copy and paste this google map to your website or blog!
Press copy button and paste into your blog or website.
(Please switch to 'HTML' mode when posting into your blog. Examples: WordPress Example, Blogger Example)
Input Form:Deal with this potential dealer,buyer,seller,supplier,manufacturer,exporter,importer
(Any information to deal,buy, sell, quote for products or service)
Work and change in kinetic energy - Physics Forums I know that the net work = the change in kinetic energy, delta KE But what if the object has both the potential energy and kinetic energy, for example, a
Work Energy Theorem: Delta K Calculation for Particle Moving in x . . . A particle moving in the x direction is being acted upon by a net force F (x)=Cx^2, for some constant C The particle moves from x initial =L to x final=3L What is Delta K, the change in kinetic energy of the particle during that time? I tried thih by doing the integral of F (x), replacing x with 2L (because final-initial, 3L-L)
Delta U = 0 - CHEMISTRY COMMUNITY Delta U can be zero because it means there is no change in internal energy between the initial and final states of this system For example, for an isolated system, there is no change in energy and delta U is always zero Also since delta U = w + q, if the net change of internal energy caused by doing work and heat transfer counters each other, delta U is also zero
Why is the net work done equal to the change in K. E. only? Why is it so that the net work done by a force on an object is equal to change in kinetic energy only and not any other form of energy like potential energy? Also is the work energy theorem valid for both conservative and non conservative forces
Find Thickness of Oil Film on Water, \lambda = 550 nm To find the thickness of an oil film on water that appears yellow-green at 550 nm, the refractive indices of oil (1 50) and water (1 33) are used in the equation for constructive interference The net phase difference is calculated, leading to the formula for thickness: d = (2m + 1)λ0 4nf The challenge lies in determining the appropriate integer value for m, as increasing m results in
Conceptually understanding change in potential energy with 0 net work Suppose somehow an object is moving upwards with a speed ##v##, at this point I start applying a force ##F## that is equal to its weight, so the net force on the object is zero So it will continue moving upwards with its initial speed Suppose after the height difference is ##h##, I stop
The rocket equation, one more time • Physics Forums Furthermore, what matters in variable mass problems is the rate of change ##\Delta m \Delta t## Since ##\Delta t## is always positive, when the system loses mass (as in the case of a rocket) the ratio is negative; if it gains mass (as in the case of a moving container accumulating rainwater) the ratio is positive