Recent content by Needhelp2

Yes! Sorry Q6) part ii (Blush)

So I have my final maths exam tomorrow and thought I'd do a few new past papers to brush up. Here is my problem: Q 5) part ii.... I worked out the tension in the string which was correct at 2.24N, but when I resolved vertically to find out the normal reaction force on C, I came out with...

you haven't mentioned the gain in GPE the skier would have or the work done against the friction? Where would these come in?

Hi! I am finding the work-energy principle and idea of total mechanical energy hard to apply to finding the work done against/by a particular force. For example in the question below, why is there no work done against the weight of the skier acting down the slope? Any help would be great! Thanks :D

Not really...(Sweating) I don't understand where you got the value of -284 from, and it isn't within the degree specification? (at least if it goes from -180 up to 180) Sorry for being so dim, could you possibly take it step by step to get the second value of 74.5 and explain why 134.5 isn't a...

What is a reference angle? Sorry for such a silly question!

[FONT=arial]After a long summer, I finding my new C3 homework a bit tricky, so any help would be great! Here is the question: sec(θ-150 degrees)=4 (solving for theta is greater than or equal to -180, but less than or equal to 180) So I know that sec is the reciprocal of cos so I changed the...

Ahh so my method was right, it was just a stupid arithmetic error somewhere! Thank you so much!

This question is probably impossibly easy, but I have been staring at it for half an hour now, and don't seem to be able to get the right answer :mad:, so any help would be great! A particle P moves on a straight line with constant acceleration. At t=0, P passes through the point 0 on a line...

I think it, once the values for p,r and q are found you can sub them into the equation,so it is now 4cos2x-3=0. cos2x= 3/4, so you get cos-1(3/4)=2x. therefore 2x= + or - 0.723,5.560 and 7.006. Hence x would equal half of each of these values? It seems a lot easier now I have all the values!

Haha yes it was really part 2 that was the problem :) If the line joining the peaks together is y=1 and the one joining the troughs together is y=-7, would that not mean that the axis of oscillation was y=-3? Sorry for all the questions- I am just quite confused! And in response to...

surely the amplitude of the curve is 4? (1+7/2), and perhaps the period is [FONT=arial]π? although I am not too sure what you mean by the line the curve oscillates? thank you so much for the help by the way! according to the markscheme this is the method to work it out- 1=p-r -7=-p-r and...

I was wondering if anyone could help with this- I've never seen a question like this before and don't know exactly how to tackle it...(Sweating) any help would be greatly appreciated!

Thank you so much! It came up in my Mechanics mock, so the fact I couldn't do it worried me slightly... But the simple solution makes me feel a bit better :)

Three horizontal forces of magnitudes 8N, 15N and 20N act at a point. The 8N and 15N are at right angles to each other. The force 20N makes an angle of 150 degrees with the the 8N force and an angle of 120 degrees with the 15N force. State the greatest and least possible magnitudes of the...