i Need help for isaacphysics assignment please!!!!! ASAP! Watch

poonshark
Badges: 2
Rep:
?
#1
Report Thread starter 3 years ago
#1
1) A current of 5.5mA5.5mA is flowing through a resistance of 220kΩ Find the potential difference (in volts) across the resistance.

5.5 mA =5,500,000 A , 220 kohms =220,000 ohms. P.D= IxV, = 5,500,000 X 220,000 = 1.21 X 10^12 Volts
i got : 1.21x10^12. but it says sig figs are incorrect? WHERE have i gone wrong???
0
reply
Smithenator5000
Badges: 9
Rep:
?
#2
Report 3 years ago
#2
(Original post by poonshark)
1) A current of 5.5mA5.5mA is flowing through a resistance of 220kΩ Find the potential difference (in volts) across the resistance.

5.5 mA =5,500,000 A , 220 kohms =220,000 ohms. P.D= IxV, = 5,500,000 X 220,000 = 1.21 X 10^12 Volts
i got : 1.21x10^12. but it says sig figs are incorrect? WHERE have i gone wrong???
Hello there,

Firstly, if you have transcribed the problem correctly, then  5.5 mA represents  5.5 \times 10^{-3} A ; the correct way to read  mA is 'milli-amperes'. If you instead meant to type  5.5 MA , then your interpretation would be correct.Your interpretation of the resistance is correct and you can now use Ohm's law,  V = IR to calculate the potential difference.

 V = IR

 I = 5.5 \times 10^{-3} A

 R = 2.20 \times 10^{5} \Omega

 \therefore V = (5.5 \times 10^{-3})(2.20 \times 10^{5}) = 1210 V = 1200 V (2 significant figures)

I think Isaac Physics wants you to quote your answer to two significant figures, as this is the minimum number of significant figures that have been given to you in the question.

I hope that this has been helpful.

Smithenator5000.
0
reply
poonshark
Badges: 2
Rep:
?
#3
Report Thread starter 3 years ago
#3
(Original post by Smithenator5000)
Hello there,

Firstly, if you have transcribed the problem correctly, then  5.5 mA represents  5.5 \times 10^{-3} A ; the correct way to read  mA is 'milli-amperes'. If you instead meant to type  5.5 MA , then your interpretation would be correct.Your interpretation of the resistance is correct and you can now use Ohm's law,  V = IR to calculate the potential difference.

 V = IR

 I = 5.5 \times 10^{-3} A

 R = 2.20 \times 10^{5} \Omega

 \therefore V = (5.5 \times 10^{-3})(2.20 \times 10^{5}) = 1210 V = 1200 V (2 significant figures)



I think Isaac Physics wants you to quote your answer to two significant figures, as this is the minimum number of significant figures that have been given to you in the question.

I hope that this has been helpful.

Smithenator5000.
Thank you, that helps a lot! I didn't realise the unit was milli-amperes
0
reply
Smithenator5000
Badges: 9
Rep:
?
#4
Report 3 years ago
#4
(Original post by poonshark)
Thank you, that helps a lot! I didn't realise the unit was milli-amperes
You're very welcome All the best with the assignment.
0
reply
poonshark
Badges: 2
Rep:
?
#5
Report Thread starter 3 years ago
#5
(Original post by Smithenator5000)
You're very welcome All the best with the assignment.
Sorry to bother you but there is a question on my assignment which i have never been taught, i am a bit confused how it works?

I am given this table-

Name:  Picture1.png
Views: 596
Size:  49.2 KB

and these questions -

Name:  Picture1.png
Views: 626
Size:  44.9 KB
Attached files
0
reply
Smithenator5000
Badges: 9
Rep:
?
#6
Report 3 years ago
#6
(Original post by poonshark)
Sorry to bother you but there is a question on my assignment which i have never been taught, i am a bit confused how it works?

I am given this table-

Name:  Picture1.png
Views: 596
Size:  49.2 KB

and these questions -

Name:  Picture1.png
Views: 626
Size:  44.9 KB
Hello there,

I am surprised that you haven't been taught this; it is pretty important. For the first part, it wants you to be able to write the units on the left in an equivalent form. All of these units can be expressed in terms of the seven 'base units'. These are:

Name:  si base.PNG
Views: 558
Size:  12.7 KB
source:http://physics.nist.gov/cuu/Units/units.html

You can find the base unit representations by using a formula that you know and treating the units like variables. For example, the Newton  N is a unit of force. One equation involving force is  F = ma , a form of Newton's Second Law. This equation implies that  [F] = [m][a] , where  [i] represents the units of quantity  i . We can use this to find the base unit representation of the Newton.

