Physics HL/SL Help

[DropBoite]

Hello everyone, could you please help me with a question?

It's about a thundercloud and the question is: The potential difference between the thundercloud and the ground before discharge is 2.5 × 108 V. Determine the energy released in the discharge.

Some of you might be familiar with it if you've solved PPs, what has me puzzled is that the Markscheme will only give me 3 Marks if i Don't calculate the average voltage, why on earth? There wasn't anything implying that I should do that and it was TOTALLY out of the blue! IBO Nincompoops!

Second question I have is, the image attached shows a picture from a school test, I was supposed to fill up the 'Type of Field' Column and I got it wrong An explanation for your choice would be ideal.

I hope you can help, thank you!

Edited April 18, 2012 by DropBoite

[Grassroot]

Hello everyone, could you please help me with a question?

It's about a thundercloud and the question is: The potential difference between the thundercloud and the ground before discharge is 2.5 × 108 V. Determine the energy released in the discharge.

Some of you might be familiar with it if you've solved PPs, what has me puzzled is that the Markscheme will only give me 3 Marks if i Don't calculate the average voltage, why on earth? There wasn't anything implying that I should do that and it was TOTALLY out of the blue! IBO Nincompoops!

Second question I have is, the image attached shows a picture from a school test, I was supposed to fill up the 'Type of Field' Column and I got it wrong An explanation for your choice would be ideal.

I hope you can help, thank you!

Based on what u have presented here it is insufficient to solve the problem

W=Vq apparently the amount of charge here is unknown....

I think this should come from a Paper 2 long question i.e. Section B I assume

Could you please paste the whole question?

[HiggsHunter]

Hello everyone, could you please help me with a question?

It's about a thundercloud and the question is: The potential difference between the thundercloud and the ground before discharge is 2.5 × 108 V. Determine the energy released in the discharge.

There must be some information missing from the question! In addition to the potential difference between the thundercloud and ground, you would need to know its electrical charge to determine the energy released in the discharge.

Second question I have is, the image attached shows a picture from a school test, I was supposed to fill up the 'Type of Field' Column and I got it wrong An explanation for your choice would be ideal.

I hope you can help, thank you!

Only a gravitational field will affect the uncharged particles A and E. Of the positively charge particles C and D initially moving normal to the direction of the field, the force would be normal to the direction of a magnetic field, but in the direction of an electric field. In the case of the negatively charged particle B, the force would be opposite to the direction of an electric field.

[DropBoite]

Hello everyone, could you please help me with a question?

It's about a thundercloud and the question is: The potential difference between the thundercloud and the ground before discharge is 2.5 × 108 V. Determine the energy released in the discharge.

Some of you might be familiar with it if you've solved PPs, what has me puzzled is that the Markscheme will only give me 3 Marks if i Don't calculate the average voltage, why on earth? There wasn't anything implying that I should do that and it was TOTALLY out of the blue! IBO Nincompoops!

Second question I have is, the image attached shows a picture from a school test, I was supposed to fill up the 'Type of Field' Column and I got it wrong An explanation for your choice would be ideal.

I hope you can help, thank you!

Based on what u have presented here it is insufficient to solve the problem

W=Vq apparently the amount of charge here is unknown....

I think this should come from a Paper 2 long question i.e. Section B I assume

Could you please paste the whole question?

I know, but even if I procured the paper and took the time to post the entire thing, it would be taken down and copyrighted material isn't allowed on the forum, I hope you understand my predicament, I want to but I can't. So if you could be so nice as to refer to paper 2 section from the year November 2010 SL, it would be cool. That's only applicable to the cloud question btw.

In the image attached is a vector diagram with the 'connecting'/derived vector being the hypotenuse. you will notice that as the vectors Pini and Pfinal are facing away (not nose to tail) this is vector subtraction.

I'm confused as I drew the 'connecting' vector the otherway around (nose at the y axis) as compared to the one in the markscheme, will this make a logically equivalent vector or not?

Edited April 18, 2012 by DropBoite

[HiggsHunter]

I know, but even if I procured the paper and took the time to post the entire thing, it would be taken down and copyrighted material isn't allowed on the forum, I hope you understand my predicament, I want to but I can't. So if you could be so nice as to refer to paper 2 section from the year November 2010 SL, it would be cool. That's only applicable to the cloud question btw.

