P1 CIE physics doubts

[astarmathsandphysics]

22. each spring now support 1.5*2=3 times the weight
30charge=average current*time=(20mA+100mA)*8=480mC
40.2 ups and 1 down cos charge must add to1e

[sweetie]

okay...
if each spring now has extension 3x , so since they both r in parallel, isnt the total suppose to be 3x/2??

and for Q30 why dont we consider the difference between the 2 values??

[astarmathsandphysics]

The weight is doubled so the extension is doubled

[sweetie]

10 approach speed=u1+u2
separation speed=v2-v1
they are equal for elastic collisions  D
11.res for both particles using F=ma
for 2g particle 2g-T=2a (1)
for 6g particle T-6=6a (2)
(1) +(2) 2g-6=8a so a/(2*9.8-6)/8=1.7 B
32)1/R=1/100+1/10+1/10+1/10+1/10+1/10+1/10=61/100
R=100/61 B

But , for Q11 how do we know that the acceleration of  the 2 boxes r the same???

[nid404]

okay...
if each spring now has extension 3x , so since they both r in parallel, isnt the total suppose to be 3x/2??

W=kx
k=W/x
now when one spring is removed, value of k reduced to 2/3 of it's previous...which is 2W/3x
2W=2w/3x x e
e=3x

[astarmathsandphysics]

cos the boxes are always in contact

[sweetie]

W=kx
k=W/x
now when one spring is removed, value of k reduced to 2/3 of it's previous...which is 2W/3x
2W=2w/3x x e
e=3x

am sorry nid, but i didnt get  how u got k as 2W/3x?

[nid404]

am sorry nid, but i didnt get  how u got k as 2W/3x?

See all the together had a k= W/x   right?


when one spring is removed from the 3, 1/3rd of the k is eliminated...right? k reduce by 1/3 ...so remaining k= 2/3 X W/x

got it?

[sweetie]

yup..

Thank You

[nid404]

yup..

Thank You

no problem

[sweetie]

nov07
Q35, 26, 25and 6

[astarmathsandphysics]

26. the electric field points in the direction of the force on a positive charge. The charge on an electron is negative so the force points the other way, to the left.
25. 4th sinx=nw/d<1
If n=3 sinx=0.707 so w/d =0.707/3
nw/d<1 if n=4 but not n=5
6.A deflection +ve when current =0 so A or B
deflection increases more atr start so A

[sweetie]

Sir, can u xplain Q25 and 6 again plz?

[astarmathsandphysics]

as soon as you find w/d=0.707=0.236 use sinx =n*0.236
this has to be less than 1 to give a solution for x
If n=4 sin x =0.944 solutions exist
n=5 sin x =1.18 no solutions

[sweetie]

got it now,
cud u plz xplain Q35 too???