<p>i got one</p>
<p>mars has a mass 1/10 of earth and diameter 1/2of earth. the acceleration of a falling body near surface of mars is?
answers 4 m/s^2</p>
<p>another</p>
<p>A hollow metal sphere of radius R is positively charged. of the following distances from the center of sphere, which location will have greatest electric field strength? answer is 5R/4</p>
<p>there is a a picture</p>
<p>two identical conducting spheres are charged to +2q and -Q respectivly, and are separated by a distance D (much greater than the radii of spheres) as shown. the magnitude of the force of attraction on the left sphere is F1. after the two spheres are made to touch and then are reseparated by distance d, the magnitude of the force of the left sphere is F2. which relationship is correct</p>
<p>answr is F1= 8F2</p>
<p>=+2Q ----------------- -Q
d</p>
<p>I'm really confused by the stuff you're posting, and i've seen those questions before. They're from an old exam, 1980 somethin... anyway
Shouldn't you post the choices along with them?</p>
<p>k here are the choices then the mars question
A) .25
B) .5
c) 2
d) 4
E) 25
(they are all in m/s^2)</p>
<p>the sphere question
A) 0( CENTER of sphere)
B) 3r/4
C) 5r/4
d) 2r
E) none of the above</p>
<p>the 3rd question</p>
<p>a) 2f1- f2
b) F1=F2
C) F1=2F2
D) F1=4F2
E) F1=8F2</p>
<p>Does anyone have the 1988 exam? Please! I am willing to scan you 2003 Calculus AB & BC. PLEASE!!!!!!!!!!!!!!!!!!!!!!!111111111111111111111111111</p>
<p>I might, i'd have to check
but.. calc is kinda finished now</p>
<p>can someone explain the F1 and F2 relationship one...? i dont understand that.....well i get that F1 is greater (because of the greater diff in the charges) but after you touch the two spheres..what happens?...they become neutral...some of the charge from the +2Q goes to the -Q...>.<</p>
<p>You're given the charges, and distance doesn't change, so all that matters it the (Q1*Q2) part of the equation.</p>
<p>Before, F1 = (+2)(-1) = -2 --> F1 = 2</p>
<p>When the spheres touch, the charges equalize between the two spheres, so you average the charges:
((+2) + (-1))/2 = 1/2 --> each sphere has a charge of Q/2</p>
<p>Now plug back into the equation for electric force, once again ignoring the distance and the constant:
(1/2)*(1/2) = 1/4 = F2</p>
<p>F2<em>x = F1
1/4</em>x = 2
x = 8
--> F1 = 8F2</p>
<p>oooh thanks..that was simple enough...</p>
<p>and does anyone have the answers to the 1988 or 1984 MC exams?</p>
<p>bump for the other 2 unanswered questions</p>
<p>a question on one of the 2004 mc questions.</p>
<ol>
<li>The operating efficiency of a 0.5A, 120V electric motor that lifts a 9kg mass against gravity at an average velocity of 0.5m/s is most nearly</li>
</ol>
<p>(A) 7%
(B) 13%
(C) 25%
(D) 53%
(E) 75%</p>
<p>E baby. The answer is e</p>
<p>crichesill, your first question with the gravity, i think it is not the diameter, i think by diameter you mean radius because otherwise the problem wouldn't work. For your second question, i would need to see the diagram.</p>
<p>can you explain why the answer is (e)?</p>
<p>Finally! One that I can answer.</p>
<p>efficiency = (Power Output) / (Power Input)</p>
<p>efficiency = (P = Fv) / (P = IV)</p>
<p>efficiency = (F[gravity]*v) / (IV)</p>
<p>e = ([9kg<em>10 m/s^2]</em>0.5 m/s) / (0.5A * 120V)</p>
<p>e = 45/60 = .75 = 75%</p>
<p>M17 got that sucka</p>
<p>what do you multiply your free response score by and what do you multiply your mc score by to get your raw score?</p>
<p>english its diameter. thats not a typo. i think they are trying to confuse us by saying diameter coz u know thers an extra calculation to convert it to radius so i guess its 1/4 radius of earth</p>
<p>multiply free response score by (1)
multiply mc score by (54/45)</p>
<p>114-180 = 5
86-113 = 4</p>
<p>that's all i remember</p>
<p>aznballerlee how did u get 2004 MC? is that released?</p>
<p>anyone have an answer on the second question i posted</p>