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For the method you used what you've done is take the distance between y=x² and y=x³ and rotated that around the axis. What that doesn't take into account is the distance the strips are from the axis of rotation: Imagine a strip one unit high from (1,0) to (1,1). When...

If you click 'quote' on my post you should be able to see how. Otherwise type: & ----then---> int ---then---> ; without any spaces. Check out this post by Xayma: http://www.boredofstudies.org/community/showthread.php?t=7127&page=3

For the second question remember that if 0 < x < 1 then x³ < x² hence you need to subtract the volume of the rotation of y=x³ from y=x² Assume that all the following integrals are between 0 and 1: V = &pi;&int; x⁴ dx - &pi;&int...

For the first one, remember the general formula for internal division of an interval: When dividing the points (x₁,y₁) and (x₂,y₂) into the ration m:n the coordinate is: (mx₂ +nx₁)/(m+n) , (my₂...

Physics rocks my socks. N.B. HSC 'physics' is not physics.

I don't know what happened, but somehow I ended up doing 4U math, phys and chemistry. In the end they were relatively interesting and pretty easy to get the marks in. I almost took french and modern history and music was one of my subjects until the start of year 12 and it wasn't until the first...

It's basically one of those questions where you have to cheat. When I write this, just assume that all integral signs and integrated brackets are between 0 and 1. I_n = &int; x(1 - x³)ⁿ dx use int. by parts where &int; u'v = uv - &int; uv' , where u' = x...

EDIT: bah, I dunno. I'm too tired to make choices.

4. &int; cos&theta;/(2 - cos&theta; ) d&theta; let t = tanθ/2 ---> dθ = 2dt/(1+t²) cosθ = (1-t²)/(1 + t²) Integral becomes = &int; [(1-t²)/(1 + t²)]/[2 - (1-t²)/(1 + t²)] x 2/(1+t²) dt collect 2 -...

Have a go at some of the others and remember you can constuct a right angled triangle with sides (1) and (t) and hypotenuse &radic;(1 + t²) to work out values of sin(x/2) and cos(x/2) ----> hence you can work out cosx, sinx and tanx since: cosx = cos²(x/2) -...

2. &int; dx/(1-sinx) let t = tanx/2 ---> dx = 2dt/(1+t²) sinx = 2t/(1 + t²) int. becomes &int; 1/[1 - 2t/(1 + t²)].2dt/(1+t²) = &int; 1/[(1 - 2t + t²)/(1+t²].2dt/(1+t²) = &int; 2dt/(1 - 2t + t²) =...

1. &int; d&theta;/cos&theta; let t = tan&theta;/2 ---> d&theta; = 2dt/(1+t²) cos&theta; = (1-t²)/(1 + t²) int. = &int; (1+t²)/(1-t²). 2/(1+t²) dt = &int; 2/(1+t)(1-t) dt = &int; 1/(1-t) + 1/(1+t) dt = -ln(1-t) + ln(1+t)...

Both Geostationary and Geosynchronous orbits have a period of 24 hours however a Geostationary satellite must have an equatorial in order to remain 'stationary' above the Earth's surface, otherwise its orbit would be Geosynchronous (where it still has a period of 24 hours but it is not...

Shafqat's method works pretty decent but if you need to use t-results: t = tan(x/2) dt/dx = 1/2.sec²(x/2) dx = 2dt/[1 + tan²(x/2)] dx = 2dt/(1 + t²) cosx = (1-t²)/(1+t²) sinx = 2t/(1+t²) hence &int; (1 +cosx)/(1...

Just because it's question two doesn't mean it should be easy :p. Don't stress about it. The question I disagreed with most on my chemistry exam was question 1... and it was multiple choice!

I'm not sure how you got those values but remember that: Gravitational potential energy = -GMm/r where G = 6.67 x 10^-11, M= 5.98 x 10^24, m= mass of the object and r= distance from the earth. GPE at distance of 7.8x10^6 m = -7.6705 x 10^9 J (my bad, I think I said 10^8 above) GPE at...

I have a sneaking suspicion that the satellite wont reach the earth's surface at a speed of 10^8 times c :p. A speed close to that of light would be unlikely in itself, but what you've got is something else altogether ;).

1) F_G = Gmm/r^2, I'll assume that G = 6.67 x 10^-11and that m_e = 5.98 x 10^24 kg ---> F_G = 983.4 N 2) F_C= F_G a_c = a_g = 983.4/150 = 6.556 ms^-2 3) a_c = v^2r = 6.556 ms^-2 ----> v = sqrt(6.556 ms^-2 x7.8x10^6 m) = 7,151 ms^-1 4) E_p = -Gmmr = -7.6705 x 10^8 J = -7670.5 MJ...

Bah, it's not enabled in the physics forum yet. Sub is for subtext. I'll quickly neaten up but I gotta go have dinner.

1) FG = Gmems/r2, I'll assume that G = 6.67 x 10-11 and that me = 5.98 x 1024 kg ---] FG = 983.4 N 2) FC = FG ac = ag = 983.4/150 = 6.556 ms-2 3) ac = v2/r = 6.556 ms-2 ----] v = √(6.556 ms-2x7.8x106 m) = 7,151 ms-1 4) Ep = -Gmems/r = -7.6705 x 10[sup]8[sup] J = -7670.5 MJ 5)...