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PDF Editor FAQ

What are some rare photos of Pakistan?

NOTE: I’m facing some technical difficulties uploading Photos. Therefore, the answer is done in 3 Parts. Links given below.PART 1#1 - Political:[A] - ZA Bhutto with Libya’s Gaddafi[B] - Yahya with Yasir Arafat[C] - President of Pakistan greeted in Washington[D] - Yasir Arafat [Palestine], Mujeeb-Ur-Rehman [Bangladesh], Z.A Bhutto and Col. M Qaddafi [Libya] at Shalimar Gardens in Lahore on 23rd February 1974[E] - Mountbatten Address to Pakistan[F] - MA Jinnah to the opening of State Bank of Pakistan[G] - DAWN 15th Aug, 1947[H] - Ayub Khan with Indian Premier. Habib Bank sign board in Urdu.[I] - US President Nixon in Taxila, Pakistan[J] - A Memorable picture of Lord and Lady Mountbatten with Mohammad Ali Jinnah[K] - King Faisal II of Iraq visiting 501 Workshop, Rawalpindi[L] - Mandela in Pakistan[M] - Lady Diana in Pakistan[N] - Obama in Karachi[O] - W Bush in Pakistan[P] - Queen Elizabeth in Valley Swat, Pakistan[Q] - Pope John Paul II in Pakistan[R] - Bhutto along with various Muslim heads of states praying at Lahore’s Badshahi Mosque in Lahore during the 1974 International Islamic Conference.[S] - Indira Gandhi met ZA Bhutto[T] - OIC Summit, Lahore[U] - 1968 - Crown Prince Hassan Bin Talal, younger brother of King Hussein of Jordan smiles with his bride Sarvath Ikramullah of Pakistan at their wedding in Karachi.PART 2:Part 2 of ‘What are some rare Photos of Pakistan?’PART 3:Part 3 of ‘What are some rare Photos of Pakistan?’

If [math]F=ma[/math] and [math]E=mc^2[/math], does [math]F=E[/math] if [math]ma=mc^2[/math]?

F=ma makes m=F/aE=mc2 makes m=E/c2so F/a=E/c2 or c2F/a=EBut what would you be comparing?Rotary movement (A) against energy that is released when atoms are split (B).I don't think you can put those 2 together since A does not imply B or the other way around.

Aren't [math]\mathbf{F}=m\mathbf{a}[/math] and [math]E=\frac{1}{2}mv^2[/math] equivalent statements of Newton's 2nd law of motion?

They are not equivalent. F = ma gives you enough information to actually calculate the trajectory of a particle. However, the equation for the energy in terms of velocity doesn't give you enough information.For example, for a particle in a magnetic field, we can write[math]\mathbf{a} = \mathbf{F}/m = \frac{q}{m}(\mathbf{v} \times \mathbf{B})[/math]so if we know the magnetic field everywhere, and we know the initial position and velocity, we can integrate this equation to compute the entire trajectory of the particle.Now, the energy of a particle in a magnetic field is still[math]E[/math][math] = \frac{1}{2} mv^2[/math]Since the energy is the same as it is without a magnetic field, but now we know the force is different and the trajectory is different, obviously, knowing the energy alone doesn't give you enough information to find the trajectory.If you want to use energy as the fundamental quantity that determines motion rather than force, you actually can do this, but you need to express the energy not in terms of velocity but in terms of a variable called canonical momentum. When you do this, then the equation contains enough information to determine the equations of motion. (Using just the velocity doesn't give you enough information, as I said above.) This is the foundation of Hamiltonian mechanics.

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