Physics, 04.12.2020 20:50 c4thegoat

Which was not listed as a hazard of mining.

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Answer from: Quest

here we know that ball will drop down by total distance 3.2 m

so here we know that

$v_{yi} = 0$

$\delta y = - 3.2 m$

now using kinematics

$\delta y = v_{yi} t + \frac{1}{2}at^2$

$-3.2 = 0 -\frac{1}{2}(9.8) t^2$

$t = 0.81 s$

now in the same time it will cover horizontal distance of 0.75 m

$\delta x = 0.75m$

so speed of the ball will be

$v = \frac{delta x }{\delta t}$

$v = \frac{0.75}{0.81} = 0.93 m/s$

Answer from: Quest

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.

Answer from: Quest

It indicates no change in temperature and the process of matter changing its state

Answer from: Quest

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

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

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

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