A closed container is partially filled with water. Initially, the air above the water is at atmospheric pressure (1.01 × 10 5 Pa) and the gauge pressure at the bottom of the water is 2500 Pa. Then additional air is pumped in, increasing the pressure of the air above the water by 1500 Pa. (a) What is the gauge pressure at the bottom of the water? (b) By how much must the water level in the container be reduced, by drawing some water out through a valve at the bottom of the container, to return the gauge pressure at the bottom of the water to its original value of 2500 Pa? The pressure of the air above the water is maintained at 1500 Pa above atmospheric pressure.
A closed container is partially filled with water. Initially, the air above the water is at atmospheric pressure (1.01 × 10 5 Pa) and the gauge pressure at the bottom of the water is 2500 Pa. Then additional air is pumped in, increasing the pressure of the air above the water by 1500 Pa. (a) What is the gauge pressure at the bottom of the water? (b) By how much must the water level in the container be reduced, by drawing some water out through a valve at the bottom of the container, to return the gauge pressure at the bottom of the water to its original value of 2500 Pa? The pressure of the air above the water is maintained at 1500 Pa above atmospheric pressure.
A closed container is partially filled with water. Initially, the air above the water is at atmospheric pressure (1.01 × 105 Pa) and the gauge pressure at the bottom of the water is 2500 Pa. Then additional air is pumped in, increasing the pressure of the air above the water by 1500 Pa. (a) What is the gauge pressure at the bottom of the water? (b) By how much must the water level in the container be reduced, by drawing some water out through a valve at the bottom of the container, to return the gauge pressure at the bottom of the water to its original value of 2500 Pa? The pressure of the air above the water is maintained at 1500 Pa above atmospheric pressure.
Need help on the following questions on biomechanics. (Please refer to images below)A gymnast weighing 68 kg attempts a handstand using only one arm. He plants his handat an angle resulting in the reaction force shown.A) Find the resultant force (acting on the Center of Mass)B) Find the resultant moment (acting on the Center of Mass)C) Draw the resultant force and moment about the center of mass on the figure below. Will the gymnast rotate, translate, or both? And in which direction?
Please help me on the following question (Please refer to image below)An Olympic lifter (m = 103kg) is holding a lift with a mass of 350 kg. The barexerts a purely vertical force that is equally distributed between both hands. Each arm has amass of 9 kg, are 0.8m long and form a 40° angle with the horizontal. The CoM for each armis 0.5 m from hand. Assuming the lifter is facing us in the diagram below, his right deltoidinserts 14cm from the shoulder at an angle of 13° counter-clockwise from the humerus.A) You are interested in calculating the force in the right deltoid. Draw a free body diagramof the right arm including the external forces, joint reaction forces, a coordinate system andstate your assumptions.B) Find the force exerted by the right deltoidC) Find the shoulder joint contact force. Report your answer using the magnitude and directionof the shoulder force vector.
Chapter 12 Solutions
University Physics with Modern Physics, Volume 2 (Chs. 21-37); Mastering Physics with Pearson eText -- ValuePack Access Card (14th Edition)
Applications and Investigations in Earth Science (9th Edition)
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