Recent content by NTUENG

Hmm, i tink trigo. substitution may work. U can let u^4 = \cos 2\theta , du = \frac{1}{4}(\cos 2\theta)^{-\frac{3}{4}} (-2 \sin 2\theta). Since \cos 2\theta = 2 \cos^2 \theta - 1 , the bottom of the integral simplifies to 1 + 2 \cos^2 \theta - 1 = 2 \cos^2 \theta . But the top can be quite...

Hi christwwt, Electrons with parallel spin would experience greater repulsion than those with opposite spin. This is also the reason why electrons occupying the atomic subshells(1s,2s,2p,3s etc) are always paired in opposite spins.

Static equilibrium of rigid bodies Hi prettyinpink, For this question, u may need to use the equations \sum F_x = 0 , \sum F_y = 0 and \sum \tau = 0 . Let the tension in the support cable be T and the normal reaction force at P be R. Then, using \sum F_y = 0 , u'll get T sin 32 + R sin...

Hi giant016, I had the same problem as u when i started learning "Rotational motion of rigid bodies about a fixed axis". I also pondered for a long time over the 'moment of inertia' thing. I looked up my textbook and various sources to find out more about it. I did not want to start my...

Hi mashjavad, Looks like integration by parts doesn't solve the integral. It takes the form (e^-x sin ax)/x. I used mathematical software to evaluate and obtained exponential integrals with complex numbers in the result. I tink this involves incomplete gamma functions or numerical methods of...

Dun like partial fractions?? Nevermind... Hi m3cklo, If u dun want to break the integral into partial fractions, u can use a trigo substitution instead. Let y = sin t. Then, dy = cos t dt. Simplify using the substitution. U should get sec t as the integral. Integrate it and u get ln (sec t +...

Hi Robierob, When u compute f(x)-g(x), u should get x² - cos x on which u can apply Newton's method. Actually u can ensure that your guess is not far from the actual root by finding when your function changes from positive to negative or vice versa. There will be a root when the function...

Hi ddr, The substitution should be x= √3 sin t instead. Then, dx = √3 cos t dt. Remember to change the limits for the trigo substitution. Use your trigo identities to evaluate the final integral. Hope that helps.

Hi ddr, Perform polynomial algebraic division and integrate the result directly.

Dear berkeman, Thanks for the reminder as I'm still new to this forum. I'll follow the forum rules the next time i reply to or post a new message. It's been such a wonderful year for me in my local university in Singapore. Nice to meet u. Where are u from?? Regards Lin Ruijin

Hi cokezero, I saw your question while doing some questions on rotational motion. For (b), << complete solution removed by berkeman >> For part (c), << complete solution removed by berkeman >> For part (d), << complete solution removed by berkeman >> Hope that helps...

Hi chenny, This is a rotational motion problem. By Newton's 2nd law, Net force= ma. F- Ff = ma. Since tangential acceleration = radius x angular acceleration, F- Ff = m (0.3m x 0.5 rad/s²). When block starts to slip, Ff = (meu)N = 0.53mg. Manipulate Newton's 2nd law, u get F = ma = m dv/dt...

Hi fsm, U need to know how to interpret graphs and the conservation of mechanical energy to do this. Ok, initially, the total energy of the particle is 5J. At point B, it has lost some P.E. and what's left is 2J. Some of the P.E. has been converted into K.E. which is 3J. U know the mass of...

Hi Mikefitz, It has been a while since i touched my thermodynamics but nevertheless i'll try to help. For part (a), U need to know that 1 atm = 1.01 x 10^5 Pa and 1 litre = 1000 cm³ = 0.001 m³ Work done from A-E (isobaric) = P dV = 2 x 1.01 x 10^5 Pa x (8-4) x 0.001m³ = 808 J Work done...

U used the term 'rolls without slipping'. I thought of using rotational mechanics to solve it but u are not given the radius of the spherical shell. Thus, i shall treat it like a case of linear motion down the slope. Resolving forces and applying Newton's 2nd law on the shell, we have Normal...