Asailboat moves north for a distance of 10.00 km when blown by a wind from the exact south with a force of 5.00 x 10^4 n. how much work was done by the wind?

In a thunderstorm, charge builds up on the water droplets or ice crystals in a cloud. thus, the charge can be considered to be distributed uniformly throughout the cloud. for the purposes of this problem, take the cloud to be a sphere of diameter 1.00 kilometer. the point of this problem is to estimate the maximum amount of charge that this cloud can contain, assuming that the charge builds up until the electric field at the surface of the cloud reaches the value at which the surrounding air breaks down. this breakdown means that the air becomes highly ionized, enabling it to conduct the charge from the cloud to the ground or another nearby cloud. the ionized air will then emit light due to the recombination of the electrons and atoms to form excited molecules that radiate light. in addition, the large current will heat up the air, resulting in its rapid expansion. these two phenomena account for the appearance of lightning and the sound of thunder. take the breakdown electric field of air to be eb=3.00ă—106n/c. part a estimate the total charge q on the cloud when the breakdown of the surrounding air is reached. express your answer numerically, to three significant figures, using ďµ0=8.85ă—10â’12c2/(nâ‹…m2) .

The ratio of lift to drag l/d for a wing or airfoil is an important aerodynamic parameter, indeed, it is a direct measure of the aerodynamic efficiency of the wing. if a wing is pitched through a range of angle of attack, l/d first increases, then goes through a maximum, and then decreases. consider an infinite wing with an naca 2412 airfoil. estimate the maximum value of l/d. assume that the reynolds number is 9x10^6.

The electric field between square plates of a parallel-plate capacitor has magnitude e. the potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. the magnitude of the electric field between the plates is now equal to a)e b)e/4 c)e/2 d)4e e)2e