h9_6a - éNarmi Fame, WA CM“?sz Ram, (Leak/\VJ r...

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Unformatted text preview: éNarmi Fame, WA CM“?sz Ram, (Leak/\VJ r A" .f “at, INF?!» <5? M4474?“ w M cap? koaLsiKLa: "M W! (1/erst 40¢ Mu (SE 1512, W—i’ {i towfié‘fi-A—i"\ / in w?— owLJ M243 +0 (swan, m Camrgy.‘ «activaqu ‘6)" Cueng jg 3.1%“ \l'fikkkes. f - a fix {LHSM '- 81?) ~ M 61¢: ,f '_ ?.G {1 vii), “1/; 53‘” ${sgraigf¢¢v¢fltijfim/m.mdh {bah 0w Fmis Tm. “Mimi gm; fan“;ng aim-a: few“ Srlkrmijulr 310% 1.5:? WW5. {v.12 is aim/35 pram“, in, M M33 5% in mm. Emmm%m%mM&%®%C%@ m‘L aim gait-Q1 x Qfifil [Emmi Maw M44434 MAI Lg’hk igng drf Jirecéd mi visa “QR/(2(6- ‘Baxk w {:Dt, 01 m wafltvfieqid muting/Aim '. 5W (he/Lew Simer (was: MM Erma «m Joan/Ung wgeflwé M ihM‘L‘J m Sim-w {mall-v mfim 11‘) MS ‘f‘b’Liiii 47% WW 1“, mg” L? m MW Wm} M- m flaw pew; 4v okLOW/‘i' \ELLAQ ca; A‘ m I. I S fi-Qfihoafi 3-1413}; did-3L \IQifi L‘Il ‘ [’ZHL)I r 2. M ac: T?” 1-: m 1' 3:1 K W: W wxl‘MM W(M\ ( whrqeék dfiCMIY fOWUE’i‘] is - I "GN So m M j— : 6'3"“ 2;, 2 Git-r 3 “if “ll/‘1' G;an // Fm“ (5'44 (2(th V2 :— I T 9117] PM 5 f m flggmri—mijw (“-0me M “a? IWi “Mimi “*9 ’I‘M mar; 49? 63% (Uri; [Last A {u {LJL M, few-1 who»: N" t‘ w‘j M‘ N“ 3°54. L/WW gJWJM \ WW? 4:1) LR? “N'mfi 'W‘“ («J-1 «a UAjr‘iO‘Sy 15 “Q Cami we C-‘M fiu‘j 0L ’L mci = Offfwg" M - W9“ v: ‘0 W 3’ Mafia} A .5 z N5 M-Mj = “3537;” (Mtjmfiu‘i “Cm” 11" POM-é {8 fifbfi (L x” v1 . E: mugs; 1%9’?’ (?],L£T£:f—q3W/» ( "‘ T T ‘ / S"“A‘j f” ‘3'?“ T: [1125 ¢ ML? - ‘ - us?— I ‘ I 7 M *" 94m“ WELng {‘93 f ) N n/ I d“ V. ingot/MR ,, N p ,_, 5 2““3 ‘ 71?“ N) 5180094 //; L '3 ‘ L“ m 5w» fif {flag at” M (AW-493$ (205d 3&5: T; ’f_ " 2 NA. IFSW :5) r- _ Part A Which of the following quantities represent mass? A. 12.0 31’ B. 0.34 g c. 120 D. 1600 w“ E. 0-34 m F. 411 gm ANSWER: BC Explanation.- Those are the only ones with units of mass Part B Which of the following quantities would be acceptable representations of weight? A. 12.0 a}; B. 0.34 g c. 120 kg D. 1600 3‘3" E. 0.34 fis- F. 411 ANSWER: AD Explanation: Weight is afarce and is measured in newtons (or kilonewtons, meganewtons, etc.) or in pounds (or tons, megatons, etc.). Using the universal law of gravity, we can find the weight of an object feeling the _ _ . w w amass? gravitatlonal pull of a nearby planet. We can write an expresswn ' , where Wis the weight of the object, Qis the gravitational constant, mis the mass of that object, Mis mass of the planet, and iris the distance from the center of the planet to the object. If the object is on the surface of the planet, iris simply the radius of the planet. Part C The gravitational field on the surface of the earth is stronger than that on the surface of the moon. If a rock is tranSpOrted from the moon to the earth, which properties of the rock change? ANSWER: weight only Explanation: Mass is constant. It is anaflected by the force of gravity, so only the weight changes. Part D An object is lified from the surface of a spherical planet to an altitude equal to the radius of the planet. As a result, what happens to the mass and wieght of the object? ANSWER: mass remains the same; weight decreases Explanation: The radius is increased, and the radius appears in the denominator of the equation for the force of gravity, therefore the force must decrease as the radius is increased. ...
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h9_6a - éNarmi Fame, WA CM“?sz Ram, (Leak/\VJ r...

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