Recent content by trew

I'm continuing to struggle when it comes to objects that are dropped at different times. At height h, a ball is dropped at rest. Also at height h, another ball is dropped one second later. For the first ball, time=t. For the second ball, time=t+1...but this is wrong? The answers show it...

So when x>3, this will be on the right hand side of the point R making it clockwise, but why do I assume that the boy stands on the right hand side of R?

So I let x be the distance from P. I then considered moments about R and got this: 1) (40g)(x-3) = 40gx - 120g and this is anticlockwise 2) (30g)(0.5) = 15g and this is anticlockwise 3) (2.5g)(2) = 5g and this is clockwise And then by equating clockwise and anticlockwise I get this: 40gx -...

Part (a) was fine. For part (b), I started off with finding the position vector of bird A and bird B after 4 seconds, and this was: Position vector of A after 4 seconds: -4i + 11j Position vector of B after 4 seconds: (-8 + 4p)i + (9+8p)j But after this I'm lost. Usually I'd have some idea...

I thought that since he micrometer is to 2 significant figures (0.35 and 0.01mm) that the final answer should also be to 2 sig.figs, thus answer A. But the final answer is C, 3%. Can someone explain why?

For part (b), I counted the number of squares underneath the 2 day sample to be roughly 33 'whole' squares (the 5by5 tiny squares is 1 whole sqaure). I then equated 33 whole square = 0.35 MJ to calculate 1 whole square to be 0.0106 MJ. I then counted the number of whole squares underneath the...

Thanks for the explanations. I think I'm beginning to see the bigger picture. What would help is I can get my understanding of centre of gravity right. You can see here that the persons centre of gravity is low so he is harder to push over. And every resource (so far) that I've come...

I appreciate your post but this is what I don't understand. What does it mean to have centre of gravity high or low? I get that a 4x4 car has a high centre of gravity so it's easier to topple over, but I understand this intuitively and based on real world experiences and nothing else. But what...

Hey, This isn't a homework question I need help with but more of a concept I can't seem to grasp. I understand that Centre of Gravity is: the point on a body where all the weight can be considered to act. And I understand how to find this point on different types of bodies. However, this...

Also, I like using sin0=opp/hyp and cos0=adj/hyp to work out these forces but since I can't see how that triangle is formed I can't use this method. For me it's the best method since I can learn what's going intuitively as opposed to assuming that cos is associated with the x-direction and sin...

Ok I'm back a day later and I still don't get it o_O I looked at this video: which has pretty much the same question and I still can't visualise the triangle with the 25N force being the hypotenuse. I've tried many combinations but I can't form a triangle with the 25N and it's two components.

Ok so I did that and this is what I got: so I can't see how I would still get 25cos10 for the hypotenuse side? Or have I done this triangle wrong?

Thanks for your help so far but tbh I'm still lost. Here's what I've done so far: So I've resolved the first part into 2gsin10 and 2gcos10 but after looking at the problem for the past hour I still can't see how I can break down the 25N into two components.

I'm okay with the concept of resolving into two components. BUT, with re: to resolving perpendicular to the acceleration, I don't understand how 25cos80 comes into? Surely it should be included when I'm resolving in the direction of the acceleration?

Homework Statement Homework EquationsThe Attempt at a Solution I managed to find dy/dx as follows: But I'm having difficulty finding the second derivative. I've looked at examples using the chain rule but I'm still confused. Would someone mind shedding some light on this for me?