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Physics, Chemistry and Biology: Post your doubts here!

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to increase the sensitivity of a thermometer you have to increase the length of the capillary tube and and to increase the range u have to have a finer capillary tube and a larger bulb right? Anything for linearity? no right?
 
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to increase the sensitivity of a thermometer you have to increase the length of the capillary tube and and to increase the range u have to have a finer capillary tube and a larger bulb right? Anything for linearity? no right?
Sensitivity:
To increase, decrease diameter of capillary tube or use a liquid with higher expansivity.
Range:
Opposite to sensitivity
Increase diameter of capillary tube or use liquid with lower expansivity
Linearity:
If a thermometer isn't linear then all you can do is replace the liquid with a linear liquid
 
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http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Physics (0625)/0625_w10_qp_31.pdf

Anyone can please explain how to answer the first and last questions? please.

question 1: draw a perpendicular lyn, n measure 60 degrees frm da lyn 2 da left. measure 30 degrees frm da lyn 2 da ryt. xtend da lyns. since u hv 2 use da scale given, use a compass 2 measure 5cm, n cut an arc on da ryt lyn frm da point @ which da 2 lyns meet. measure 8.7cm, n cut an arc on da left lyn. frm da left arc, measure 5cm n draw a large arc. frm da ryt arc measure 8.7cm n draw anothr large arc. da 2 arcs shud intersect. join da point of intersection 2 da previous 2 arcs. dis shud form a parallelogram. connect da point @ which da first 2 lyns meet 2 da point @ which da 2 arcs intersect. measure da lyn, n usin da scale, calculate it into newtons. dats da value of da resultant.

Q 11) wen a dc voltage, dere will b a horizontal lyn @ da top of da screen. wen an ac voltage is used, dere will b waves formed. draw @ least 3. wen an ac voltage is used, n if a diode is connected, dere will b half of da waves draw in da previous screen. ie; only da top portions OR only da botttom portions of da waves will b visible.
 
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accelerationg anode only accelerates the electrons ... it has high dc voltage connected across it ! focuscing anode makes the electrons in a beam ! usally they both r intergrated n made into one
thanx!!
 
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6 b)(i) it says it's almost parallel to AE, so the angle of incidence is 90 degrees
6 b)(ii) at critical angle the refracted ray is at 90 degrees while here the incident ray is 90 degrees. So that refracted ray in this case is the critical angle which is 43 degrees.
6 b)(iii) refractive index= sin i/sin r so sin 90/sin 43 = 1.47
 
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it says the ray is ALMOST parallel to AE - that means it has an angle of incidence of ALMOST 90.
Mark scheme says anything between 88-90 is acceptable, so I'd write 89.
The critical angle occurs when the ray going from more optically dense medium to less optically dense medium is refracted with an angle of refraction of 90*. b)i) Is that but opposite - you're looking at the refracted ray of angle 90* as an incident ray of angle 90* - so the critical angle is 43*
refractive index = 1/sinC where C is the critical angle
so n=1/sin43 = 1.466
 
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Sensitivity:
To increase, decrease diameter of capillary tube or use a liquid with higher expansivity.
Range:
Opposite to sensitivity
Increase diameter of capillary tube or use liquid with lower expansivity
Linearity:
If a thermometer isn't linear then all you can do is replace the liquid with a linear liquid
thanks alot :D
 
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6 b)(i) it says it's almost parallel to AE, so the angle of incidence is 90 degrees
6 b)(ii) at critical angle the refracted ray is at 90 degrees while here the incident ray is 90 degrees. So that refracted ray in this case is the critical angle which is 43 degrees.
6 b)(iii) refractive index= sin i/sin r so sin 90/sin 43 = 1.47
THANK U SOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO MUCH
it says the ray is ALMOST parallel to AE - that means it has an angle of incidence of ALMOST 90.
Mark scheme says anything between 88-90 is acceptable, so I'd write 89.
The critical angle occurs when the ray going from more optically dense medium to less optically dense medium is refracted with an angle of refraction of 90*. b)i) Is that but opposite - you're looking at the refracted ray of angle 90* as an incident ray of angle 90* - so the critical angle is 43*
refractive index = 1/sinC where C is the critical angle
so n=1/sin43 = 1.466

THANK U SOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO MUCH
 
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Can someone explain the left-hand and right-hand rules?
left hand rule is used to determine the direction of force if given direction of current and magnetic field. On your left hand raise your thumb(will represent force) ,make your first finger face with the direction of magnetic field, your second middle finger should be perpendicular to the first finger and facing the direction of current, the ndirection where your thumb looks is the direction of force. i found a video on youtube showing you what i just said hope it helps,
 
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