Lower Secondary Science
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Can anyone recommend Physics home tutor who can teach Upper Sec Pure Physics, please?
PM me.
TQ -
Vivian22:
That was a very clear explanation. Thank you! :goodpost:
I'll try to answer this question and minimise calculations as much as I can. (Not easy to type in Math in forums).dioprem:
[quote=\"Vivian22\"]Hi,
I have two Physics question to ask. Can someone please help? Thanks in advance!
![](\"http://i41.tinypic.com/9ier77.png\")
http://i41.tinypic.com/9ier77.png\">
The bulbs are rated at 6V, 12W. In other words, it would require a potential difference of 6V across each bulb for the bulb to work normally. And when it does so, it dissipates 12W of power. With that information, we know that the bulb has a resistance of 3 ohms. (P=V^2/R). With this information, lets look at each option.
A) If the current though L1 is 2A, it would imply that the current through L2 is 2A as well. This would mean that a current of 4A is passing through L3. Using V=RI, it would imply that the potential difference across L3 is 12V, which is not possible since the e.m.f of the battery is only 6V. So this cannot be the answer.
B) If the potential difference across L2 is 3V, then using V=RI would show that the current through L2 is 1A. Similarly the current through L1 would also be 1A since it is in parallel. This would mean that the current through L3 is 2A. Using V=RI again, this would mean that the potential difference across L3 is 6V, which cannot be the case as the e.m.f of the battery is 6V, which would mean that the potential difference across L1 and L2 is 0V, which is not possile.
C) The effective resistance of the whole circuit is 4.5 ohms. Since the e.m.f of the battery is 6V, the current drawn from the battery is 1.33A. Hence by using P=IV, the power drawn from the battery is 8W. So this answer is also worng. (Also by doing this from the beginning, you would know from circuit analysis the the current drawn from the battery is the same as the current through L3. You could have used this method to eliminate A and B as the answers as well.
D)Using P=(I^2)R, the power dissipated by L3 is 5.33W. Similarly, the current drawn by L1 = current drawn by L2 = 1.33/2. Hence the power dissipated by L1 and L2 is less.
Hope this helps.[/quote]No worries.
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Can someone help me with this question please?
![](\"http://i39.tinypic.com/33w135t.png\")
Thanks! -
Vivian22:
The answer is c.Can someone help me with this question please?
Thanks!
(1) For a wire in a magnetic field,
If you Start with a current and want to know the direction of the force, you use Flemming's LEFT hand rule.
If you start with a force and want to know the direction of the current, you use Flemming's RIGHT hand rule.
The question says a wire is moved, so you are starting with a force. Applying Flemming's Right hand rule to this situation shows that X is a South Pole. So right away c is the only choice.
(2) You will not get a current for the portion of the wire that you push parallel to the direction of the magnetic field. But the magnetic field curves back , so portions of the wire upward and downward in the vertical are not exactly perpendicular. However, due to the symmetry of the problem, it looks like you induce currents in the opposite direction for the above case and the underneath case, so it looks like these cancel out and you are left with no current.
(3) would be more complete if it included the phrase \"at the same velocity\". If you move the magnet at the same velocity with which you move the wire, then you get the same value of current. -
Dr.Daniel:
:thankyou: I'm still quite unclear about the explanation for (2), could you please explain it again perhaps in a simpler way? :oops:
The answer is c.Vivian22:
Can someone help me with this question please?
Thanks!
(1) For a wire in a magnetic field,
If you Start with a current and want to know the direction of the force, you use Flemming's LEFT hand rule.
If you start with a force and want to know the direction of the current, you use Flemming's RIGHT hand rule.
The question says a wire is moved, so you are starting with a force. Applying Flemming's Right hand rule to this situation shows that X is a South Pole. So right away c is the only choice.
(2) You will not get a current for the portion of the wire that you push parallel to the direction of the magnetic field. But the magnetic field curves back , so portions of the wire upward and downward in the vertical are not exactly perpendicular. However, due to the symmetry of the problem, it looks like you induce currents in the opposite direction for the above case and the underneath case, so it looks like these cancel out and you are left with no current.
(3) would be more complete if it included the phrase \"at the same velocity\". If you move the magnet at the same velocity with which you move the wire, then you get the same value of current.