Unfortunately I only have HL past Physics papers, not SL ones. Searching on the web, I found the paper you cited in Spanish (not my language!), but not the markscheme. If I have understood the Spanish correctly, the charge is given as 35 Coulombs, in which case the energy released in the discharge would be 0.5 x 35 x 2.5 x 10^8 Joules = 4.375 Gigajoules. Is that not the answer given in the markscheme?

[HiggsHunter]

W=Vq apparently the amount of charge here is unknown....

Common sense might lead you to expect that when the thundercloud is charged to voltage V with charge q it has energy Vq Joules. But in fact the energy stored when the cloud/earth capacitance C charges is 1/C times the integral from 0 to q of q.dq, which is (1/C) x (qˆ2)/2, or only Vq/2 Joules.

[HiggsHunter]

In the image attached is a vector diagram with the 'connecting'/derived vector being the hypotenuse. you will notice that as the vectors Pini and Pfinal are facing away (not nose to tail) this is vector subtraction.

I'm confused as I drew the 'connecting' vector the otherway around (nose at the y axis) as compared to the one in the markscheme, will this make a logically equivalent vector or not?

It depends on the question. If the vector subtraction is Pfinal minus Pini, the connecting vector is shown in the correct direction in your diagram. If the nose were at the y axis, the vector would be Pini minus Pfinal.

[Award Winning Boss]

I'm being really stupid here...

A small boat in still water is given an initial horizontal push to get it moving. The boat gradually slows down. Which of the following statements is true for the forces acting on the boat as it slows down?

A. There is a forward force that diminishes with time.

B. There is a backward force that diminishes with time.

C. There is a forward force and a backward force both of which diminish with time.

D. There is a forward force and a backward force that are always equal and opposite.

Why is it B?

Is it because the forward force isn't constant, which means that the backwards force had to equal the forward force at the start. ( I think it's newton's 3rd law) so as it slows down the backwards force (air resistance) decreases to 0. Is that the reason why?

[HiggsHunter]

The boat slows down as its initial momentum is transferred to the water by the drag. Since all the components of drag diminish as the boat slows, B is true.

[Drake Glau]

You could also think of it as the momentum from the boat forcing the water to speed up make it's FORWARD force increase so it's backward force must decrease

[DropBoite]

I know, but even if I procured the paper and took the time to post the entire thing, it would be taken down and copyrighted material isn't allowed on the forum, I hope you understand my predicament, I want to but I can't. So if you could be so nice as to refer to paper 2 section from the year November 2010 SL, it would be cool. That's only applicable to the cloud question btw.

Unfortunately I only have HL past Physics papers, not SL ones. Searching on the web, I found the paper you cited in Spanish (not my language!), but not the markscheme. If I have understood the Spanish correctly, the charge is given as 35 Coulombs, in which case the energy released in the discharge would be 0.5 x 35 x 2.5 x 10^8 Joules = 4.375 Gigajoules. Is that not the answer given in the markscheme?

That's the answer, but I don't get why you multiplied by 0.5, that's exactly where I got stumped I got 8.something Gigajoules. And the MS says 0.5 is for average Voltage, but what's you explanation?

Thanks for the vectors help!

Attached is a graph of Current against Voltage, the paper asks me to determine the emf of the battery, the battery is made of two cells FYI. From the graph a shot in the dark told me intersection must mean something, thus I answered 4V which is partially correct, but apparently " realization that the voltage must be 4.0 V across each resistor;

and so emf is 8.0 V;" is the correct answer.

My question is what does the intersection really translate to/ imply for the conductros X and Y?- How can an intersection symbolize they both have that voltage across them? I don't get it!

Additionally the times 2 thing, I'd assume at 4V intersection = the sum of their voltage, but I guess I was wrong. what is wrong in my reasoning there?

Question 2(Also based on the graph): calculate the total power dissipated in the circuit. I actually got this right, but I don't agree with my reasoning for using I=0.8A, could anyone tell me why that is the current we work with for dissipation?

Thank you for the prompt replies so far Higgs Hunter!

Edited April 19, 2012 by DropBoite

[HiggsHunter]

That's the answer, but I don't get why you multiplied by 0.5, that's exactly where I got stumped I got 8.something Gigajoules. And the MS says 0.5 is for average Voltage, but what's you explanation?

