Physics, 26.07.2019 18:30 mamshie

When air pressure increases the liquid in a mercury barometer?

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Answer from: hannahkharel2

It forces the mercury to rise, being pushed up the tube by pressing down on the dish.

I hope I could help :)

Answer from: kuddlebugsmommy

rises

Explanation:

The pressure depends on the height of mercury column raised. It is given by the formula

P = h d g

where, h is the height of mercury, d be the density of mercury and g be the value of acceleration due to gravity.

The density of mercury is constant and the value of acceleration due to gravity is constant, so as the pressure increases the value of height also increases. Thus, the height of liquid in the mercury barometer rises.

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.

Answer from: Quest

Answerthe force between the charges will be f/9 (one-ninth of force f)explanationcoulomb's law relates the magnitudes two point charges q₁ and q₂, the distance, r, between the charges, and the electrostatic force, f, between the two charges: $f = \dfrac{k q_1 q_2}{r^2}$ k represents coulomb's constant, k = 8.99×10⁹ n·m²/c².note that if we have r = 3r, then $\begin{aligned} f & = \dfrac{k q_1 q_2}{r^2} \\ f_{3r} & = \dfrac{k q_1 q_2}{(3r)^2} \\ & = \dfrac{k q_1 q_2}{3^2 \cdot r^2} \\ & = \frac{1}{9} \cdot \boxed{\dfrac{k q_1 q_2}{r^2} } \\ & = \frac{1}{9} f & & \left(\text{\footnotesize since $ f = \tfrac{k q_1 q_2}{r^2}$}\right) \end{aligned}$