It has been proposed that we could explore mars using inflated balloons to hover just above the surface. the buoyancy of the atmosphere would keep the balloon aloft. the density of the martian atmosphere is 0.0154kgm3(although this varies with temperature). suppose we construct these balloons of a thin but tough plastic having a density such that each square meter has a mass of 5.00 g. we inflate them with a very light gas whose mass we can ignore. (a) what should be the radius and mass of these balloons so they just hover above the surface of mars? (b) if we released one of the balloons from part (a) on earth, where the atmospheric density is 1.20kgm3, what would be its initial acceleration assuming it was the same size as on mars? would it go up or down? (c) if on mars these balloons have five times the radius found in part (a), how heavy an instrument package could they carry?

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

where is unknown. because the position is parabolic, there is symmetry about the midpoint at , and we know that

so that the position function is

which means the velocity is

then the total distance traveled is

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at and ends at , so its displacement is 0.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j