Physics, 26.03.2020 02:01 khalilwhite75

A long U-tube contains mercury (density = 14×103 kg/m3). When 10cm of water (density =1 .0 ×103 kg/m3) is poured into the left arm, the mercury in the right arm rises above its original level by

Answers: 2

Show answers

Answers

A long U-tube contains mercury (density = 14 × 103 kg/m^3). When 10 cm of water (density = 1.0 × 103 kg/m^3) is poured into the left arm, the mercury in the right arm rises above its original level by:

A. 0.36 cm B. 0.72 cm C. 14 cm D. 35 cm

Option B is the right choice as the rise in the mercury level is of 0.72 cm.

Explanation:

Given:

Density of the mercury, $\rho_m$ = $14\times 10^3\ kg.m^-^3$

Density of the water, $\rho_w$ = $1\times 10^3\ kg.m^-^3$

Water poured on the left arm of the U-tube, = $10\ cm$

According to the question:

The mercury in the right arm rises above its original.

We have to find this rise, h_2 .

Lets take a horizontal reference line on the U-tube.

Note:

In a static fluid there is no horizontal variation of pressure.

So,

Considering P_1 = P_2 on left and right of the manometer.

Hydro-static pressure = $\rho gh$

⇒ $P_1=\rho_w\times g\times h$

⇒ $P_2=\rho_m\times g\times h_2$

⇒ Equating both.

⇒ $\rho_w\times g\times h =\rho_m\times g\times h_2$

⇒ $h_2=\frac{\rho_w\times g\times h}{\rho_m\times g}$

⇒ $h_2=\frac{\rho_w\times h}{\rho_m}$

⇒ Plugging the values.

⇒ $h_2=\frac{1\times 10^3\times 10}{14\times 10^3}$

⇒ $h_2=\frac{10}{14}$

⇒ h_2=0.714 cm ≈ 0.72 cm

So the rise of the mercury in the right arm is of 0.72 cm.

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

false

hope this : d

Another question on Physics

Physics, 22.06.2019 13:50

Need only answer if you think you know it

Answers: 2

Answer

Physics, 23.06.2019 02:30

Why are nonmetals more likely to react with eachother by sharing electrons

Answers: 3

Answer

Physics, 23.06.2019 03:30

Ineed with my physics homework i can’t find it on google

Answers: 3

Answer

Physics, 23.06.2019 05:00

Consider the phase change taking place in the picture. you could use all but one statement to describe what is happening here. that is: a) temperature is increasing. b) kinetic energy is increasing. c) the space between the particles remains the same. d) the attractive force between the particles is decreasing.

Answers: 1

Answer

You know the right answer?

A long U-tube contains mercury (density = 14×103 kg/m3). When 10cm of water (density =1 .0 ×103 kg/m...

Questions