Recent content by mtong

"If you drop a basketball each time it bounces it is half the original height/energy that you inputted" The height at which the ball bounces is determined by the air resistance and other forms off energy loss. Even if this was so it makes no sense to make the general claim: "The laws of...

Hi Gelsamel Epsilon. I just came through the High School (in Australia) and I did not do all that much work. I did however get into the course I wanted a degree in (Electrical engineering / Physics) and am currently quite happy. Compared to the states the job prospects for physics are...

Well before you go and get medicated I suggest you seek a diagnosis from a professional and go from there.

Correct me if I’m wrong but is you wish to test you new theory on a simulator won’t this be inherently doomed as any simulator will be based on current theory?

Energy is just a human concept devised to help with our analysis of the physical world.

c would be the constant of proportionality. eg. if y directly varies with x then the relationship between y and x will be y=kx; Where k is a constant whose value is determined to fit a set of data. http://www.purplemath.com/modules/variatn.htm In the case above the constant will take the...

Drag is directly proportional to velocity squared. Hence Drag=C * V^2 where c is a constant and v is the velocity of the body relative to the fluid. The usual form is Drag=1/2 CpAv^2. Where C is the drag coefficient, p is the Greek letter rho representing the density of the fluid and A the...

Thanks to all for your help! I was able to get the correct answer using the methods outlined in my first post. It seems that my calculator was using a different a approximation for the permittivity of free space. :mad:

I understand that electrical breakdown will occur at the sphere ie. where the electric field is greatest and where the radius of curvature is least. However i am unable to calculate at what potential the sphere will breakdown. I tried Unfortunately this method failed and I am unsure why...

Hi, I have recently been reading through / self learning some material on Electrostatics and I came across this question. The minimum electric field that can be supported by dry air at atmospheric pressure is about 10^5 volts/m. What is the maximum potential difference to the Earth for a...

In regards to the first post i was under the impression that it answer was C. That is dy/dx=du/dx*dy/du, where u is e^(2x) . As in the case of the derivative of cos( e^(x) ), which is - 2e^(2x)*sin (e^(2x)).

I would consider it a “projectile motion” problem; however one must take into the acceleration that the gravitation field of the star would provide, in both the direction of the star and in the direction of motion attributed to the moving planet.

In my opinion the package would have an upwards velocity on 30m/s and be accelerationg downwards at g.

Where abouts in aus do you live? Where are you studing at?

I should have made myself more clear, this was in response to CarlB's solution, using special relativity.