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Biology; Chemistry; Physics: Post your doubts here!

Suchal Riaz

Suchal Riaz

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Saad Mughal said:
Can anyone help me with part (f) and (g)?
P.S. The picture quality is poor, uploaded in a hurry.
If you can't read them then just see the diagram,
The parts are,
(f) The man now walks past A towards the left-hand of the plank. What is the upward force from the trestle at B at the instant the plank starts to tip?
(g) How far is the man from A as the plank tips?
I need explanations for both parts please.
Click to expand...

(f)
i m not fully sure but what i think is that when plank has not yet tipped, the force on b will be sum of weight divided by 2 as for one object to be in equilibrum the resultant force and moment must be the same. as plank will tip the moment have anticlockwise resultant from A as pivot but forcec still must be same as plank is still resting on B as well as A.
(g)the sum of anticlockwise is bigger than clockwise moment. it means 120*2=240 Nm anticlockwise and 480*x>240Nm it means x>0.5m is limit after which plank will tip
 
Saad Mughal

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Suchal Riaz said:
(f)
i m not fully sure but what i think is that when plank has not yet tipped, the force on b will be sum of weight divided by 2 as for one object to be in equilibrum the resultant force and moment must be the same. as plank will tip the moment have anticlockwise resultant from A as pivot but forcec still must be same as plank is still resting on B as well as A.
(g)the sum of anticlockwise is bigger than clockwise moment. it means 120*2=240 Nm anticlockwise and 480*x>240Nm it means x>0.5m is limit after which plank will tip
Click to expand...
For, part (f) and (g) I had precisely the same answers but the problem is that for (f), the answer is 0 (no force acting upwards) and for (g) the answer is 0.5 m not >0.5 m or equivalent. These are weird questions (stephen pople's book), I'll ask my teacher. Thanks Anyway! :)
 
Suchal Riaz

Suchal Riaz

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Saad Mughal said:
For, part (f) and (g) I had precisely the same answers but the problem is that for (f), the answer is 0 (no force acting upwards) and for (g) the answer is 0.5 m not >0.5 m or equivalent. These are weird questions (stephen pople's book), I'll ask my teacher. Thanks Anyway! :)
Click to expand...

i was thinking that maybe answer is zero but i was reluctant to say it. it means that as plank is just about to tip, it puts no force on B therefore according to 3rd law there is not force by B upwards. it depends on at what time u see. if u see when it is still resting it will be i guess half of both forces but as it is going to tip, there will be no force applied by plank as it going to move away from it not towards it(against gravity so no weight acting on B)
 
Suchal Riaz

Suchal Riaz

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Saad Mughal said:
I read it off the internet and from the Stephen Pople book.
I'll explain it to you in another way.
Bistable circuit consists of two NOR gates (simple bistable circuit, the circuit diagram is irrelevant) that are cross-coupled. It may work in a fancy manner but its task is ultimately to keep one STABLE (constant) output (0 for eg.) for the two inputs (1,1 for e.g). If we change one of the inputs (for example if the inputs are Q and R and we change R) then the output will change to the other STABLE option (i.e. 1 in this case). If we then change Q then the output will revert back to 0. Hence, R is used to set the output and Q is used to reset it. It exhibits the signs of memory since eventually it reverts back to its original output hence it remembers its original output (i.e. 0 in this case). It is used in computer circuits for storing binary digits.

Astable circuits have no stable output, it consists of multiple resistors, capacitors and NOT gates (the circuit diagram is irrelevant) that continuously go back and forth (0 then 1 then 0 then 1 then 0). It is hence known as a multivibrator. If we increase the resistance and/or capacitance then the frequency of the back and forth changes decreases (time duration b/w them increases). It is used in ticking clocks and flashing lights.

That's all you need to know according to the syllabus. Hope you get it now. :)
Click to expand...

thnx. hope fancy things dont come in exams
 
Suchal Riaz

Suchal Riaz

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usama321 said:
i think i'll just pass them :D thanks by the way


by the way here, i thought these gates at least required two inputs initially. I wonder how they would work, if they had to start from scratch. i mean their second input is both from the output of the other one.
Click to expand...
circles are buzzing around my head from the time i look to them, i dont know where it starts...o_O
 
Saad Mughal

Saad Mughal

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Suchal Riaz said:
i was thinking that maybe answer is zero but i was reluctant to say it. it means that as plank is just about to tip, it puts no force on B therefore according to 3rd law there is not force by B upwards. it depends on at what time u see. if u see when it is still resting it will be i guess half of both forces but as it is going to tip, there will be no force applied by plank as it going to move away from it not towards it(against gravity so no weight acting on B)
Click to expand...
Thanks. I got it. The weight is at the center and not B hence the upward force = 0.
 
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nidz

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Saad Mughal said:
NH4NO3 (heated) ----> 2H2O (g) + N2O (g)
They've told us that the ammonium nitrate decomposes to steam and gas X so just write down the equation,
NH4NO3 (heated) ----> 2H2O (g) + ____
Deduce the gas by comparing on both sides.
Hope that helps :)
Click to expand...
Wow thanks a lot! :D
 
Saad Mughal

Saad Mughal

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nidz said:
Can someone ans this question? It's from june 2012 p22
Copper(II) chloride can be prepared by the reaction between copper(II) carbonate and
hydrochloric acid.
(i) Construct the ionic equation for this reaction.

Since copper carbonate is insoluble it should not be written in ionic form, right?
But according to the examiner, the ionic equation should be CO3+ 2H ---> CO2 + H2O......
Click to expand...
Well, I'm not quite sure of this, but from what I know, the hydrogen being more reactive will replace the copper in the solution (that will lead to copper (II) chloride being formed), and hydrogen carbonate is soluble so it will break down into ions CO3- + H+ ----> CO2 + H2O (balanced eq. is CO3 + 2H+ ---> CO2 + H2O).
The H+ from the Acid reacts with the Carbonate ion to form carbon dioxide and water and a salt (Copper II chloride) since these are the general products from the Acid + Metal Carbonate reaction.
 
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