SAT II Physics (Gary Graff) Episode 1 Part 5 docx

... SUMMARYPeterson’s SAT II Success: Physics 98The 250 N boulder shown in the diagram above possesses PE,which can be stated in equation form below.PE = mgh = wt h= ( 250 N) (50 m)= 12 ,50 0 JIf the boulder ... rearranging yields: vv=××()×+×=×−667 10 610 64 10 3 81 10 59 11 24 65 . .Nmkgkgmm22 11 06m/sWEIGHTLESSNESSThe strength of the earth’s gravitational field decreases ... sincossincostanθθθθθNEWTON’S LAWS OF MOTIONPeterson’s SAT II Success: Physics 11 8• Diagram D is the same as Diagram B except that the trailing partsof the two waves are superimposed on one another.•...

... describe the image.Solution 11 1fpq=+Rearrange and substitute: 11 1 1 8 11 12 5 05 1 10 75 fpqqqq−=−=−===cm 20cmcm cmcm 13 .25cm . The image is located 13 . 25 cm from the lens on the ... length of 3.6 cm. Describethe image.Solution 11 1 11 1 36 1 37 1 277 027fpqfpqq=+−=−=−rearranges to 1 cm cmcm cm. 1 cm 4cm===qq 1 0 25. The image is located 4 cm from the mirror. ... convex mirror. 11 1 11 1 36 1 37 1 277−=+−−=−−=−−fpqfpqqrearranges to 1 cm cmcm. 0027 1 304 3 29cm 1 cm cm==− = −qq CHAPTER 3Peterson’s: www.petersons.com 12 9Concave lenses...

... of a set of parallel resistors is found in thefollowing manner: 11 11 123RRRRt=++ 11 2 1 5 1 10 1 40 1 5 2 1 0 25 1 8 25 12 1RRRRtttt=++ +=+++==ΩΩ Ω ΩΩΩΩ ΩΩ ..The value of ... series). 11 111 1 11 2 1 3 1 4 1 5 1 6 1 5 12 3 45 CCCCCCCfffffCttt=++++=++++=µµµµµµ. fffff fCft++++= .33 25 2 16 67 14 5 µµµ µµCHAPTER 5 Peterson’s SAT II Success: Physics 17 0SERIES AND ... R3 – R4 parallel set of resistors: 11 11 11 10 1 8 1 4 1 25 4 12 3RRRRRRRtttt=++=++==ΩΩΩΩΩ.CHAPTER 5 Peterson’s SAT II Success: Physics 18 4When two current-carrying wires are...

... )−−°++ = +°−26 13 1 25 19 5 19 50 836 8((, . ),. . .,26 13 1 25 26 13 1 25 19 50 836 8 19 5 28 0 81 JJJJCtJCtJff+=+°+°==°=°= 856 328 0 81 32 8.,.JCtJJtCtfffIt ... tJJoulesgCgtfof==−=°−°(),.()( )()( 12 50 04 18 4 300 20CCJJoulesgCtf),.()( )()(,=° 12 50 0 12 55 2 25 1 004 12 50 0 25 10 4 12 55 237JJJJoulesgCtf),,.()( ... halved (to 5 mm)?SolutionFKqqrFmCCCm==×××−−()()(. )(. )(. ) 12 2922 19 19 2 910 16 10 16 10 01 N=ו×=×−−−FmmF2 304 10 23 10 28 2224..N 11 04That’s...

... coasting.vatvvfff===(. ). 12 5 17 5 m/s ) (14 sm/s2Now we find the displacement while the train coasts, using:s = v ts = (17 .5 m/s) (50 s)s = 8 75 mThe equation that best fits the third part of the problem ... the already known F 1 . The calculation ofthe torques is:TF FTwt wt=+()( ) 11 There are two torque-producing entities.==•+•=•+•=•()(. )10 1 2 5 10 11 Nm N mN m 1N mNmTTShould ... operation svt ato=+ 1 22 is suitable for part one. Rememberthat the train started from rest, so the (vot ) term is zero. The workingequation is: sat ss=== 1 2 1 2 12 5 14 12 2222( . )(...

... exactly 18 0° opposite the direction thebird must fly. Take the resultant vector and add (or subtract) 18 0° toor from the vector’s direction. 18 0° + 31 = 211 °The bird must fly 58 3 m @ 211 ° from ... www.petersons.com 63Rope b extends 35 into the third quadrant .Side y = (side r) (sin 35 ) = (16 N) ( .57 ) = 9.1NSide x = (side r) (cos 35 ) = (16 N) (.82) = 13 .1NBoth components of rope b are ... with thex componentsROPERope aRope bRope cRope dY axis+ 7.8N 9.1N +19 .4N 26.1NX axis+29.1N 13 .1N+ 5. 2N + 15 .0N–8N in x direction+36.2N in y directionAfter combining the x components...