In HERTZ (rounded to 3 sig figs), approximately how frequently does this 1) device cycle through the full range of total charge magnitudes (from its maximum, through o to its minimum, and all the way back through o to its initial/maximum)? DON'T FORGET TO PROVIDE SOME KIND OF DIAGRAM! In seconds, find how much TIME will elapse from t=0 until precisely the 2) second instant the current is running at a maximum value but in the negative direction (the second time, as measured from t=0, that the current reaches 1= -15 Amperes). You may round to something like 3 sig. figs. 3) a) In Amperes per second, find the instantaneous rate at which the current changes with respect to time at the very first instant the current value is 1=0 Amperes. You may round to something like 3 sig figs. b) At this moment (described above), explain whether the current is growing increasingly negative, increasingly positive, decreasingly positive or decreasingly negative. Thoroughly and convincingly explain how you know. HINT for both the above. Whenever we talk about an 'instantaneous rate, we are talking about a derivative, no? Yes. 4) Assume that the whole phenomenon/experiment is re- done with all the same given equations and values EXCEPT: This time lo = -45 Amperes (instead of +15 Amperes). Provide answers to ALL the above three questions under these new conditions. You can re-calculate and/or provide verbal explanations, but you must fully explain how each of your above three answers does or does not change. 6) TRUE or FALSE and SHOW WHY!: 1 = +45 cos [377t + (pi)] is "A SOLUTION" to the Dif. Equation AND the new condition lo = -45 A

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Imagine an electrical current which comes out of an American wall outlet in a smoothly
alternating pattern of high current in one direction down in magnitude all the way through
zero to equally high current in the other direction, and so on forever. This alternating
current, is measured in Amperes and designated with a variable called / fluctuates
AROUND an EQUILIBRIUM value of 0 Amperes according to the following equation (for
which time is measured in seconds):
d²1
dt2
= -1421291.
Imagine, further, some wall device for which the current at t=0 (the initial and
maximum current) is l = +15 Amperes.
In HERTZ (rounded to 3 sig figs), approximately how frequently does this
1) device cycle through the full range of total charge magnitudes (from its
maximum, through o to its minimum, and all the way back through o to its
initial/maximum)? DON'T FORGET TO PROVIDE SOME KIND OF
DIAGRAM!
In seconds, find how much TIME will elapse from t=0 until precisely the
2) second instant the current is running at a maximum value but in the negative
direction (the second time, as measured from t=0, that the current reaches 1= -15
Amperes). You may round to something like 3 sig. figs.
3)
a) In Amperes per second, find the instantaneous rate at which the
current changes with respect to time at the very first instant the
current value is 1-0 Amperes. You may round to something like 3 sig
figs.
b) At this moment (described above), explain whether the current is
growing increasingly negative, increasingly positive, decreasingly
positive or decreasingly negative. Thoroughly and convincingly explain
how you know.
HINT for both the above. Whenever we talk about an 'instantaneous
rate', we are talking about a derivative, no? Yes.
4) Assume that the whole phenomenon/experiment is re-done with all the same
given equations and values EXCEPT: This time / = -45 Amperes
(instead of +15 Amperes). Provide answers to ALL the above three
questions under these new conditions. You can re-calculate and/or provide
verbal explanations, but you must fully explain how each of your above three
answers does or does not change.
6) TRUE or FALSE and SHOW WHY!:
1 = +45 cos [377 t + (pi)] is "A SOLUTION" to
the Dif. Equation AND the new condition / = -45 A
Transcribed Image Text:Imagine an electrical current which comes out of an American wall outlet in a smoothly alternating pattern of high current in one direction down in magnitude all the way through zero to equally high current in the other direction, and so on forever. This alternating current, is measured in Amperes and designated with a variable called / fluctuates AROUND an EQUILIBRIUM value of 0 Amperes according to the following equation (for which time is measured in seconds): d²1 dt2 = -1421291. Imagine, further, some wall device for which the current at t=0 (the initial and maximum current) is l = +15 Amperes. In HERTZ (rounded to 3 sig figs), approximately how frequently does this 1) device cycle through the full range of total charge magnitudes (from its maximum, through o to its minimum, and all the way back through o to its initial/maximum)? DON'T FORGET TO PROVIDE SOME KIND OF DIAGRAM! In seconds, find how much TIME will elapse from t=0 until precisely the 2) second instant the current is running at a maximum value but in the negative direction (the second time, as measured from t=0, that the current reaches 1= -15 Amperes). You may round to something like 3 sig. figs. 3) a) In Amperes per second, find the instantaneous rate at which the current changes with respect to time at the very first instant the current value is 1-0 Amperes. You may round to something like 3 sig figs. b) At this moment (described above), explain whether the current is growing increasingly negative, increasingly positive, decreasingly positive or decreasingly negative. Thoroughly and convincingly explain how you know. HINT for both the above. Whenever we talk about an 'instantaneous rate', we are talking about a derivative, no? Yes. 4) Assume that the whole phenomenon/experiment is re-done with all the same given equations and values EXCEPT: This time / = -45 Amperes (instead of +15 Amperes). Provide answers to ALL the above three questions under these new conditions. You can re-calculate and/or provide verbal explanations, but you must fully explain how each of your above three answers does or does not change. 6) TRUE or FALSE and SHOW WHY!: 1 = +45 cos [377 t + (pi)] is "A SOLUTION" to the Dif. Equation AND the new condition / = -45 A
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