Recent content by MrEnergy

since it's a painting the "actuality" is not even concerned. But try the dark shadow thing i said right up. edit: whatever you do with the painting, you will get a ratio of lenghts, so how can you predict the actuall? The lady may be 1,50m or 1,80m and the rooms length would slightly...

Do you see the dark shadow and lighter shadow at the paintings and the open chests right side ?? The long and lighter shadow is caused by the first set of windows. The windows at back can reach where the front ones cannot so that makes it lighter. As you can see there is a darker shadow also. I...

oh sorry i forgot! bad mistake:P Anyways I have remade my answer and found a very long one. All the integrals might need to change into sum!Don't know if it's correct... but this is what i found: http://img84.imageshack.us/img84/9630/spaceshipum6.th.jpg http://g.imageshack.us/thpix.php...

I would advise you to choose an engineering career such as; electrical or computer engineering and double major with physics or maths if you can. You can also be an engineer and research theoretical sciences, as engineering also provides you much about theoretical sciences alone, opens your way...

E_{f}=∫_{R}^{R+h}[G⋅((M⋅m)/(r²))dr] ===> (integral from R to R+h and E_{f}=Fuel energy) E_{f}=G⋅M⋅m⋅(h/(R(R+h))) is it the solution?? By calculating the fuel energy in terms of joule and defining R, you can find the maximum height i guess...

It is the friction that causes the ball to roll. So, without friction the ball would only slide without rolling. To give an example, if we had an artificial waterfal with an angle "alfa", and we have put a beach ball on it, and if we have released; then we would observe that, as the water is...

Magnets, which have a stabilized magnetic field, does not lose or get energy from anywhere. For example lifting an object or pushing one; if you close the magnet to an object and if the magnet pushes it, then the energy required to push the object is given by your fingers. And you can not...

...after you solve a little bit, you get; 137/2=F3*Sin(a) so for a=>90 degrees sin is max and F3 is min. Min F3 is 137/2...And a is 90 degrees... and magnitude of F1 is; F1=Cos30*137

I mean't i didn't work for it:P Just used a program.

Ok, at least now I think that I did right. And yeah I don't get the 0,427 number...

Sakha, after 65 seconds second car's velocity becomes 12m/s^2, and they are not supposed to crash with same velocities. So that your t2=77 second is impossible... The right answer is 13 seconds(I could'nt solve but you can use a program to get the exact number). After 13 secs they crash...

If the second car had a velocity of 12m/s^2, then they wouldn't crash, so we try to see if they do; 25 - 0,2t = 12 => t=65s in 65 seconds we see that the cars actually crash. So its time to see where they actually crash.. so in "t" seconds after the break, the second cars velocity will...

Firstly thank you for replying, secondly, i also did the same thing that you did, but the answer given on the book was different to the one that I've found...That's why I've asked.. The book gave an answer like =0,427(m/s^2)(mi/h*s) but that equation didn't give the exact... Any ideas??

My question seems rather simple, but I could't have the right answer. Hope you can help me about it. Thanks! Determine a convertion factor that, upon multiplication, changes an acceleration in miles per hour times second (mi/h*s) to meter per second squared (m/s^2).