A.  Oblate\nB. Slope\nC. Weight\nD. Contour
204774
b-although gradient is deals with curves(like quadratic graphs) and slopes with straight lines
500476
1D\n2D (acceleration can be negative in which case you mihgt call it deceleration)\n3B
482177
a) zero momentum, as velocity on top is zero. p=mv\n     p=0.100kg * Om/s\n\nb) Vf = Vi + at\n      0=15+(-10)t\n     t=1.5ec (to reach maximum height)\n\nDivide by 2 for halfway up .i.e 0.75ec\n\nVf = Vi + at\nO=15 + (-10)(0.75)\nVf = 7.5m/s\n\np=mv   (p = momemtum)\n\np = (0.100kg)(7.5m/s)\np = 0.75kg*m/s, upwards (Direction is important)
620821
That would be C.  \n\nThese links should help you understand this question.
53165
The answer is obviously (A). The bar may shrink a little, but it can't disappear completely, as the options B, C and D imply!!\nHence, by process of elimination, the answer is (A). :-)
351449
The correct answer is B.
537696
if acceleration is on the x axis, the slope represents 1/mass\n\nas\n\nslope = x/y = acc/force\nand\nforce=mass x acc\nso\nacc/force=1/mass\n\nif force is on the x axis, the slope represents mass\n\nas\n\nslope = x/y = force/acc\nand\nforce=mass x acc\nso\nforce/acc=mass
367965
1.) B\n2.) D\n3.) C\n4.) E\n5.) E\n6.) E\n7.) A
441131
1) increases\n2) decreases\n3) remains the same \n4) increases\n5) decreases\n6) increases
320800
fraction line, quotient line, ratio line
146655
a d............or a corner..........or a straight line
14287
d. all of the above.\n:]
352710
The chute slows the FALL!  Right? \nF\nA\nL\nL\nI\nN\nG\nChute\nStill falling but, at a slow decent.  You don't rise.
548282
a.) newtons\nb.) km/h\nc.) f(x) = 0\nd.) x = f(0) + c
651181
Loose weight.\nCurls.
66623
Circular. Imagine that you have a weight on a string, and you whirl it around over your head. That's the basic gist of it. However, in real life, orbits are elliptical.
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ii) \nacc. of center of mass = linear acc. of the body\nacc of pt. of contact = centripetal acc.=r*w^2\ni)\nacc. of center of mass = g sin a + acc. due to any othe force (a = angle of incline)\nacc of pt. of contact = centripetal acc. =r*w^2