Mathematics: Post your doubts here!

[ZainH]

Ohlord, I can't think atm
Can someone solve these 2 simultaneous equations for me.

5.5sinx = wgcosx
And
5.5cosx + wgsinx = 7.3

 

[D0cEngi]

Plz help me with question 7 of this ppr. Plz Plz Plz.
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_41.pdf

 

[mominzahid]

It's given you the maximum velocity, at max velocity acceleration is 0.

So, since we have the equation for velocity , we need to change it to acceleration by differentiating it.
This gives us , 12ot-3t^2
We take this as 0 and find the value of t

120t-3t^2 = 0
3t(40t-4) = 0
t= 40

Now that we have a value of t , replace it in the original equation for velocity and take velocity as 6.4 since thats it's max.
k(60t^2-t^3) = 6.4
k(32000) = 6.4
k= 6.4/32000
k= 0.0002

Hope you get it x)


You're welcome.

Got Itt...
Thanks ..
Can u please explain q3? same paper..

 

[Pie-man]

Well maybe you can get it if you imagine it happening in your head...

WOW....that sounded deep

 

[haha101]

Yeah

 

[Pie-man]

Plz help me with question 7 of this ppr. Plz Plz Plz.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_41.pdf

here is the whole idea (or how i did it) just draw a vertical line at point B to get a letter Z, and by using trigonometry get the angles so you can make a resolution, then Tension c and Tension a will be equal to each other (with the sine and cosine of course) and 8 would be equal both added together.

then get the tensions by substitution between both equations.

 

[haha101]

here is the whole idea (or how i did it) just draw a vertical line at point B to get a letter Z, and by using trigonometry get the angles so you can make a resolution, then Tension c and Tension a will be equal to each other (with the sine and cosine of course) and 8 would be equal both added together.

then get the tensions by substitution between both equations.

man i cant understand a thing Please any diagram or something

 

[GorgeousEyes]

Please juun012 v.42 no.5 with steps pleaase .

 

[ZainH]

Got Itt...
Thanks ..
Can u please explain q3? same paper..

Oh that's easy just make 2 simultaneous equations.
Use s=ut+1/2at^2

So your first equation is: 55=5u+12.5a
And second one is: 120=10u+50a

Solve them simultaneously, you should get a=0.4 and u=10 m.s-1

 

[Esme]

Ohlord, I can't think atm
Can someone solve these 2 simultaneous equations for me.

5.5sinx = wgcosx
And
5.5cosx + wgsinx = 7.3

sinx/cosx =wg/5.5
tanx * 5.5 = wg

Substitute this in the second equation
5.5cosx + sinx( tanx * 5.5) = 7.3
after simplifying the brackets, you'll get
5.5cos^2 x + 5.5 sin^2 x = 7.3 cosx
5.5( cos^2 x + sin^2 x) =7.3 cosx
5.5/7.3 = cosx
x= 41.4

now wg= tan41.4 * 5.5

 

[Esme]

WOW....that sounded deep

I think i hear sarcasm

 

[GorgeousEyes]

Ohlord, I can't think atm
Can someone solve these 2 simultaneous equations for me.

5.5sinx = wgcosx
And
5.5cosx + wgsinx = 7.3

I think i work this out but can u tell me the final answer ?

 

[Esme]

man i cant understand a thing Please any diagram or something

This question is hard to explain...

 

[Esme]

I think i work this out but can u tell me the final answer ?

x=41.4
w=4.8

 

[ZainH]

Please juun012 v.42 no.5 with steps pleaase .

First part is easy, since you know tension is a force use F= mass x acceleration

For S1, only weight hanging is 3kg so: F= 3x10 = 30N
For S2, both weights are hanging , so: F=5x10 = 50N

Part 2 is a bit tricky, but easy to understand.

When the masses fall , they'll have forces acting on them in different directions. For mass A it's weight will be the only thing acting downwards. Tension in S1and air resistance will be acting against it, so we can make an equation with this information.

Fnet= 30-T-1.6=3a

When mass B falls, it's weight and tension in string S1 will be acting downwards, where as air resistance will be acting upwards. It's equation is:
Fnet= 20+T-4=2a

Now solve them simultaneously,

The T's will cancel out.

30-1.6=3a
20-4=2a

44.4=5a
a= 8.88 m.s-2

Now just put your value for a back into one of the equations to get T.
20+T-4 = 2(8.88)
T = 1.76N

Hope you got it x)

 

[haha101]

This question is hard to explain...

atleast tryy !

 

[GorgeousEyes]

x=41.4
w=4.8

yeaah , I did it till 55/73 .. well done

 

[Pie-man]

I think i hear sarcasm

what? sarcasm? ME? NEVER....

 

[Pie-man]

atleast tryy !

ok here is what i mean. thats how i'd solve it

 

[GorgeousEyes]

First part is easy, since you know tension is a force use F= mass x acceleration

For S1, only weight hanging is 3kg so: F= 3x10 = 30N
For S2, both weights are hanging , so: F=5x10 = 50N

Part 2 is a bit tricky, but easy to understand.

When the masses fall , they'll have forces acting on them in different directions. For mass A it's weight will be the only thing acting downwards. Tension in S1and air resistance will be acting against it, so we can make an equation with this information.

Fnet= 30-T-1.6=3a

When mass B falls, it's weight and tension in string S1 will be acting downwards, where as air resistance will be acting upwards. It's equation is:
Fnet= 20+T-4=2a

Now solve them simultaneously,

The T's will cancel out.

30-1.6=3a
20-4=2a

44.4=5a
a= 8.88 m.s-2

Now just put your value for a back into one of the equations to get T.
20+T-4 = 2(8.88)
T = 1.76N

Hope you got it x)

Youu are great mash'allah thank youuu