A loaded spring launches a 2.50 kg block, using a force of 450 N. If the change in momentum is 12.0 kg*m/s, how long was the block in contact with the spring?

37.5 seconds

Explanation:

The given parameters are;

The mass of the block on the spring, m = 2.50 kg

The force with which the loaded spring launches the block, F = 450 N

The change in momentum of the block, Δp = 12.0 kg·m/s

We have;

Let the force with which the block was launched = The net force,

By Newton's second law of motion, we have;

F = = Δp × Δt

Where;

Δt = The time the block is in contact with the spring

Therefore;

By plugging in the values for and Δp, we have;

The time duration the block is in contact with the spring, Δt = 37.5 seconds.

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.

float.

archimedes' principle

denser than water sinks in water.

denser than mercury would sink in it