Concept explainers
(a)
The length of the water column in the right arm of the U-tube.
(a)
Answer to Problem 14.22P
The length of the water column in the right arm of the U-tube is
Explanation of Solution
Given info: The cross sectional area
The formula to calculate the variation of pressure of the liquid with respect to the depth of the fluid is,
Here,
Write the expression for the volume of the water column.
Here,
The formula to calculate the density of the water is,
Here,
Substitute
Rearrange the above equation for the
Substitute
Conclusion:
Therefore, the length of the water column in the right arm of the U-tube is
(b)
The rise in the length of the mercury column in the left arm of the U-tube.
(b)
Answer to Problem 14.22P
The rise in the length of the mercury column in the left arm of the U-tube is
Explanation of Solution
Given info: The cross sectional area
Consider the height of the height of the left tube be
The volume of mercury on the left arm of the tube is equal to the right arm of the tube.\
Here,
Rearrange the above equation.
The pressure of the right tube is,
The pressure of the left tube is,
Here,
The pressure on the both the arm of the tube is same.
Compare the equation (1) and (2).
Substitute
Substitute
Conclusion:
Therefore, the rise in the length of the mercury column in the left arm of the U-tube is
Want to see more full solutions like this?
Chapter 14 Solutions
Physics For Scientists And Engineers, Technology Update, Loose-leaf Version
- Mercury is poured into a U-tube as shown in Figure P15.17a. The left arm of the tube has cross-sectional area A1 of 10.0 cm2, and the right arm has a cross-sectional area A2 of 5.00 cm2. One hundred grams of water are then poured into the right arm as shown in Figure P15.17b. (a) Determine the length of the water column in the right arm of the U-tube. (b) Given that the density of mercury is 13.6 g/cm3, what distance h does the mercury rise in the left arm?arrow_forwardThe inside diameters of the larger portions of the horizontal pipe depicted in Figure P9.45 are 2.50 era. Water flows to the right at a rate of 1.80 104 m3/s. Determine the inside diameter of the constriction. Figure P9.45arrow_forwardAn incompressible, nonviscous fluid is initially at rest in the vertical portion of the pipe shown in Figure P15.61a, where L = 2.00 m. When the valve is opened, the fluid flows into the horizontal section of the pipe. What is the fluids speed when all the fluid is in the horizontal section as shown in Figure P15.61b? Assume the cross-sectional area of the entire pipe is constant. Figure P15.61arrow_forward
- A 10.0-kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended from a scale and immersed in water as shown in Figure P15.24b. The 12.0-cm dimension is vertical, and the top of the block is 5.00 cm below the surface of the water. (a) What are the magnitudes of the forces acting on the top and on the bottom of the block due to the surrounding water? (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.arrow_forwardA fluid flows through a horizontal pipe that widens, making a 45 angle with the y axis (Fig. P15.48). The thin part of the pipe has radius R, and the fluids speed in the thin part of the pipe is v0. The origin of the coordinate system is at the point where the pipe begins to widen. The pipes cross section is circular. a. Find an expression for the speed v(x) of the fluid as a function of position for x 0 b. Plot your result: v(x) versus x. FIGURE P15.48 (a) The continuity equation (Eq. 15.21) relates the cross-sectional area to the speed of the fluid traveling through the pipe. A0v0 = A(x)v(x) v(x)=A0v0A(x) The cross sectional area is the area of a circle whose radius is y(x). The widening pan of the pipe is a straight line with slope of 1 and intercept y(0) = R. y(x) = mx + b = x + R A(x) = [y(x)]2 = (x + R)2 Plug this into the formula for the velocity. Plug this into the formula for the velocity. v(x)=A0v0(x+R)2arrow_forwardFigure P15.47 shows a stream of water in steady flow from a kitchen faucet. At the faucet, the diameter of the stream is 0.960 cm. The stream fills a 125-cm3 container in 16.3 s. Find the diameter of the stream 13.0 cm below the opening of the faucet. Figure P15.47arrow_forward
- The inside diameters of the larger portions of the horizontal pipe depicted in Figure P9.45 are 2.50 era. Water flows to the right at a rate of 1.80 104 m3/s. Determine the inside diameter of the constriction. Figure P9.45arrow_forwardReview. The lank in Figure P14.15 is filled with water of depth d = 2.00 m. At the bottom of one sidewall is a rectangular hatch of height h = 1.00 m and width w = 2.00 in that is hinged at the top of the hatch, (a) Determine the magnitude of the force the water exerts on the hatch, (b) Find the magnitude of the torque exerted by the water about the hinges.arrow_forwardA 10.0-kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended from a scale and immersed in water as shown in Figure P14.11b. The 12.0-cm dimension is vertical, and the top of the block is 5.00 cm below the surface of the water. (a) What are the magnitudes of the forces acting on the top and on the bottom of the block due to the surrounding water? (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block. Figure P14.11 Problems 11 and 12.arrow_forward
- The small piston of a hydraulic lift (Fig. P15.6) has a cross-sectional area of 3.00 cm2, and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude F1 must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN? Figure P15.6arrow_forwardA hollow copper (Cu = 8.92 103 kg/m3) spherical shell of mass m = 0.950 kg floats on water with its entire volume below the surface. a. What is the radius of the sphere? b. What is the thickness of the shell wall?arrow_forwardOil having a density of 930 kg/m3 floats on water. A rectangular block of wood 4.00 cm high and with a density of 960 kg/m3 floats partly in the oil and partly in the water. The oil completely covers the block. How far below the interface between the two liquids is the bottom of the block?arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
- College PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning