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PHY 201
Your 'cq_1_08.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_08.1_labelMessages **
A ball is tossed upward with an initial velocity of 25 meters / second. Assume that the acceleration of gravity is 10 m/s^2 downward.
• What will be the velocity of the ball after one second?
answer/question/discussion: ->->->->->->->->->->->-> :
v_f = v_0 + a*’dt = 25 m/s + (-10 m/s^2)*1 s = 15 m/s
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
v_f = v_0 + a*’dt = 25 m/s + (-10 m/s^2)*2 s = 5 m/s
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
v_f - v_0 = 5 m/s - 25 m/s = -20 m/s
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
‘ds = (v_0+v_f)/2 * ‘dt = (25 m/s + 5 m/s)/2 * 2s = 15 m/s * 2s = 30 m
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• What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion: ->->->->->->->->->->->-> :
v_f = v_0 + a*’dt = 25 m/s + (-10 m/s^2)*3 s = -5 m/s
v_f = v_0 + a*’dt = 25 m/s + (-10 m/s^2)*4 s = -15 m/s
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• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: ->->->->->->->->->->->-> :
Max height will be when v_f = 0. We use second equation solved for ‘dt.
‘dt=(v_f - v_0)/a = (0 - 25 m/s)/-10 m/s^2 = -25 m/s / -10 m/s^2 = 2.5 s
Then the first equation
‘ds = (v_0+v_f)/2 * ‘dt = (25m/s / 2) * 2.5 s = 12.5 m/s * 2.5 s = 31.25 m
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• What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
answer/question/discussion: ->->->->->->->->->->->-> :
at t=4, v_f=-15 m/s v_ave = (25 m/s + -15 m/s)/2 = 10 m/s / 2 = 5 m/s
‘ds = v_ave*’dt = 5 m/s * 4 s = 20 m
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
v_f = v_0 + a*’dt = 25 m/s + (-10 m/s^2)*6 s = -35 m/s
v_ave = (20 m/s + -35 m/s)/2 = -15 m/s / 2 = -7.5 m/s
‘ds = -7.5 m/s * 6 s = - 45 m so 45 m below its initial position
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&#Good responses. Let me know if you have questions. &#
critiqued_student work modified 140303__________
Your areas, and hence the units in which you have reported work and energy to this point, should be Newton * mm (the units of the first graph are Newtons of force vs. millimeters of length) and Newton * cm (the units of your calibration graphs should be Newtons of force vs. cm of length).
Newton * mm and Newton * cm are both valid units of energy, but neither is a standard unit.
There are two standard units for energy. One is the Newton * meter, also called a Joule, and you are likely familiar with this unit. The other is the dyne * cm, also called an erg.
• How many mm are in a meter? How many Newton * mm are therefore in a Newton * meter or Joule? Give your comma-delimited answers in the first line.
• By what number would you therefore multiply a quantity given in N * m in order to get the same quantity in N * mm? And by what number would you multiply a quantity in N * mm to get the same quantity in N * m or Joules? Give your comma-delimited answers in the second line.
• How many cm are in a meter? How many Newton * cm are therefore in a Newton * meter or Joule? Give your comma-delimited answers in the third line.
• By what number would you therefore multiply a quantity given in N * m in order to get the same quantity in N * cm? And by what number would you multiply a quantity in N * cm to get the same quantity in N * m or Joules? Give your comma-delimited answers in the fourth line.
• Starting in the fifth line give an outline of the reasoning used to compare these units.
Your answer (start in the next line):
1000mm=1M, 1N=1J
I have no clue on the N to meters, and the N to cm… can you explain??? Maybe multiply by 1000 for mm and 100 for cm??
100cm=1M
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Newtons don't convert to meters or centimeters. Newtons measure force, millimeters and centimeters measure distance.
However the unit of force * distance is the Newton * mm, and this unit would convert to Newton * meters. The conversion in this case would involve the factor 1000, since it takes 1000 mm to make a meter.
Do think of the following and make sure your answer to the question is consistent with your answers to these questions:
What is bigger, a Newton * millimeter or a Newton * meter?
Which would you have more of in a given quantity, the bigger unit or the smaller unit?
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#$&* mm in m & N mm in N m, mult N m to N mm ^ N mm to N m, cm in m & N cm in N m, multipliers
F = m a, so units of force are equal to units of mass, multiplied by units of acceleration.
A Newton is a kg * m / s^2, and a dyne is a gram * cm / s^2.
• How many grams are in a kg, and by what number would you multiply a quantity in kg m/s^2 to express the same quantity in gram * m / s^2? Give your comma-delimited answers in the first line.
• How many cm are in a meter, and by what number would you multiply a quantity in gram m/s^2 to express the same quantity in gram * cm / s^2? Give your comma-delimited answers in the second line.
• By what number would you therefore multiply a quantity in kg m/s^2 to express the same quantity in gram cm / s^2? Answer in the third line.
• By what number would you multiply a quantity in gram cm / s^2 to get the same quantity in kg m/s^2? Answer in the fourth line.
• Starting in the fifth line explain your reasoning.
Your answer (start in the next line):
1000g=1kg, 500
100cm=1m, 10
The third and fourth questions I do not understand.
#$&* g in kg & mult kg m/s^2 to g m/s^2, cm in m & mult g m/s^2 to g cm/s^2, mult kg m/s^2 to g cm/s^2, vice versa
From the above you should see that a dyne is 1/100,000 N, or 10^-5 N, while a Newton is 100,000 dynes, or 10^5 dynes.
Recall that a Joule is a Newton * meter, and an erg is a dyne * centimeter.
• How many ergs does it therefore take to make a Joule?
• By what number would you multiply work in Joules to get work in ergs?
• By what number would you multiply work in ergs to get work in Joules?
Give your comma-delimited answers in the first line.
Explain your thinking starting in the second line.
Your answer (start in the next line):
100, 10, 10
The first number I am sure about but not the next two. I would assume they would be less than 100 and in intervals of 10 though.
#$&* ergs in J, mult J to erg, mult erg to J
How much energy, in Joules, does your first rubber band store when it is at the length required to support 10 dominoes?
How many ergs is this?
Give your two answers in comma-delimited format in the first line below, and your explanation starting in the second line:
Your answer (start in the next line):
.53N, 5.3
.53*10=5.3 ergs
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energy in J at 10 dom lgth, same in ergs
Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
• Approximately how long did it take you to complete this experiment?
Your answer (start in the next line):
2.5 hours
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You've done some good work, especially in the early part of this exercise.
However you will need to make a number of revisions.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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