I HAVE VERY HIGH EXPECTATIONS FROM YOU TOMM (PHYS P6)

[EmanAkhuZeina]

EVERYBODY I HAVE VERY HIGH EXPECTATIONS TOMMOROW FROM YOU....... I HOPE A ************************************************ FOR ALL
THE EXAM GOD WILLING WILL BE EASY : FOCUS ON ELECTRICITY...LIGHT (REFR,REFL)...MOMENTS....GRAPHS AND PRECAUTIONS
AND I HOPE U ALL DO WELL ...I HAVE VERY HIGH EXPECTATIONS FROM YOU.

BEST REGARDS AND IM 24/7 AVAILABLE
YOURS FAITHFULLY
MRS. SALMA EMAN IJBID

AND PLS RELAX WE WANT CALM NERVES!!!!!!
I WILL ALWAYS BE HERE FOR ANY QUESTIONS/HELP/GUIDANCE/ADVICE
BTW I AM AN ACADEMIC QUALITY CONTROLLER IN THE MATTHEW ACADEMY

GOOD LUCK :beer:

 

[Divalicious]

Can you remind me of the precautions with the lens experiment? The one with an illuminated object, a lens and a screen all in a row. And what're the properties of the image formed? Is it always at the lens focus? Is it inverted or enlarged? I always keep forgetting those things.
Also, what're precautions to avoid overheating of a wire? Can I say reducing current, or cooling with a fan or something?
Thanks in advance, would be greatly appreciated ^^

 

[unplugged]

Can you remind me of the precautions with the lens experiment? The one with an illuminated object, a lens and a screen all in a row. And what're the properties of the image formed? Is it at always to lens focus? Is it inverted or enlarged? I always keep forgetting those things.
Also, what're precautions to avoid overheating of a wire? Can I say reducing current, or cooling with a fan or something?
Thanks in advance, would be greatly appreciated ^^

Make sure the lens and the object are always the same height. That the lens , object and screen are perpendicular to the surface. To use a darkened room. To avoid parallex by taking measurements at 90 degrees.

Image formed by a lens is always inverted.

And you lower current to avoid overheating.

 

[haochen]

@Divalicious
precautions of lens exp.:
1) the object, lens and screen should be at the same height above the bench.
2) the center line of the object, lens and screen should be perpendicular on the common axis
3) the object, lens and screen should be on the same row.

properties of image formed:
it is inverted and it depend on the position of the lens for the size.

precaution of overheating the wire:
i dont understand it ?? is it for long distance
then use a step up transformer to increase the voltage so the current will be low so the heat loss will be less therefor the wire wont burn or overheat.

hope u got them

 

[EmanAkhuZeina]

Ok, with lens 1. ALL must be collinear in arranged with centers aligned in the same position
2.all must be stood vertically
image is always real and inverted doesnt mmatter if enlarged

precautions are: Never keep circuit on when not in use
use high resistance variable resistor to prevent high over currents at the beginning
anything else

REMEMBER I HAVE VERY HIGH EXPECTATIONS FROM YOU

 

[EmanAkhuZeina]

U ARE ALL FROM A VERY GOOD SOCIETY

 

[haochen]

and thx to unplugged
i remembered

when checking anything for its perpendicular position always use a set square to prove it

 

[haochen]

U ARE ALL FROM A VERY GOOD SOCIETY

thx

 

[unplugged]

Ok, with lens 1. ALL must be collinear in arranged with centers aligned in the same position
2.all must be stood vertically
image is always real and inverted doesnt mmatter if enlarged

precautions are: Never keep circuit on when not in use
use high resistance variable resistor to prevent high over currents at the beginning
anything else

REMEMBER I HAVE VERY HIGH EXPECTATIONS FROM YOU

erm , if the distance between the object and the lens is less than the focal length , then the image is virtual and magnified ;p.