Yes, I can understand your surprise, since the result is a little counterintuitive. As I mentioned in my reply to Grassroot, you might expect that when the thundercloud is charged to voltage V with charge Q that it has energy QV Joules. But in fact the energy stored when the cloud earth capacitance C charges is 1/C times the integral from 0 to Q of Q.dQ, which is (1/C) x (Qˆ2)/2, or only 0.5 x QV Joules.

Thanks for the vectors help!

Thank you for the prompt replies so far Higgs Hunter!

I'm very happy to help when I can. But the administrators prefer that if you find a post useful you just click the green button beside it, rather than post a written acknowledgement!

[HiggsHunter]

Attached is a graph of Current against Voltage, the paper asks me to determine the emf of the battery, the battery is made of two cells FYI. From the graph a shot in the dark told me intersection must mean something, thus I answered 4V which is partially correct, but apparently " realization that the voltage must be 4.0 V across each resistor;

and so emf is 8.0 V;" is the correct answer.

To understand the question correctly, would it be possible for you to post the circuit diagram, or to let me know where I can find it on the web?

Thanks!

[DropBoite]

Haha! I already repped you, but! Here is the circuit diagram.

[DeBrogliez]

Haha! I already repped you, but! Here is the circuit diagram.

Yea the diagram circuit is important here.

Because if they were connected in parallel there will be a different story.

Assuming that they are connected in series; then the current throughout the circuit must be the same.

Also, it is given in the question that the power dissipated in each of the two resistors is the same.

Since Power = I*V, or Power = I^2*R. And now we know that both power & current are the same. Then resistance and voltage must be the same as well (according to the equation).

That's why you take the intersection of the graph, because you wanna know that "same voltage" occurs at what current?

Which happens to be 0.8A in this case

Then, things are easier.

ϵ = I(R1+R2) = 0.8*(5+5) = 8V

Power = I^2*(R1+R2) = 0.8^2*(5+5) = 6.4W

Anyways, hope I helped

[HiggsHunter]

Also, it is given in the question that the power dissipated in each of the two resistors is the same.

Thanks for the additional information about the question, DeBrogliez. Perhaps there is a lesson here? To be able to solve the problems it's really important to take note of all the information given in the questions!

[HiggsHunter]

Haha! I already repped you, but! Here is the circuit diagram.

Thanks, DropBoite. I didn't know that, because when I click on Reputation I get the message "Your member group is not allowed to view received reputation".

[DropBoite]

Hello! i have a question, the class test I gave has a question that goes " The magnitude of the electrostatic force between the proton and electron in a hydrogen atom is FE. The magnitude of the gravitational force between them is FG."

Answer the teacher gave me to the Fe part: identifies Fe or (1.6 x 10^-19)²/ epsilon₀ x 4 pi r²

I'm really confused by this as I'm an SL student and this seems a BIT MUCH to ask stupid IBO, Also this atleast at seemingly bears no semblance to any formula we've been taught!

[DropBoite]

Haha! I already repped you, but! Here is the circuit diagram.

Yea the diagram circuit is important here.

Because if they were connected in parallel there will be a different story.

Assuming that they are connected in series; then the current throughout the circuit must be the same.

Also, it is given in the question that the power dissipated in each of the two resistors is the same.

Since Power = I*V, or Power = I^2*R. And now we know that both power & current are the same. Then resistance and voltage must be the same as well (according to the equation).

That's why you take the intersection of the graph, because you wanna know that "same voltage" occurs at what current?

Which happens to be 0.8A in this case

Then, things are easier.

ϵ = I(R1+R2) = 0.8*(5+5) = 8V

Power = I^2*(R1+R2) = 0.8^2*(5+5) = 6.4W

Anyways, hope I helped

You lost me at Power and Current are the same, care to elucidate?

[HiggsHunter]

You lost me at Power and Current are the same, care to elucidate?

Since R1 and R2 are connected in series, the same current I flows through each resistor. Also the question apparently states that the power dissipated in each resistor is the same, which means that V1=V2 since power W=VI.

According to the graph provided, the only current at which V1=V2 is 0.8A, when V1=V2=4v and the battery is supplying V1+V2=8v. (Note that this is only equal to the emf of the battery if it has negligible internal resistance).

At 8v and 0.8A, the total power dissipated in the resistors is 8 x 0.8 = 6.4W.