Enigma

So can I find the initial velocity by putting zero into the velocity equation, and then find the time by using the initial velocity in the equation?

Hi, I'm Enigma. I would have introduced myself earlier, but only just found where! Anyway, I'm 16 and currently studing Physics, Further Maths and ICT at A-level. I like playing the piano, composing music, listening to music (well anything to do with music ), I enjoy sports and reading, and the subjects I'm studying. I'm fascinated about space and the world.

Enigma

I've managed to answer 3 parts of the following question, now I'm just stuck on the last part. Please can someone tell me how to go about this. Thanks.

The displacement of a particle at time t seconds after it passes through a fixed point is s m, where s = 4.8t + 0.06t^2 - 0.004t^3

Write down expressions in terms of t for

a) the velocity

v = 4.8 + 0.12t - 0.012t^2

b) the acceleration

a = 0.12 - 0.024t

c) Find the value of s when the acceleration of the particle is zero.

0 = 0.12 - 0.024t

0.12/0.024 = t

t = 5

s = (4.8 x 5) + (0.06 x 5^2) - (0.004 x 5^3) = 25.4

d) Find the distance travelled by the particle from the point where it reaches its maximum velocity to the point where its velocity is half its initial velocity.

I'm thinking that when the particle has reached its maximum velocity, the acceleration will be zero, so the time will be 5 seconds, but I don't know how to finish the question.

Many thanks,

Enigma

Okay, I think I understand. I just hope I don't get a really hard question in the exam. Thank you very much

OKay, I get this:

P cos30 = 0.5R

R = 29.4 - P sin30

P cos30 = 0.5(29.4 - P sin30)

P cos 30 - 14.7 - 0.5P sin30

P cos 30 + 0.5P sin30 = 14.7

P(cos30 + 0.5P sin30) = 14.7

P = 13.2N

But I don't understand how you got the equations out!

After some thought:

R = 29.4 + P sin30

P cos30 = 0.5R

P cos30/0.5 = R

Substitute into top equation

P cos30/0.5 = 29.4 + P sin30

Multiply right-hand-side to get rid of fracion

P cos30 = 14.7 + 0.5(P sin30)

P cos30 = 14.7 + 0.5p + 0.25

P cos30 - 0.5p = 14.95

P(cos30 - 0.5) = 14.95

P = 14.95/(cos30 - 0.5)

P = 40.8N

Is that correct?

Thanks,

Enigma

I have broken up P into its components, but I don't know where to go from there.

Horizontal = P cos30

Vertical = P sin30

Does P cos30 = 14.7?

Thanks,

Enigma

I used to recycle glass, but it was too far away to go to the recycling point all the time. If I could, I would recycle everything, because I try to do my bit for the environment as I save on electricity and water as well when I can. At the moment, I recycle paper and oragnic waste.

Such a philosophical question...

 

Most of the arguments against life in the Universe on probabilistic grounds are flawed on two points: (1) they assume life must be "like us"' date=' and (2) they are based on an assumption about the size and population of the Universe.

 

Spritually I'd say the Universe IS alive. How else do you explain consciousness?

More practically, though even if we assumed that the chances of life arising around some star are only 1/1,000,000,000,000,000, there are still more than enough stars to cover the probability.

 

I often consider that life exists as a symmetrical complement to entropy.

 

Has anyone else ever puzzled over the fact that we can mathematically generate something like Sierpenski's Triangle using at least three different methods: IFS, cellular automata, and discrete fractal iteration of line segments; all of which are iterative (note: DNA is iterative)? Does life describe a fractal shape between 3 and 4 dimensions? Is the existence of information quantitative? Why does DNA have to be present for there to be "life"? Would any self-replicating, iterative process of such fractal dimension result in "life"?

 

What is "life" anyway?[/quote']

 

 

Yes, exactly.

 

I consider all living cells to be life, so I believe that life definately exists elsewhere in the universe.

Could someone please help me with this question? I've had a go, but I don't really understand it.

A block of mass 3kg is at rest on a rough horizontal plane.

The block is acted on by a horizontal force of magnitude 14.7N. Given that the block is on the point of sliding, find the coefficient of friction between the block and the plane.

Friction = coefficient of friction x normal reaction

14.7 = coefficient x (3 x 9.8)

14.7 = coefficient x 29.4

Coefficient of friction = 0.5

The horizontal force is now replaced by a force of magnitude P N acting downards at 30º to the horizontal. Given that the block is again on the point of sliding, find the value of P.

The red is my working out - I'm not sure if it is correct. I don't know how to do the last part, so please could someone point me in the right direction.

Many thanks,

Enigma