 

[haochen]

Ok, with lens 1. ALL must be collinear in arranged with centers aligned in the same position
2.all must be stood vertically
image is always real and inverted doesnt mmatter if enlarged

precautions are: Never keep circuit on when not in use
use high resistance variable resistor to prevent high over currents at the beginning
anything else

REMEMBER I HAVE VERY HIGH EXPECTATIONS FROM YOU

erm , if the distance between the object and the lens is less than the focal length , then the image is virtual and magnified ;p.

when the image is form on the mirror it is real and laterally inverted.
but in a lens it is not real and inverted (upside down)

hope its correct

 

[Divalicious]

Thanks alot unplugged, EmanAkhuZeina and haochen ^.^
I remembered all the precautions know

 

[unplugged]

when the image is form on the mirror it is real and laterally inverted.
but in a lens it is not real and inverted (upside down)

hope its correct

An image formed on a mirror is virtual and laterally inverted, because the lines never converge and the image appear to form behind the mirror.
From a lens it's vertically inverted and real unless the object is closer to the lens than the focal lens , then the lines won't converge and the image is virtual.
At 2f the image appears at the proper size , more than that it's smaller , less than that it's larger.
That's how the book says it .

 

[haochen]

ahhh
i just saw the same
(book is beside me )

 

[haochen]

when the image is form on the mirror it is real and laterally inverted.
but in a lens it is not real and inverted (upside down)

hope its correct

An image formed on a mirror is virtual and laterally inverted, because the lines never converge and the image appear to form behind the mirror.
From a lens it's vertically inverted and real unless the object is closer to the lens than the focal lens , then the lines won't converge and the image is virtual.
At 2f the image appears at the proper size , more than that it's smaller , less than that it's larger.
That's how the book says it .

thx

 

[misternervous]

can any one help me with november 2003 question 1 d plz..

 

[Hassanmahmud]

HEY PLEASE COULD YOU TELL ME WHY THE SCALE IS NOT MATCHING WITH MARKING SCHEME..........

 

[Kokeeto]

Hi all! Can you help me in this:
http://www.xtremepapers.me/CIE/Cambridge%20IGCSE/0625%20-%20Physics/0625_w05_qp_6.pdf
Question 4 - f.
Why should we put Y and Z on the reflected ray, and not on the incident ray?

And if there was a point or two that don't fit into a straight line grapgh, shall I join the other points and keep these outside, or make a curve?

Thanks alot
Good luck

 

[DowntownCrew]

HEY PLEASE COULD YOU TELL ME WHY THE SCALE IS NOT MATCHING WITH MARKING SCHEME..........

Some people says it is because of photocopying , on the original paper they will match for sure.Don't worry about that , happens to me when practicing pastpapers , just ignore it

 

[unplugged]

can any one help me with november 2003 question 1 d plz..

I don't know how to calculate this without using the ruler to measure the diamter of the half sphere at the bottom then using the formula to calculate the volume of a sphere then dividing it by half to get the area of that part. Then subtracting the radius from the length of the cylinder used earlier and recalculating the volume then subtracting the new volume and the volume of the sphere from the old volume to get the shaded area. I think it's a bit too much work.

They might have wanted an approximation by getting the volume of a rectangle at the bottom and then multiplying it by 3/4 or so.

 

[unplugged]

Hi all! Can you help me in this:
http://www.xtremepapers.me/CIE/Cambridge%20IGCSE/0625%20-%20Physics/0625_w05_qp_6.pdf
Question 4 - f.
Why should we put Y and Z on the reflected ray, and not on the incident ray?

And if there was a point or two that don't fit into a straight line grapgh, shall I join the other points and keep these outside, or make a curve?

Thanks alot
Good luck

The pins will be used to plot the reflected ray. That's the reason for the drawing. To figure out where the reflected ray will be projected to. SO you basically draw a reflected line with the same angle to the normal of the mirror and then put the pins on top of that line and make them more than 3 CM apart.

You draw the best fit line unless he asks you to draw a curve. So if there are 2 points that don't fit on the line , you just draw the line that is closest to them.