Recent content by Mongster

Ahh I apologise. It's a silly mistake I've made, shouldn't have rushed through things. Thank you all for the valuable inputs!

Apologies for the late response. Yes the answer is now correct. Thank you for the heads up, I never realized that but now it makes sense why--- with regards to their love for the angle 37deg.

Using Kepler's Formula, I tried to solve for the answer but was told that it's incorrect.

Update: Found a couple of mistakes in my second attempt at this Qn 1. My component force for Ma.g for Block A has been mixed up. Basically M.agSin37 & M.agCos37 are on the wrong sides. 2. Forgot to input Frictional Force (Fr) in Eqn1 + Eqn2. Thank you all for the valuable input!

I changed the sign convention for the pulley in accordance to the direction of motion. However I still could not obtain the answer :(

Homework Statement It appears to me that output load voltages in a Bridge Rectifier circuit with capacitor always seem to be greater than that without a capacitor. I'm aware that the function of the capacitors connected across output in rectifiers is to remove pulsation of the DC output, so...

Ahhhhh, took me some time to figure it out. Tension 2a does not equate to 5a. It equates to 49.05 - 5a. That makes much more sense. Thank you for your time jbriggs444, appreciate it greatly :)

But just as I thought I have managed to solve the problem.. Another popped up and I would like to clarify this... If T= ma T of cord horizontally is 2a while T of cord vertically would be 5a... But we know that the tensional force would have to cancel out each other somehow and what I've...

kay... So I've realized that the 2kg block moves horizontally and that the only force that it exerts in the system is the tensional force of the cord. Correcting my previous workings will give me: a= [Ft - Ft + 5(9.81)] / 5+2 = 49. 05/ 7 ≈ 7m/s^2 Which is the correct answer...

Hmm to be honest I'm unsure of what does it mean by external forces or internal. It's just so that in a question earlier which gives that two weights are being suspended by a pulley, I used this approach and it somehow worked and I'm merely trying to reapply it to this question. * perhaps...

Since the pulley will move in the direction of the 5kg block suspended, I let its force be positive, which is 5 x 9.81 Another force will be coming from the 2kg block moving horizontally in an unknown acceleration, which will be 2 x a Since this force opposes the direction of force the system is...

Homework Statement Consider the systems of 2kg and 5kg blocks. The 2kg block is resting upon a smooth frictionless horizontal surface and friction in pulley bearing is negligible. The 5kg block is suspended by a cord connected to the 2kg block. Determine the acceleration of system and tension...

Oh wait... I see the mistake now oh my, hahaha! It is really stupid... *cringing* But thanks a lot for the detailed explanation there, appreciate it really! Cheers!

See I figured that since Velocity = m/s Acceleration = m/s^2 If I have velocity divided by Acceleration ----> m/s ÷ m/s^2 = sRelevant equations Velocity --> s/t Acceleration --> (v-u)/tThe attempt at a solution My idea seems reasonable to me but somehow I...