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Physics: Post your doubts here!

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If we take the coordinates from the top of the line and calculate gradient after ensuring units consistentency : 105 km/h into m/s by 105*1000/3600 -----》 grad = [ans/4] and we get D =7.3 rounded off to 2 s.f.
 
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In finding absolute error, do we divide least count of instrument by two? Or do we take the least count as absolute error?
E.g. In metre rule will absolute error be 1mm or 0.5mm for a single reading?
 
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The sig fig in raw t would be 4. So we should take sig fig in T^2 as 4 or 5..? Don't we usually take calculated values to 3 sig figs?
Any help would be appreciated.
 

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In finding absolute error, do we divide least count of instrument by two? Or do we take the least count as absolute error?
E.g. In metre rule will absolute error be 1mm or 0.5mm for a single reading?
This was confusing for me too. Absolute error or absolute incertainty is the uncertainty in a measurement, which is expressed using the relevant units, 0.5mm is the estimated error ( least count/2) so you should take 1mm.
 
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The cross-section of an Olympic-size swimming pool filled with water. It is not
drawn to scale. The density of the water is 1000kg m–3... Can someone explain how the answer is 3.4 KPa and not 3.5kPa. Q20 winter 2015 p12?
Beacuse probably you are taking g as 10 instead of 9.81
 
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You're supposed to know 14 by thoery,
That a COUPLE consists of (i)Same magnitude forces acting in opposite directions (ii) Therefore producing rotational motion
opps i meant Q 15 I knew this one it quite obvious. Thanku fr helping me :) Can u try 9,37,15?
 
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opps i meant Q 15 I knew this one it quite obvious. Thanku fr helping me :) Can u try 9,37,15?
For Q9, use the equation of motion s= ut + 1/2 at^2
We are only considering vertical velocity at the moment so: s=8, u=0, a=9.81 and you have to find t. If you put these in the formula you'll get the time as 1.27s, so C is the answer.

Fpr Q15, the 4N force has no effect on the moment about the pivot since its acting in line with it.
Total anticlockwise moment= (3N x 2m) + (2sin(30) x 4 ) = 10
Total clockwise moment = 5m x F = 10
Thus, F=2N so the answer is B

Q37 is kida hard to explain, you basically have to use V=IR ultiple times. For the last two resistors the voltage across each is 1V. So the voltage across the resistor connected in parallel is also 2V. The current through this resistor must be 2A (I= V/R). Thus the current across the 2nd last horizontal resistor adds up to 3A and so the p.d across it is 3V. The last two resistors we talked about are connected in parallel to the first vertical resistor and so the p.d across it is 5V and the current flowing through it is 5A. This mean that the current flowing through he first resistor is 8A and the p.d across it is 8V. Excluuding the very first resistor, the circuit is a big parallel setup with a voltage of 5V across it. So the e.m.f must be 8V +5V = 13V so the ans is D
 
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Hi there,
I need help with this mcq.
Correct ans is D
view

Attached is the image of question 36 from 9702_W16_11

Link: https://drive.google.com/file/d/0B4LUJjAvVTcWMWVuRnRJMHBGckk/view?usp=drivesdk
 
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does anyone know where to get the physics may/june 2017 paper 41. i know it's not yet uploaded but i was hoping if anyone of u knew how to get the paper
 
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