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

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???

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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 5.5mA 5.5 mA represents 5.5×103A 5.5 \times 10^{-3} A ; the correct way to read mA mA is 'milli-amperes'. If you instead meant to type 5.5MA 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 V = IR to calculate the potential difference.

V=IR V = IR

I=5.5×103A I = 5.5 \times 10^{-3} A

R=2.20×105Ω R = 2.20 \times 10^{5} \Omega

V=(5.5×103)(2.20×105)=1210V=1200V \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.
Reply 2
Original post by Smithenator5000
Hello there,

Firstly, if you have transcribed the problem correctly, then 5.5mA 5.5 mA represents 5.5×103A 5.5 \times 10^{-3} A ; the correct way to read mA mA is 'milli-amperes'. If you instead meant to type 5.5MA 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 V = IR to calculate the potential difference.

V=IR V = IR

I=5.5×103A I = 5.5 \times 10^{-3} A

R=2.20×105Ω R = 2.20 \times 10^{5} \Omega

V=(5.5×103)(2.20×105)=1210V=1200V \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
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. :smile:
Reply 4
Original post by Smithenator5000
You're very welcome All the best with the assignment. :smile:


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-

Picture1.png

and these questions -

Picture1.png
(edited 7 years ago)
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-

Picture1.png

and these questions -

Picture1.png


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:

si base.PNG
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 N is a unit of force. One equation involving force is F=ma F = ma , a form of Newton's Second Law. This equation implies that
Reply 6
Original post by poonshark
Thanks again! I'll give this a go.


You're welcome. Do not hesitate to ask for help if required.
Reply 8
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. Picture1.png
(edited 7 years ago)
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. Picture1.png


Hello there,

Notice that the y-axis (showing current in amperes) starts from 6A 6 A , as opposed to 0 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.

current graph.PNG

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

Smithenator5000.
Reply 10
Original post by Smithenator5000
Hello there,

Notice that the y-axis (showing current in amperes) starts from 6A 6 A , as opposed to 0 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.

current graph.PNG

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? :biggrin:
Original post by poonshark
Of course! i understand now. Thanks i wouldn't have realised that. Are you a physics teacher or something? :biggrin:


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:

beat.gif
source: http://www2.ccsd.ws/sbfaculty/team8e/jecole/Math/Graphing%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?
Reply 12
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:

beat.gif
source:


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 :biggrin:

Picture1.png
(edited 7 years ago)
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.
Reply 14
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 :biggrin:


Picture1.png
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 :biggrin:


Picture1.png


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 3km 3 km ) and a rectangle (from 3km 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?
(edited 7 years ago)
Reply 16
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 3km 3 km ) and a rectangle (from 3km 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 ?
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.
Reply 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 ?
here is the graphs again in a more complete view , no extra infomation has been given

Picture2.png
Original post by poonshark
here is the graphs again in a more complete view , no extra infomation has been given

Picture2.png


This is okay. You should now split the areas into trapeziums like this:
trapeziums.PNG
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.

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