 [m] = kg

 [a] = ms^{-2}

 [F] = (kg)(ms^{-2}) = kgms^{-2}

 \therefore N \equiv kgms^{-2} \because [F] = N
You would enter this into Isaac Physics by telling it the power of each unit it lists.

Once again, I hope this has been helpful and please do not worry about bothering me.

Smithenator5000.
0
reply
poonshark
Badges: 2
Rep:
?
#7
Report Thread starter 3 years ago
#7
(Original post by Smithenator5000)
Hello there,

I am surprised that you haven't been taught this; it is pretty important. For the first part, it wants you to be able to write the units on the left in an equivalent form. All of these units can be expressed in terms of the seven 'base units'. These are:

Name:  si base.PNG
Views: 558
Size:  12.7 KB
source:http://physics.nist.gov/cuu/Units/units.html

You can find the base unit representations by using a formula that you know and treating the units like variables. For example, the Newton  N is a unit of force. One equation involving force is  F = ma , a form of Newton's Second Law. This equation implies that  [F] = [m][a] , where  </b>[i]<b> represents the units of quantity  i . We can use this to find the base unit representation of the Newton.

 [m] = kg

 [a] = ms^{-2}

 [F] = (kg)(ms^{-2}) = kgms^{-2}

 \therefore N \equiv kgms^{-2} \because [F] = N
You would enter this into Isaac Physics by telling it the power of each unit it lists.

Once again, I hope this has been helpful and please do not worry about bothering me.

Smithenator5000.
Thanks again! I'll give this a go.
0
reply
Smithenator5000
Badges: 9
Rep:
?
#8
Report 3 years ago
#8
(Original post by poonshark)
Thanks again! I'll give this a go.
You're welcome. Do not hesitate to ask for help if required.
0
reply
poonshark
Badges: 2
Rep:
?
#9
Report Thread starter 3 years ago
#9
(Original post by Smithenator5000)
You're welcome. Do not hesitate to ask for help if required.
Hi , i have to find the area under the line in the graph. I don't understand where i gone wrong? i got (25x5)/2 + 25x1 = 87.5 - 88 to 2 sf. This is the graph i have. Name:  Picture1.png
Views: 532
Size:  3.2 KB
0
reply
Smithenator5000
Badges: 9
Rep:
?
#10
Report 3 years ago
#10
(Original post by poonshark)
Hi , i have to find the area under the line in the graph. I don't understand where i gone wrong? i got (25x5)/2 + 25x1 = 87.5 - 88 to 2 sf. This is the graph i have. Name:  Picture1.png
Views: 532
Size:  3.2 KB
Hello there,

Notice that the y-axis (showing current in amperes) starts from  6 A , as opposed to  0 . I believe that the question wants you to consider the area which includes that which has been omitted from the graph. Here is a diagram of what I am trying to describe.

Name:  current graph.PNG
Views: 530
Size:  6.4 KB

You can now try calculating the area of this. Please reply if this doesn't work.

Smithenator5000.
0
reply
poonshark
Badges: 2
Rep:
?
#11
Report Thread starter 3 years ago
#11
(Original post by Smithenator5000)
Hello there,

Notice that the y-axis (showing current in amperes) starts from  6 A , as opposed to  0 . I believe that the question wants you to consider the area which includes that which has been omitted from the graph. Here is a diagram of what I am trying to describe.

Name:  current graph.PNG
Views: 530
Size:  6.4 KB

You can now try calculating the area of this. Please reply if this doesn't work.

Smithenator5000.
Of course! i understand now. Thanks i wouldn't have realised that. Are you a physics teacher or something?
0
reply
Smithenator5000
Badges: 9
Rep:
?
#12
Report 3 years ago
#12
(Original post by poonshark)
Of course! i understand now. Thanks i wouldn't have realised that. Are you a physics teacher or something?
You are very welcome. It is always worth checking for things like that. Personally, I think they should have indicated the transposed scale with a scribble on the y-axis like so:

Name:  beat.gif
Views: 496
Size:  3.0 KB
source: http://www2.ccsd.ws/sbfaculty/team8e...g%20Points.htm

I am actually an upper sixth student at the moment, so I have just completed my AS in physics and am now studying for full A-levels in physics, maths and further maths. Where are you at?
0
reply
poonshark
Badges: 2
Rep:
?
#13
Report Thread starter 3 years ago
#13
(Original post by Smithenator5000)
You are very welcome. It is always worth checking for things like that. Personally, I think they should have indicated the transposed scale with a scribble on the y-axis like so:

Name:  beat.gif
Views: 496
Size:  3.0 KB
source: [/b]

I am actually an upper sixth student at the moment, so I have just completed my AS in physics and am now studying for full A-levels in physics, maths and further maths. Where are you at?
i have just started AS level. i was assigned loads physics homework on isaac physics which is due in on monday. I am studying maths ,physics, economics , and chemistry for AS.

