For the first part, we are looking for Vf when dy=11.0
Upward is positive, downward is negative.
So <span>Vf = square root [2(-9.8)(11.0) + (18.0)^2] </span>
<span>Vf = 10.4 m/s your answer is correct.
For the part b, t is equals to the time took to reach and dy is equals to 11.0
you did, </span>11= 18t m/s-(1/2) 9.8t^2 then <span>-11 + 18t- 9.8t^2. By quadratic formula, for the way down the answer is 2.9 s while on it's way up, the answer is 0.77 s</span><span>
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At the end of one full time period, the ant has returned to where it was at the beginning of the time period. Its displacement is <em>zero</em>.
If I am to understand this question correctly this is what asks you:
If a person is riding a motorized tricycle how much work do they do?
You may ask yourself, why did I only use part of the question. Simple, the rest is not relevant to what is being asked. The weight, speed, and distance wont affect the person riding any <em><u>motorized vehicle</u></em> other than the time it takes to get from one place to another.
So to answer this question I would say:
Not much, all they really have to do is to steer and set the motorized tricycle to cruise control. Just like any rode certified vehicle.
If you have any questions about my answer please let me know and I will be happy to clarify any misunderstandings. Thanks and have a great day!
He reasoned that since parallax could not be observed for celestial objects near the sun, then the earth was stationary. This erroneous assumption was because at the time he had no way of knowing that celestial objects were so far away that their parallax angles were too small to detect.
Answer:
amount of charge on the source charge
