Introductory Physics | Forum profile

Also at the same link I see: R = 53.34 ft·lbf·°R−1·lbm−1 (for air)

Is this like a propeller deisgn or is it a VAWT (vertical axis wind turbine) or a horizontal? It would be great if the results of that experiment could be posted here. i'm making a wind turbine at home and i would like some outside data to compare with my own. i'm sure i'm not the only one that would appreciate it either. Thanks Lalbatros

DisplayAds("Right1"); This is less of a strict math problem than me thinking my online teacher is wrong. I will, however, format the question as per PF's requirements, and I think it should be in this forum because it involves my coursework. 1. The problem statement, all variables and given/known data Essentially, when should I use relativistic calculations (...

Imagine the sphere has lines of latitude and longitude drawn on it. The (differential element of ) area of a rectangle defined by these lines is ( working in polar coords ) so, using your formula Now integrate out between 0 and 2*pi, and that gives the formula for the whole shell. Of course you don't actually need to integrate - just think about it and ...

Thanks. However, in your derivation, you say nothing about the necessary orthogonal property of the transformation matrix. Just try in matlab: C(AxB) (CA)x(CB) Once with C a non-orthogonal matrix and once with C a orthogonal matrix. In first case both results will be different, in second case both results will be equal. [edit]: from the third last line till the second last line ...

DisplayAds("Right1"); Hello, I've tried solving the following question: A car collides with a tree with a velocity of 75km/h. The car was compressed 1.1m after the collission and the car's mass is 1100kg (including the driver), and the driver's mass is 71kg. a) What is the kinetic energy just before the collission? b) What is the average force that works on the ...

Hi Tranquility13! Welcome to PF! Show us what you've tried, and where your'e stuck, and then we'll know how to help you!

Hi Zukie91! Originally Posted by Zukie91 were you able to get the correct answer with this method, because i think i get what you are saying, and cannot get the problem right using any combination of m and h...

DisplayAds("Right1"); An elastic conducting material is stretched into a horizontal circular loop of radius 20 cm and placed in a magnetic field of strength 0.5 T, directed vertically downwards. a. What is the magnetic flux through the loop? b. When released, the radius of the loop decreases at an initial rate of 50 cm/s; what is the initial emf induced in the ...

Originally Posted by Number1Ballar 1. The problem statement, all variables and given/known data The 5.00-V battery in the figure is removed from the circuit and replaced by a 20.00-V battery, with its negative terminal next to point ...

Hi Zukie91! Originally Posted by Zukie91 were you able to get the correct answer with this method, because i think i get what you are saying, and cannot get the problem right using any combination of m and h...

Originally Posted by Number1Ballar 1. The problem statement, all variables and given/known data The 5.00-V battery in the figure is removed from the circuit and replaced by a 20.00-V battery, with its negative terminal next to point ...

ok i started again with this: I drew a diagram as suggested. We are starting off in the Y direction at 210ms^-1 so Yold= 210ms^-1 Ynew=210cos(14)=204ms^-1 Xold=0 Xnew=210sin(14)=50.80ms^-1 then magnitude = (sqrt) 204^2 + 50.80^2 =210ms^-1 ??? same as what we started with...... please help I cant get my head round this and I have an exam on monday!

Another one of those "thinking it out" things. Now the "Air" has been turned on. I'm thinking that the equation would change slightly (since it still wants the horizontal disposition) to: x-xo = vo cos(//)t + 1/2 at^2 Plug in the same set of variables (with the time changing due to the new resistance). x = (50)cos(45)(6.5) + 1/2(-9.8)(6.5)^2 ...and get 22.78. Looking for ...

Originally Posted by stewartcs "More concentrated or useable" is no more of a better definition than is "disorganzied". You even contradict yourself later with the following quote: . Disorder is...

3.142

Note that multiplying (airy disc diameter) x (orbital height) gives incorrect units of m^2, when we're looking for an answer in meters. Hint: what is the camera's magnification factor?

DisplayAds("Right1"); 1. The problem statement, all variables and given/known data A 100 kg man stands on the left end of a 100 kg board 2 meters long which sits on a frictionless surface as shown. Where will the left end of the board come to rest if the man walks to the right end of the board and stops? 2. Relevant equations p=mv 3. The attempt at a ...

Acceleration 9.81= -w2 A cos(wt + teta) w=2pi/1.5 A=9.81/(-w2 A cos (wt + teta) i thought of this... what puzzle me...is how to calculate the cos (wt + teta) part....was the value of t and teta????

Hi I think I finally got it. The new rms must be 500 m/s . .. . . . ? Has to be.. If it is, it is actually simple and I actually feel silly that I was so blind. But thanks a lot for your help and sticking it out