01meters=10 mm (exact) 1   a  g  9.80m s2 32.2ft/s2 for free fall Newton's 2nd Law  F m  a fk kN fs sN Equilibrium  F =0,means Fx 0 and Fy 0 Weight mg W   F   x cos Wnet KEFINAL  KEINITIAL  1 2 mv final 2  1 2 mv initial 2 GPE mgy; GPE mgy final  mgy initial EPE  1 2 kx final 2  1 2 kxinitial 2 Conservation of Energy: PE+KE=0 WNC PE+KE Pavg  W t =Fvavg Conservation of Momentum  p initial   p final where  p m  v Impulse Momentum Theorem  I   F t   p Kinetic Energy in terms of p; K p2 2m