<p>WRONG! thing of it this way. U stand on the lift with a weighing balance. the balance measure would increase as the lift goes up. essentially, the g is being increased (even though here, its a fictitious force)</p>
<p>Hence, T should decrease</p>
<p>I said g decreases as the elevator goes up because if you go up higher and higher, g decreases.
@indianmonster, what the scale reads is equal to the normal force</p>
<p>Essential, if you’re putting it that way then g should increase because if you go down INTO the earth, g actually decreases.</p>
<p>And if the increase in the scale reading is only because of normal force, then you would just fly off the scale. And we know this doesn’t happen.</p>
<p>I just checked and yes the g should increase.</p>
<p>what are you talking about @essential</p>
<p>The rule for fictious force: if frame is accelerating in one direction, the fictious force is APPLIED on the opposite direction.</p>
<p>Hence, if the elevator moves with, there is a force component on the pendulum pointing downwards, which caused the ‘overall g’ that I am talking about to increase.</p>
<p>U answer about decreasing g as we move it is bloody negligible, the elvator is not some inter space lift…=.-</p>
<p>what was the acceleration at point p?</p>
<p>does anyone remember the correct letter for the elevator question? I think I got A (forget if increase or decreasing T)</p>
<p>To above^^
Im pretty sure Ep is an upside down U. The block slowly comes to a stop at the top, so its more of a curve than a sharp line</p>
<p>But the actual increase in h decreases over time.</p>
<p>As h increases, PE does too, however, h isn’t linear…the speed of the block changes linearly, so h would be a parabola. Speed should be a V, I think.</p>
<p>speed should be a parabola, not a V</p>
<p>sorry to ask again, but what was the acceleration at point P?</p>
<p>speed should be a parabola, not a V </p>
<p>I think speed should be a V. a “v vs time” graph showing constant acceleration consists of a straight line</p>
<p>^^ to bball, what problem are you talking about?</p>
<p>Displacement is a parabola, so shouldn’t speed be a V? PE is based on displacement (height) so if speed is a parabola (2nd order) , PE would have to be an equation of the third order, which wasn’t an answer choice.</p>
<p>it was towards the end. it asked for the centripital acceleration and it gave 5 arrows. one pointed straight up, one straight down, one to the left but slightly up, one to the left but slightly down</p>
<p>What about the q vs time and current vs time graphs? where they linear or curved?
(There was a DC circuit with a capacitor and a resistor)</p>
<p>the Equivalent Resistance was 3/2 right?</p>
<p>@ basketball1 it was the left and slowly down one because it was the centripetal acceleration towards the center of the circle and the speed was decreasing aswell.</p>
<p>oh yeah, wasn’t the capitecence graph a line w/out slope?</p>
<p>@spectravoid yes</p>
<p>or I might have been wrong on the acceleration one, the centripetal acceleration was lower but that didn’t mean the acceleration had to be orientated downwards…I am not sure. As there is not only centripetal acceleration in this case, but also g force orientated down, I still believe the right answer is left and down (S-W)
At the elastic collision was the heavier object going entirely down or also to the right?</p>