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emi:
A see-saw has total length of 4m and is pivoted in the middle. A child of weight 400N sits 1.4m from the pivot . a child of weight 300N sits 1.8m from the pivot on the other side .A parent holds the end of the see-saw on the same side as the lighter child, work out the magnitude and direction of the force the parent must exert to hold the see saw level .

i answered like this ------------>

for see-saw to be in same level the clock wise moment = anti clock wise
so
400x1.4 = 300x1.8
560= 540
so subtract them from each other  560-540  =  20N downwards force .

but at the back of the book answer is 10N downward
so wats my mistake ?

and one Q from past paper
how do we do O/N 08 paper 6  Q2 part a)   & part c) iii  

they say suggest one reason other than change in temp. of wires ,why the results may not support the theory ?
the MS says :
"Resistance at connections
Internal resistance of source/other sensible suggestion"

i dont understand it

yinyang:
560=540

it means that you need 20 more Nm to balance the see-saw, the distance from te pivot is 2(holds it at the end) ,  2*x=20,  x=20/2, x=10N


for Oct/NOv

Q6 ciii> all circuit components and wires have an internal resistance but this resistance is small so we are not expected to find the value of that resistance untill our A-levels, however as an IG student you should know that there is some resistance in the connections as well as the components. 

i have attached a diagram of the cicuit.           

astarmathsandphysics:

--- Quote from: emi on May 05, 2009, 01:57:39 am ---A see-saw has total length of 4m and is pivoted in the middle. A child of weight 400N sits 1.4m from the pivot . a child of weight 300N sits 1.8m from the pivot on the other side .A parent holds the end of the see-saw on the same side as the lighter child, work out the magnitude and direction of the force the parent must exert to hold the see saw level .

i answered like this ------------>

for see-saw to be in same level the clock wise moment = anti clock wise
so
400x1.4 = 300x1.8
560= 540
so subtract them from each other  560-540  =  20N downwards force .

but at the back of the book answer is 10N downward
so wats my mistake ?

and one Q from past paper
how do we do O/N 08 paper 6  Q2 part a)   & part c) iii  

they say suggest one reason other than change in temp. of wires ,why the results may not support the theory ?
the MS says :
"Resistance at connections
Internal resistance of source/other sensible suggestion"

i dont understand it

--- End quote ---

Take moments for the parents force 20=2*F so F=10N

how do we do O/N 08 paper 6  Q2 part a)   & part c) iii 
c)i)The voltages are not equal since thie differences are bigger than 0.01V, the uncertainty in the reading. Voltage 3 is half voltage 1 within the limitations of the experiment.
ii)The battery and wires has internal resistance.

twilight:

--- Quote from: emi on May 05, 2009, 01:57:39 am ---A see-saw has total length of 4m and is pivoted in the middle. A child of weight 400N sits 1.4m from the pivot . a child of weight 300N sits 1.8m from the pivot on the other side .A parent holds the end of the see-saw on the same side as the lighter child, work out the magnitude and direction of the force the parent must exert to hold the see saw level .

i answered like this ------------>

for see-saw to be in same level the clock wise moment = anti clock wise
so
400x1.4 = 300x1.8
560= 540
so subtract them from each other  560-540  =  20N downwards force .

but at the back of the book answer is 10N downward
so wats my mistake ?

and one Q from past paper
how do we do O/N 08 paper 6  Q2 part a)   & part c) iii  

they say suggest one reason other than change in temp. of wires ,why the results may not support the theory ?
the MS says :
"Resistance at connections
Internal resistance of source/other sensible suggestion"

i dont understand it

--- End quote ---

emi:
thanks Twilight .. and astar  <3

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