Also how would you go about calculating the area under these two non-linear graphs? I tried to estimate them but i failed

Name:  Picture1.png
Views: 515
Size:  44.4 KB
0
reply
Smithenator5000
Badges: 9
Rep:
?
#14
Report 3 years ago
#14
(Original post by poonshark)
i have just started AS level. i was assigned loads physics homework on isaac physics which is due in on monday. I am studying maths ,physics, economics , and chemistry for AS.
Cool. Are you enjoying it so far? I suppose that the physics homework is supposed to be a prerequisite task. Otherwise I find it strange that your teacher assigned it to you, given that you claim that they haven't taught you some of it.
0
reply
poonshark
Badges: 2
Rep:
?
#15
Report Thread starter 3 years ago
#15
(Original post by Smithenator5000)
Cool. Are you enjoying it so far? I suppose that the physics homework is supposed to be a prerequisite task. Otherwise I find it strange that your teacher assigned it to you, given that you claim that they haven't taught you some of it.
Yeah we have't been given any lessons yet at all, but we need to have these assignements completed by the first lesson on monday. also how would you go about calculating the area under these non linear graphs? i tried to estimate them but i failed


Name:  Picture1.png
Views: 544
Size:  44.4 KB
0
reply
Smithenator5000
Badges: 9
Rep:
?
#16
Report 3 years ago
#16
(Original post by poonshark)
Yeah we have't been given any lessons yet at all, but we need to have these assignements completed by the first lesson on monday. also how would you go about calculating the area under these non linear graphs? i tried to estimate them but i failed


Name:  Picture1.png
Views: 544
Size:  44.4 KB
It depends on what information you have about them. The first one could be estimated by calculating the combined area of a trapezium (up to  3 km ) and a rectangle (from  3 km and onward). For the second and third one, exact areas can be calculated by integration (though I expect that you won't have learned about that yet). To do this however, you need to know the exact equations of the lines and besides, A-level physics does not expect this knowledge. The traditional way to estimate the areas is by splitting the area up into trapeziums and summing their areas. For each of these graphs, you should be cautious, as the some of the units have prefixes. If you have already tried this, could you post your calculations?
0
reply
poonshark
Badges: 2
Rep:
?
#17
Report Thread starter 3 years ago
#17
(Original post by Smithenator5000)
It depends on what information you have about them. The first one could be estimated by calculating the combined area of a trapezium (up to  3 km ) and a rectangle (from  3 km and onward). For the second and third one, exact areas can be calculated by integration (though I expect that you won't have learned about that yet). To do this however, you need to know the exact equations of the lines and besides, A-level physics does not expect this knowledge. The traditional way to estimate the areas is by splitting the area up into trapeziums and summing their areas. For each of these graphs, you should be cautious, as the some of the units have prefixes. If you have already tried this, could you post your calculations?
so because some have prefixes, would you have to convert the values back into standard units before calculating e.g convert uS into seconds ?
0
reply
Smithenator5000
Badges: 9
Rep:
?
#18
Report 3 years ago
#18
(Original post by poonshark)
so because some have prefixes, would you have to convert the values back into standard units before calculating e.g convert uS into seconds ?
Yes, I would advise converting the values to standard units before using them in your calculations.
0
reply
poonshark
Badges: 2
Rep:
?
#19
Report Thread starter 3 years ago
#19
(Original post by poonshark)
so because some have prefixes, would you have to convert the values back into standard units before calculating e.g convert uS into seconds ?
here is the graphs again in a more complete view , no extra infomation has been given

Name:  Picture2.png
Views: 452
Size:  71.6 KB
0
reply
Smithenator5000
Badges: 9
Rep:
?
#20
Report 3 years ago
#20
(Original post by poonshark)
here is the graphs again in a more complete view , no extra infomation has been given

Name:  Picture2.png
Views: 452
Size:  71.6 KB
This is okay. You should now split the areas into trapeziums like this:
Name:  trapeziums.PNG
Views: 554
Size:  108.9 KB
You can then calculate the area of each trapezium and add them together to get the total area. You can make the estimate more accurate by increasing the number of trapeziums.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • The University of Law
    Solicitor Series: Assessing Trainee Skills – LPC, GDL and MA Law - Exeter campus Postgraduate
    Thu, 27 Feb '20
  • University of East Anglia
    PGCE Open day Postgraduate
    Sat, 29 Feb '20
  • Edinburgh Napier University
    Postgraduate Drop-in Brunch Postgraduate
    Sat, 29 Feb '20

Do you get study leave?

Yes- I like it (289)
60.33%
Yes- I don't like it (23)
4.8%
No- I want it (132)
27.56%
No- I don't want it (35)
7.31%

Watched Threads

View All