free4spirit

Hi Everyone Hope all is well with you all, wherever in the world you are....and also happy day to you all I've got a question for my uni assignment that i'm having problems with....I just literally hit the wall....my brain doesn't seem to be able to process the question...or want too... and I have now been staring blankly at this question for the best part of the afternoon.....and i dont have a clue where to start ! also i was expecting there to be an example of this question in our book....but there wasn't sadly.... so if someone out there has any spare time and can give a girl a hand, it would be hugely appreciated !! ok..... said question; The total power output of the Sun is about 3.8 × 1026 W, and the power-producing core has a radius of about 1.4 × 105 km (about 20% of the full radius of the Sun). Using these figures and knowledge of the energy release in the proton– proton chain; estimate the power density (i.e. the number of watts per cubic metre) in the core of the Sun, and the number of helium nuclei formed per second per cubic metre. Give your answer in scientific notation and to two significant figures. (You may also need to refer to the relationship between the joule and eV, and you will also need to know that the volume, V, of a sphere with radius, r, is V = 4/3πr3 .) If someone could help me start it off or give me an example i would really really appreciate it ... ! Many many thanks, Axx

Hi all Just wondering if anyone out there can help me with one of my questions for my uni assignment? Use the following balanced chemical equation to calculate the mass of carbon dioxide produced from the burning of 3.73 kg of jet fuel which can be represented by the molecule C12H26. You should quote your answer to 3 significant figures in units of kg. The relative atomic masses (RAMs) needed to do this calculation are as follows; RAM (carbon) = 12.0; RAM (oxygen) = 16.0; and RAM (hydrogen) = 1.01. 2 C12H26 + 37 O2 = 24 CO2 + 26 H2O If anyone can please give me any advice or point me in the right direction, it would be greatly appreciated As i dont even know where to start! Thanks, Ax

yeah i know its staring me in the face......ive been at this for 4hrs now and my head is mush......so i apologise if im missing something...... but anyways i think i may have got it! well.....again see what you think Deceleration = initial velocity – final speed/ total time taken - v = u + at But we do not know t. So we know u = 49.5m s¯¹ and the final velocity v = (49.5m s¯¹ + 0) = 24.25m s¯¹ 2 So, since the mean velocity is equal to the displacement divided by time vave = D t Therefore, the time for the arrow to come to rest is t = D = 0.20m = 0.00825 s vave 24.25m s‾¹ Then lastly, for the accelerating arrow vf = at + vo So, a = (vf – vo) = (0 – 24.25m s‾¹) = -2940m s² = -2.94m s² x 10³ t 0.00825s hope it makes sense..... Ax Thanks for your help its much appreciated my head just mush now and ive still got more questions to get through...so yeahh apologies if im being slow Ax

Haha, indeed we are! Thanks to you guys! thats precisely what i have been pondering the last wee whiley.....havent been told anything else.....just the acceleration due to gravity, the max height of the arrow, the depth at which the arrow penetrates the ground and now i know the final velocity.....i have no other information....not that i can find! am at the brink of insanity!! XD lol Ax

Haha! no no dont worry! I literally just found the equation in the back of my notebook!! yeah true....but im pretty damn sure these questions are meant to trick you....well in my last assignment i got tricked for sure! but....i am still trying to figure out an equation to calculate the deceleration as the arrow hits the ground...... maybe i'll find another equation in another notebook! i might go check... thank thank you so much for your help! Annax oh and i am aware that deceleraiton is initial velocity - final speed/ total time taken...... but i do not know what its initial velocity is..... Ax

I know....but we havent done the suvat equations....theyre not even mentioned in the book....and yes dont worry i am aware of the mass! I think ive done it! well, the first part of the question anyway.....see what you think.... Find speed of arrow v as it hits the ground. Δh = 125m and we assume g = 9.81m s‾² So equations to use are; Ek = ½ mv² and ΔEg = mgΔh We assume there is no friction or air resistance. So, as there is no friction or air resistance, all the gravitational potential energy at the maximum height of the arrow will be converted into kinetic energy as it falls to the ground. ΔEg = Ek which becomes, ½ mv² = mgΔh Then I divide both sides of the resulting equation by m, to cancel m as we do not know the arrows mass. ½ mv² = mgΔh which then becomes, ½ v² = gΔh m m So then multiply both sides by 2; v² = 2gΔh Then take the square root of both side; v = √2gΔh Insert values, so; v = √2 x 9.81m s¯² x 125m = √2452.5m² s¯² = 49.5m s¯¹ thats what ive worked out....does that make sense now? fingers crossed! and cheers sooo much for sticking with me Ax sorry about the two m's there...i think you know what i mean! i just copy and pasted from star office.... Ax

Hmm i see....well im trying really very hard but i dont seem to be getting anywhere.....but i think i figured out an equation for it....:S mgh = 1/2mv^2.....not sure if that is correct, as m is mass.....and i do not have the mass of the arrow......or i am confusing myself? again thanks sooo much for your help! Ax oh and sorry for the wait....lappy decided to freeze! Thanks for helping me out well you see we havent actually done projectile motion, and it is not in our book.... if i can figure out what equation to use to calculate the speed of the arrow as it hits the ground, and also the deceleration as it penetrates the ground, i think i may be able to manage....but at the moment i just cant quite grasp how to figure it out...... no, we do not have a drawing either.... its just the equation that confuse me! :S but thanks for the help! Ax ooohhh i may have a break through! ok so taking the mgh = 1/2mv^2 that becomes v = sqrt(2gh) ? so it would be v = sqrt(2x9.80x125) = 49.5m/s ? i hope thats right and makes sense?! thanks for the help peeps Ax

Aahh thank you for that ...im still working it out....ive been re-reading the question numerous times, but have got to the point where my brain is going arghh! lol thanks again Ax

Thanks very much should be able to figure this out now.... Ax

Hi everyone Im new here...but love all the discussions going on.....I have a question for my uni assignment, that has totally stumped me....and ive never been very good at physics... So if anyone could shed any light it would be massively appreciated! Question: The arrow fired as described in part (a) reaches a maximum height of 125 m above the ground before starting to fall back downwards. When the arrow hits the ground, the archer notices the head penetrates 20.0 cm into the earth. By considering the energies involved, calculate the speed of the arrow when it hits the ground and then its deceleration as it penetrates the earth. Quote your answer to three significant figures and using scientific notation. (Assume the acceleration due to gravity is 9.81 m s− 2 and ignore any effects of air resistance.) Im not entirely sure what equation to use to figure this out....i thought maybe W = ΔE_k? but am still utterly confused.... so absolutely any help would make my day! cheers and hope you are all well Ax

Aahh, cheers soo much for that! Really helped me out The help is greatly appreciated, as we dont go into Uni for lectures... We have to figure it out ourselves... Cheers again Annax

Hey everyone, Im new to this site, and so far has been quite helpful!

Hey everyone, My name is Anna and i live in Scotland, and am studying an introductory course with the Open University. This is my first time here! Got reccommended the site by a friend, as I have an assignment to do by tomorrow and am stuck on one particular maths question! If anyone has any pointers, or can possibly help explain this to me, it would be hugely appreciated! Question : © One cubic metre (1 m3) of atmosphere at sea level contains 3.80 × 102 ppm of CO2 and 5.00 × 103 ppm of water vapour. If there is a total of 2.6 × 1025 molecules in 1 m3 of air, calculate how many molecules of CO2 and water vapour there are in 1 m3 of air. Give your answers to the appropriate number of significant figures. Any help would be fantastic! Hope youre all well too Anna