The weight of a rectangular block of low-density material is 14.5 N. With athin string, the center of the horizontal bottom face of the block is tied to thebottom of a beaker partly filled with water. When 25.0% of the block'svolume is submerged, the tension in the string is 15.0 N.(a) Find the buoyant force on the block.N

Given information:Weight of the block (w) = 14.5Volume of the block submerged (Vin) = 0.25 VWhere…

Answered: The weight of a rectangular block of…

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter15: Fluid Mechanics

Section: Chapter Questions

Problem 27P

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Similar questions

. A juniper-wood plank measuring 0.25 ft by 1 ft by 16 ft is totally submerged in water, (a) What is its weight? (b) What is the buoyant force acting on it? (c) What is the size and the direction of the net force on it?
In an immersion measurement of a woman's density, she is found to have a mass of 62.0 kg in air and an apparent mass of 0.0850 kg when completely submerged with lungs empty. (a) What mass of water does she displace? (b) What is her volume? (c) Calculate her density. (d) If her lung capacity is 1.75 L is she able to float without treading water with her lungs filled with air?
A wooden block floats in water, and a steel object is attached to the bottom of the block by a string as in Figure OQ14.3. If the block remains floating, which of the following statements are valid? (Choose all correct statements.) (a) The buoyant force on the steel object is equal to its weight. (b) The buoyant force on the block is equal to its weight. (c) The tension in the string is equal to the weight of the steel object. (d) The tension in the string is less than the weight of the steel object. (e) The buoyant force on the block is equal to the volume of water it displaces.
(a) What is the density of a woman who floats in freshwater with 4.00% of her volume above the surface? This could be measured by placing her in a tank with marks on the side to measure how much water she displaces when floating and when held under water (briefly). (b) What percent of her volume is above the surface when she floats in seawater?
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.
A wooden block floats in water, and a steel object is attached to the bottom of the block by a string as in Figure OQ15.1. If the block remains floating, which of the following statements are valid? (Choose all correct statements.) (a) The buoyant force on the steel object is equal to its weight. (b) The buoyant force on the block is equal to its weight. (c) The tension in the string is equal to the weight of the steel object. (d) The tension in the string is less than the weight of the steel object. (e) The buoyant force on the block is equal to the volume of water it displaces.
A spherical weather balloon is filled with hydrogen until its radius is 3.00 m. Its total mass including the instruments it carries is 15.0 kg. (a) Find the buoyant force acting on the balloon, assuming the density of air is 1.29 kg/m3. (b) What is the net force acting on the balloon and its instruments after the balloon is released from the ground? (c) Why does the radius of the balloon tend to increase as it rises to higher altitude?
A man of mass m = 70.0 kg and having a density of = 1 050 kg/m3 (while holding his breath) is completely submerged in water, (a) Write Newtons second law for this situation in terms of the mans mass m, the density of water , his volume V, and g. Neglect any viscous drag of the water, (b) Substitute m = V into Newtons second law and solve for the acceleration a, canceling common factors, (c) Calculate the numeric value of the mans acceleration, (d) How long does it take the man to sink 8.00 m to the bottom of the lake?
A tank with a flat bottom of area A and vertical sides is filled to a depth h with water. The pressure is P0 at the top surface. (a) What is the absolute pressure at the bottom of the tank? (b) Suppose an object of mass M and density less than the density of water is placed into the tank and floats. No water overflows. What is the resulting increase in pressure at the bottom of the tank?
A spherical weather balloon is filled with hydrogen until its radius is 3.00 m. Its total mass including the instruments it carries is 15.0 kg. (a) Find the buoyant force acting on the balloon, assuming the density of air is 1.29 kg/m3. (b) What is the net force acting on the balloon and its instruments after the balloon is released from the ground? (c) Why does the radius of the balloon tend to increase as it rises to higher altitude?
A uniform wooden board of length L and mass M is hinged at the top of a vertical wall of a container partially filled with a certain liquid (Fig. P15.81). (If there were no liquid in the container, the board would hang straight down.) Three-fifths of the length of the board is submerged in the liquid when the board is in equilibrium. Find the ratio of the densities of the liquid and the board.
A 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)2

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