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Concerning statement (2) only,
For the portion of the wire moved in the same direction as the magnetic field there will be no induced current. Only portions of the wire that have a component of the magnetic field perpendicular to the motion of the wire will have induced current. I’ll try to describe 3 dimensions. Imagine a 3-D system of x being horizontal, y being vertical and z being in and out of the page. So the portion of the wire at y=0 is being pushed parallel to the magnetic field, along the z axis, meaning right along the magnetic field lines. That has no induced current. If you can see that much, then this is probably enough to get to the right answer in the problem.
If you start to think that points on the wire above and below the magnet many not be exactly parallel to the magnetic field, then what I am saying is whatever induction occurs below the plane of the magnet is opposite in direction to whatever induction occurs above the plane of the magnet. -
Dr.Daniel:
Oh I get it now. Thank you!Concerning statement (2) only,
For the portion of the wire moved in the same direction as the magnetic field there will be no induced current. Only portions of the wire that have a component of the magnetic field perpendicular to the motion of the wire will have induced current. I'll try to describe 3 dimensions. Imagine a 3-D system of x being horizontal, y being vertical and z being in and out of the page. So the portion of the wire at y=0 is being pushed parallel to the magnetic field, along the z axis, meaning right along the magnetic field lines. That has no induced current. If you can see that much, then this is probably enough to get to the right answer in the problem.
If you start to think that points on the wire above and below the magnet many not be exactly parallel to the magnetic field, then what I am saying is whatever induction occurs below the plane of the magnet is opposite in direction to whatever induction occurs above the plane of the magnet. -
Hi,
Can someone now help me with this Chemistry question? How do we derive at the answer? Thanks in advance!![](\"http://i44.tinypic.com/35n4py8.png\")
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Vivian22:
The winner is B. Nitrogen needs three more electrons to form a complete outer shell and I see that symbol Q has three lines coming from it. Each one of those lines represents a bond or shared electron. Symbol R has 4 lines connected to it, meaning it has four bonds. Elements that do that are in the Carbon, Silicon group (vertical column) , so R could be Silicon. And symbol V has a single bond, so it could be Hydrogen, which needs one more electron to complete the valence shell.Hi,
Can someone now help me with this Chemistry question? How do we derive at the answer? Thanks in advance!
Choice A is no good because there is no way Q is Fl. Fl just needs one bond to get to the noble gas (full valence shell) configuration. Also there is no way V is Si because it is only forming one bond. Si will form 4.
Choice C is no good because Q is not Oxygen. Oxygen will have two bonds for two shared electrons.
And D is no good because Q is not Si for reasons stated above.
I just received a neat chemistry model kit and sometimes have students put together some molecules in class. Its great because each element has a different color and there are a certain number of holes drilled in wooden spheres so that you know how many bonds each element makes. You connect the wooden spheres with little wooden rods. Also the holes are drilled at the correct bond angles so that you can see the proper shape of each molecule. -
Dr.Daniel:
Thank you! That was a great explanation :goodpost:
The winner is B. Nitrogen needs three more electrons to form a complete outer shell and I see that symbol Q has three lines coming from it. Each one of those lines represents a bond or shared electron. Symbol R has 4 lines connected to it, meaning it has four bonds. Elements that do that are in the Carbon, Silicon group (vertical column) , so R could be Silicon. And symbol V has a single bond, so it could be Hydrogen, which needs one more electron to complete the valence shell.Vivian22:
Hi,
Can someone now help me with this Chemistry question? How do we derive at the answer? Thanks in advance!
Choice A is no good because there is no way Q is Fl. Fl just needs one bond to get to the noble gas (full valence shell) configuration. Also there is no way V is Si because it is only forming one bond. Si will form 4.
Choice C is no good because Q is not Oxygen. Oxygen will have two bonds for two shared electrons.
And D is no good because Q is not Si for reasons stated above.
I just received a neat chemistry model kit and sometimes have students put together some molecules in class. Its great because each element has a different color and there are a certain number of holes drilled in wooden spheres so that you know how many bonds each element makes. You connect the wooden spheres with little wooden rods. Also the holes are drilled at the correct bond angles so that you can see the proper shape of each molecule.
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