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no problem!Thank you so much..
I know the integration part, I was stuck in making ysq the subject.. :/
and Jazak Allah..
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no problem!Thank you so much..
I know the integration part, I was stuck in making ysq the subject.. :/
and Jazak Allah..
Farru lol.. u seriously wanna test my math huh? ok.. gimme a while to gather my thoughts
I'm sorry Farru i gotta go study chem.. my maths exam is all the way on the 13th of June..i'm very sorry i can't help u
the link from the rule "Final Energy = Initial Energy + Driving force - Resistance force" ? or the one "Driving force - Resisting force = mass x acceleration"?
so you get the final and initial energy using "K.E. [1/2 mass x velocity(squared) ] + P.E. (mass x gravity [10] x height)" and substituting the mass, height and velocity as given at each point.
if given a driving force or resistant force you put them in the equation "F.E. = I.E. + DF - RF"
integrate ∫1/2(cos 2x - cos 4x) dx
all i know is if you have 3 different forces on an object you can use lamiBut I thought we use Lami's rule when we have coplanar forces.
And how can you find angle CP2X first since we have 7.3/sin90 = 5.5/sinAP1X = W/sinCP2X
you did sinCP2X= (5.5xsin90)/7.3 right ? But this should be for angle AP1X. because 5.5 is its opposite force.
I'd be grateful if you showed me your workings possibly with a diagram. Thanks a lot
I'm sorry Farru i gotta go study chem.. my maths exam is all the way on the 13th of June..i'm very sorry i can't help u
Sorry again...ugh.. i feel s bad... *sigh*....
integrate ∫1/2(cos 2x - cos 4x) dx
1/2∫cos 2x - cos 4x dx
= 1/2[-sin2x/2 + sin4x/4]
= -1/4 sin 2x + 1/8 sin 4x
now sub the values of x and u will get the answer given
Thanks! Jazaki Allahu Khairits okay buddy..
my math paper on 20th may.. :/ Best of luck btw.
well in 6 i) you use the F.E. = I.E. as there is no driving force (he didn't mention one) nor resisting force (as it is a smooth plane)What if it's just a particle(with no driving force) going upwards or downwards a rough slope? Can you explain the conditions for both motions. For example sometimes while going down a rough slope the speed increases, but in some cases it also decrease and comes to rest, like in w_12 qp 41, question 6
i did this like a few pages back :
i did this like a few pages back :
so here is what i did (the correct one):
1- i got the equation for Sp "1/2 x a x (t+2)(squared)"
2- started to get rid of the brackets "1/2 x 1.75 x (t(squared) + 4t +4)
= 0.875(t(squared) + 4t + 4)
= 0.875t(squared) + 3.5t + 3.5
3- got an expression for Sq "1/2 at(squared)"
4- subtracted them from each other to get the distance in-between to be equal to 4.9
0.875t(squared) +3.5t +3.5 - 1/2 x 1.75 x t(squared)
= the "0.875t(squared) is crossed with the "1/2 x 1.75 x t(squared) leaving the equation to be "4.9=3.5t + 3.5"
3.5t=4.9-3.5
t=1.4/3.5 = 0.4
yeah that i have no idea why i did that same mistake the first time i did it .. i was actually waiting to see if you would tell me :/thanks .... but why do we minus distance of P from that of Q ? why not the other way round?
yeah that i have no idea why i did that same mistake the first time i did it .. i was actually waiting to see if you would tell me :/
because P is further away and has covered more distance than q . it it were the other way around then we would get a negative distance which is impossible ( I am talking about distance not displacement )thanks .... but why do we minus distance of P from that of Q ? why not the other way round?
thanks .... but why do we minus distance of P from that of Q ? why not the other way round?
because P is further away and has covered more distance than q . it it were the other way around then we would get a negative distance which is impossible ( I am talking about distance not displacement )
BTW i did it another way
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