Physics help

[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]

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

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]

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

[emi]

thanks Twilight .. and astar  <3

[twilight]

welcumzz

[emi]

another Q ..
how do we calculate energy stored ?

here is a Question :

A bungee jumper of mass 60kg jumps from a bridge tied to an elastic rope which becomes taut after he falls 10m. Consider the jumper when he has fallen another 10m and is traveling at 15m/s

a) work out how much energy us stored in the rope, take g=10 m/s2 and ignore air resistance .


plz some one help me out

[twilight]

i think its this way.

if u consider the mechanical energies( gravitational potential energy, elastic potential energy, kinetic energy)
total mechanical energy at the moment he jumps = total mechanical energy when he has fallen 20 m
             KE    +    EPE   +     GPE                      =        KE              +    EPE       +     GPE
             0     +      0    +     mxgxh                    =      1/2 mv²          +    EPE       +      0
                                      60 x 10 x 20              =    0.5 x 60 x 15²    +   EPE
                                           12000                  =        6750              +   EPE

therefore     EPE( i.e. energy stored in the rope)   =        12000 - 6750         =         5250 J

[emi]

Thanks i didn't think of that

[twilight]

welcumzz .. r sure this is IGCSE ??
bcoz i studied this in mechanics A level

[emi]

i have this IGCSE guide for physics and after each topic there are Questions , and these are from them .

[twilight]

aha .. gud 
because i cudnt remember anything like that in Physics IGCSE ...
By the way i got A* in Physics and Maths

[astarmathsandphysics]

another Q ..
how do we calculate energy stored ?

here is a Question :

A bungee jumper of mass 60kg jumps from a bridge tied to an elastic rope which becomes taut after he falls 10m. Consider the jumper when he has fallen another 10m and is traveling at 15m/s

a) work out how much energy us stored in the rope, take g=10 m/s2 and ignore air resistance .


plz some one help me out

Conservation of Energy
Energy to start=0
so 0=mgh+EPE+1/2mv²
so 0=60*10*-20+EPE+1/2*60*15² so EPE=12000-6750=5250J

[emi]

thanks 

and twilight thats awesome, i wish i can be like you  =]

[nid404]

Guys I'm stuck with this magnetic deflection thingy og radioactive particles. Can anyone just help me understand what is meant by -the magnetic field acts into the plane of the paper. Moreover this makes it even harder to understand the deflection. Please tell me how I know if the deflection is into the paper or out of the paper. I'm screwed. I need help.

Just for your information
-The mark scheme for june 2006, paper 3, last question says that alpha particles move into the paper and beta out of the paper. Please refer to this question if possible.
-The textbook (IGCSE physics by Tom Duncan)says the deflection can be found from the left hand rule, taking negative charge moving to the right as equivalent to positive( conventional) current to the left.

I just don't understand it
 
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