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PHY 121
Your 'cq_1_21.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A ball is tossed vertically upward and caught at the position from which it was released.
Ignoring air resistance will the ball at the instant it reaches its original position be traveling faster, slower, or at the same speed as it was when released?
answer/question/discussion: ->->->->->->->->->->->-> n:
Since gravity is conservative, the speeds would be identical if air resistance is not a factor
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What, if anything, is different in your answer if air resistance is present? Give your best explanation.
answer/question/discussion: ->->->->->->->->->->->-> n:
Some of the speed would be lost due to air friction decreasing the total `ds that it traversed leaving acceleration due to gravity with less distance to reclaim speed and still being slowed on its way down by air resistance. It would be slower than when it was released.
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&#Very good responses. Let me know if you have questions. &#
PHY 121
Your 'cq_1_21.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A typical automobile coasts up a typically paved incline, stops, and coasts back down to the same position.
When it reaches this position, is it moving faster, slower or at the same speed as when it began? Explain
answer/question/discussion: ->->->->->->->->->->->-> n:
It is moving slower. When it is coasting up the incline friction is acting in the direction opposite of motion and the KE will decrease until it stops. When the car begins moving down the incline, friction will again act in the direction opposite of motion and will slow the car even more.
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&#Very good responses. Let me know if you have questions. &#
If no other force is exerted parallel to the incline, what will be the cart's acceleration?
answer/question/discussion: ->->->->->->->->->->->-> n:
F_net = -24.5N
F_net = a*m
A = F_net/m
A = -24.5N/5kg
=-4.9N/kg
=-4.9kg*m/sec^2/kg
=-4.9m/sec^2
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&#Very good responses. Let me know if you have questions. &#