Two red blood cells each have a mass of 4.15×10−14 kg4.15×10−14 kg and carry a negative charge spread uniformly over their surfaces. the repulsion arising from the excess charge prevents the cells from clumping together. once cell carries −2.20 pc−2.20 pc of charge and the other −3.30 pc−3.30 pc , and each cell can be modeled as a sphere 7.80 μm7.80 μm in diameter. what minimum relative speed would the red blood cells need when very far away from each other to get close enough to just touch? ignore viscous drag from the surrounding liquid.

in this image, you can see solar radiation going through ocean water.

no image, but probably absorption

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.

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.

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures