Question

J fluid K fluid i fluid h4 h3 h2 h1 A B

          J fluid
K fluid
i fluid
h4
h3
h2
h1
A
B
        
J fluid
K fluid
i fluid
h4
h3
h2
h1
A
B

Added by Sandra M.

Close

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
In the system given in the figure, h1 = 30 cm, h2 = 10 cm, h3 = 70 cm, h4 = 20 cm, and the densities of the fluids ρi = 1000 kg/m^3, ρj = 1.25 kg/m^3, ρk = 13600 kg/m^3. Pressure in line A, PA = 10 bar. Find the pressure in line B in Pascal. J fluid K fluid h4 n h3 h2 vi fluid
Close icon
Play audio
Feedback
Powered by NumerAI
Danielle Fairburn Kathleen Carty
David Collins verified

Dominique Jan Tan and 57 other subject Physics 101 Mechanics educators are ready to help you.

Ask a new question

*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.

*

Recommended Videos

-
find-the-pressure-due-to-the-fluid-at-a-depth-of-76-mathrmcm-in-still-a-water-leftrho_w100-mathrmg-m

Find the pressure due to the fluid at a depth of $76 \mathrm{~cm}$ in still $(a)$ water $\left(\rho_{w}=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)$ and $(b)$ mercury $\left(\rho=13.6 \mathrm{~g} / \mathrm{cm}^{3}\right)$. (a) $P=\rho_{w} g h=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=7450 \mathrm{~N} / \mathrm{m}^{2}=$ $7.5 \mathrm{kPa}$ (b) $P=\rho g h=(13600 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} \approx$ $1.0$ atm

Schaum’s Outline of College Physics

find-the-pressure-due-to-the-fluid-at-a-depth-of-76-mathrmcm-in-still-a-water-leftrho_w100-mathrmg-2

Find the pressure due to the fluid at a depth of $76 \mathrm{~cm}$ in still $(a)$ water $\left(\rho_{w}=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)$ and (b) mercury ( $\left.\rho=13.6 \mathrm{~g} / \mathrm{cm}^{3}\right)$. (a) $P=\rho_{w} g h=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=7450 \mathrm{~N} / \mathrm{m}^{2}=7.5 \mathrm{kPa}$ (b) $P=\rho g h=\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} \approx 1.0 \mathrm{~atm}$

Schaum’s Outline of College Physics

using-a-u-tube-manometer-to-measure-gauge-pressure-of-fluid-density-p820-kgm3-and-the-manometric-fluid-is-mercurywith-a-relative-density-of-136what-is-the-gauge-pressure-if-a-h103m-andh207m-92217

Adi S.


*

Recommended Textbooks

-
University Physics with Modern Physics

University Physics with Modern Physics

Hugh D. Young 14th Edition
achievement 1,570 solutions
Physics: Principles with Applications

Physics: Principles with Applications

Douglas C. Giancoli 7th Edition
achievement 1,500 solutions
Fundamentals of Physics

Fundamentals of Physics

David Halliday, Robert Resnick , Jearl Walker 10th Edition
achievement 1,959 solutions

*

Transcript

-
00:01 Good day, the topic is about hydrostatic pressure.
00:04 We note that a certain column of liquid will exert a pressure, a certain depth from its surface that is solved as p equals raw times g times h.
00:15 P is the pressure in pascal, raw is the density in kilogram per meter cube, which is the density of the fluid, and that g is the acceleration due to gravity in meters per second squared, and height is the and h is the height or the depth.
00:30 Of from the from the surface of the fluid so let us suppose we wish to find the pressure at a certain depth of 76 centimeter below the surface of a water and b mercury with a given densities so let us start by expressing the densities in kilogram per cubic meter so we may illustrate that using the density of water so that 1 ,000 kilogram is equal to, or rather 1 ,000 gram is equal to 1 kilogram.
01:10 And that 100 centimeter makes 1 meter.
01:15 So we'll cube, take the cube of this, since our cm is also raised to 3.
01:24 So this gives us 1 times 100 cube over 1 ,000.
01:30 Or this is equal to 1 ,000 kilogram per cubic meter...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever