Ap Physics Giancoli 5Th Edition Answers

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1,930 video remedies for all continual difficulties in Giancoli"s 7th Edition and also 1,681 remedies for a lot of continuous problems in the sixth Edition.

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g"s and we"ll convert that into meters per second squared. So, we have actually 9.8 meters per second squared per g; the g"s cancel offering us 58.8 meters per second squared. We know that centripetal acceleration is v squared over r and also we"ll solve this for r to find out what radius of curvature this course need to have so that the acceleration is 6g"s. So, we"ll times by r over a c and also the a c cancels on the left and the r cancels on the ideal leaving us with r amounts to v squared over centripetal acceleration. So that"s 233.33 meters per second squared separated by 58.8 meters per second squared offers us about 930 meters is the radius of this loop. And then for part B, we usage that free-body diagram to talk around the lift force upwards minus gravity downwards has to equal ma, that"s gonna be the net pressure and a is centripetal acceleration which we have already calculated is 58.8 meters per second squared. So, the lift force is the apparent weight of the pilot; it"s the normal pressure that they endure being pushed up by the seat and we"ll include F g both sides below to settle for the lift pressure. And lift pressure is m times acceleration plus gravity so that"s ma c plus mg and also you have the right to element out the m"s and you acquire 78 kilograms times 58.8 meters per second squared plus 9.8 meters per second squared and let"s figure out what that provides. 78 times 58.8 plus 9.8 renders about 5400 with two substantial figures. That"s the apparent weight at the bottom of the loop. At the optimal of the loop, the pilot will certainly will certainly have gravity downwards and also then it"s type of a misnomer, it"s kind of confmaking use of to call this a lift pressure bereason it"s actually gonna be a force that"s pushing dvery own in the direction of the ground to keep the aircraft going downwards in this loop. So, it goes dvery own along this path yet I"ll simply save calling it a lift force; it"s the pressure you know that the wings are, you know, the air is pushing on the wings and also then the wings push the plane dvery own via this force. OK So they are both pointing in the exact same direction in the direction of acceleration and also so we will speak to the radial direction towards the facility positive. So that renders gravity positive and also the lift force positive equates to mass times acceleration; acceleration is v squared over r because we deserve to watch the acceleration is in a circle… it"s radial so it"s v squared over r and substituting for gravity is mg. And then we fix for this lift pressure and also we"ll subtract mg from both sides here and also we gain lift force is mv squared over r minus mg; variable out the m and you get m times v squared over r. 78 kilograms times 233.33 squared over 925.9 and also then minus 9.8 and also we gain about 3800 newtons is the effective weight at the height of the loop.">
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