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Physics: Post your doubts here!

Unicorn

Unicorn

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confused123 said:
its always mutual benefit. grow up dude. yeah the teacher didn't knew the correct section in the forum to write that..
Click to expand...

umm honey he said it is great to see physics students studying hard for their subject -_-

anyways lets just forget this
 
confused123

confused123

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Unicorn said:
umm honey he said it is great to see physics students studying hard for their subject -_-

anyways lets just forget this
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yeah but if you see other forums there are sections for teachers to advertise and anyone can offer teaching. ok lets 4get it. going off topic

r u sure that electric field and potential energy question is in the syllabus? can you check syllabus.
 
Unicorn

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confused123 said:
yeah but if you see other forums there are sections for teachers to advertise and anyone can offer teaching. ok lets 4get it. going off topic

r u sure that electric field and potential energy question is in the syllabus? can you check syllabus.
Click to expand...

just did and yes it is except they don't have this weird equation
 
DragonCub

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Unicorn said:
What is the relationship between potential energy and electric field force? the equation is f= dU/dx where U is potential energy but what is x??
Click to expand...
x = distance from the centre.
Work done is equal to force × distance, this is the formula. :p
In fact, when force changes with distance (a function of distance), the work should be calculated by integrating force × dx. On a force-against-distance diagram, the area under the line/curve is work done.
Usually, force is constant no matter how distance changes, the graph is then a horizontal line. That's why the formula W = Fx comes from. :D
 
DragonCub

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NokiaN95638 said:
Guys help me with this question:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_2.pdf

Q no. 1b
Click to expand...
The diagram can be drawn using the Parallelogram Rule.
capture1.png
To find the resultant magnitude, there are two ways:
- you have a rule don't you? Use it to measure the relative length of the forces. Force magnitude ratio equals diagram length ratio. Use the ratio to calculate. This is the "unprofessional" way but it works. :p
- use purely mathematics to calculate the force. Resolve the two forces and you get for the 6N, x-component is 6 cos40, y-component 6 sin40. The 8N, x-component 8, y-component 0. So the resultant x-component = 8 + 6 cos40, y-component 6 sin40.
Resultant magnitude, using Pythagorean Theorem, is sqrt [(8 + 6 cos40)^2 + (6 sin40)^2].
Answer should be 13.2N. :)
 
Unicorn

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DragonCub said:
x = distance from the centre.
Work done is equal to force × distance, this is the formula. :p
In fact, when force changes with distance (a function of distance), the work should be calculated by integrating force × dx. On a force-against-distance diagram, the area under the line/curve is work done.
Usually, force is constant no matter how distance changes, the graph is then a horizontal line. That's why the formula W = Fx comes from. :D
Click to expand...

i didn't really get that. I know that W= F x s

but the equation said dU/dx where is U is potential energy where is x in potential energy to be differentiated??
 
DragonCub

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Unicorn said:
i didn't really get that. I know that W= F x s

but the equation said dU/dx where is U is potential energy where is x in potential energy to be differentiated??
Click to expand...
It is not x to be differentiated but Ep (potential energy) to be, with respect to x.
Think about the relationship between velocity and displacement. Velocity is defined as the rate (rapidness) of change of displacement, or dx/dt. Here the rate of change is in terms of time. So it's dx/dt
Likewise, force is an indicator of how rapidly work done responds to changes in distance. The rapidness is in terms of distance, so it's dW by dx.
 
DragonCub

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Soulgamer said:
Help in question number Question.5 (C) (i) (ii)

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_21.pdf


M.S
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_ms_21.pdf

Please explain in a bit detail and not just an answer which is the exact replica of the marking scheme.
Please and Thankyou! :)
Click to expand...
(1)Electric field lines point from positive side towards negative side, so P experiences a rightward force while N a leftward force.
capture.png
(2)Torque is a kind of moment. It is the moment of a couple.
Here the forces on two charges form a perfect couple.
The formula for torque (T): T = F d [d = vertical distance between the forces]
Use charge and electric field strength to calculate F:
F = 1.6E^-19 × 5.0E^4 = 8.0E^-15 N
Use the length PN and sine of the angle to calculate d:
d = 2.8E^-10 × sin30 =1.4E^-10 m
Last, calculate the torque:
T = 8.0E^-15 × 1.4 E^-10 = 1.12E^-24 N m
 
Unicorn

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DragonCub said:
It is not x to be differentiated but Ep (potential energy) to be, with respect to x.
Think about the relationship between velocity and displacement. Velocity is defined as the rate (rapidness) of change of displacement, or dx/dt. Here the rate of change is in terms of time. So it's dx/dt
Likewise, force is an indicator of how rapidly work done responds to changes in distance. The rapidness is in terms of distance, so it's dW by dx.
Click to expand...

Ohhhh Thanks a lot :D ♥
 
S

Soulgamer

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DragonCub said:
(1)Electric field lines point from positive side towards negative side, so P experiences a rightward force while N a leftward force.
View attachment 10316
(2)Torque is a kind of moment. It is the moment of a couple.
Here the forces on two charges form a perfect couple.
The formula for torque (T): T = F d [d = vertical distance between the forces]
Use charge and electric field strength to calculate F:
F = 1.6E^-19 × 5.0E^4 = 8.0E^-15 N
Use the length PN and sine of the angle to calculate d:
d = 2.8E^-10 × sin30 =1.4E^-10 m
Last, calculate the torque:
T = 8.0E^-15 × 1.4 E^-10 = 1.12E^-24 N m
Click to expand...

Thank you so much! :)
 
A.ELWY 7

A.ELWY 7

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yes i found that this F= -(dU/dx) is in the syllabus but i didnt understand it...what is the d and x
 
N

NokiaN95638

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Idolfanatic95 said:
In Q3 b (iv) why do we use 4.1/sin30 to calculate distance?
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_22.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_ms_22.pdf
In Q6 c (i) how do we calculate the phase difference? and how come it's 0 rad. ?
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_23.pdf
Click to expand...

S11 22;
We know that frictional force always acts along the plane. Never at an angle to it. Therefore the frictional force acts along the line AB. Since we know
Work done= Force x distance. The distance AB is found by using 4.1/sin30. Hope it helps

S11 23:
Since the wave is a stationary wave all particles between 2 nodes are in phase. Since X abd Y are in phase the phase difference is 0 (as they are between 2 nodes)

Hope it helps you
cheers
 

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