Okay I think I understand. The direction of something's velocity does not always equal the direction of the forces acting on it. Though, I'm still kinda unsure about connecting this clue with finding the acceleration of the system/tension in the string. I managed to find the value for the...
Okay, am I right in thinking that when P slips over the pulley the string will become slack anyway and hence the tension disappears? Cause going back to the equation for the resultant force acting on Q (##2g-2T=2a##) and plugging in ##T=0## yields an acceleration of g, which is the answer they...
If Q somehow disappeared, due to the string breaking, then the string would no longer exert any tension on P. So P would move downwards under the action of its weight alone?? so with an acceleration of g?? I'm not too sure about what's happening where Q is present.
Homework Statement
The diagram shows a particle P lying in contact with a smooth table top 1.5m above the floor. A light inextensible string of length 1m connects P to another particle Q hanging freely over a small smooth pulley at the edge of the table. The mass of each particle is 2 kg, and P...
Ahh I think I see now. Given that R2 is 100 degrees from the horizontal, and drawing in a vertical line perpendicular to the horizontal at the foot of ladder, its clear that the angle R2 to the vertical will be given by 100-90 .So R2 to the vertical is 10 degrees and that's where the 10 degrees...
Thanks for the response - I think I can see now how the smaller diagrams fit in with the bigger one. The only thing is I don't get how they knew that the angle between R2 and the vertical line drawn is 10 degrees. How?
Homework Statement
A uniform ladder of weight W rests with its top against a rough wall and its foot on rough ground which slopes down from the base of the wall at 10 degrees to the horizontal. Resolve, horizontally and vertically, each of the forces acting on the ladder.
Homework...
Right okay I see my mistake now. It's the ##+2## in the denominator that complicates things right? Say if it were just ##\frac{7}{5sin(\theta-36.9^{\circ})}## then you could simplify that to ##\frac{7}{5}cosec(\theta-36.9^{\circ})##.
Right so going back to the correct form of the function...
Homework Statement
Express $$4sin\theta-3cos\theta$$ in the form $$rsin(\theta-\alpha)$$
Hence find the maximum and minimum values of $$\frac{7}{4sin\theta-3cos\theta+2}$$
State the greatest and least values.
Homework EquationsThe Attempt at a Solution
Okay so putting it in the...
I'm incredibly sorry but I forgot to add that the textbook reckons the answer is a rate of increase of 5.5 per day. So we've got the 6.5 - but isn't this a weekly rate of increase? Sorry I think I'm a bit confused here. Anyhow, if I understand the question correctly, it wants me to find the rate...
Hi - I think I get at what you're saying. The graph isn't a pure exponential so an exponential model would only be approximate and get less accurate as the time increases. However I'm still quite lost with respect to the part c).
c) Find the rate at which the numbers are growing 10 weeks after...
Okay after thinking a bit I reckon I got part b) of the question sorted now.
So breaking down ##y=ab^x## to make it easier to understand:
##y=## number of stick insects
##a=## initial number of stick insects (amount that I start with)
##b=## rate of growth (the ##1.5## growth factor I found...
Homework Statement
These questions require you to see a graph - which can be accessed from the link below
43.
a). Find the growth factor between the numbers of stick insects in consecutive time intervals of two weeks. Hence find a relationship between ##x## (time in days) and ##y## (number of...
Ahh right okay I think I understand. I think I was getting massively confused before. Let me go from near the beginning again. Thanks for bearing with me all this time :)
So we have ##y=(2-p)x^2+6x+1-2p## and using the discriminant, I've found those values of ##p## for which...