
We fully submerge an irregular 3 kg lump of material in a certain fluid. The fluid that would have been in the space now occupied by the lump has a mass of 2 kg. (a) When we release the lump, does it move upward, move downward, or remain in place? (b) If we next fully submerge the lump in a less dense fluid and again release it, what does it do?

To find:
a) Movement of block when we release the irregular lump in a certain fluid.
b) Movement of block when we release the lump in a less dense fluid.
Explanation of Solution
1) Concept:
To float the object in fluid, buoyant force must be greater than or equal to the weight of object.
2) Formulae:
Buoyant force
3) Given:
Mass of irregular submerged lump
Mass of fluid
4) Calculations:
We have
Buoyant force
Weight of lump
We know that to float an object in a fluid, the buoyant force must be greater than or equal to the weight of the object.
i.e.
a) As the mass of the submerged lump is greater than the mass of fluid, when the lump is released, it moves downwards.
b) If another fluid has less density, buoyant force will be less. Hence, the lump will move downwards due to greater weight than buoyant force.
Conclusion:
By comparing the buoyant force and the weight of object for each case, we can conclude whether an irregular lump of material floats or moves downwards.
Want to see more full solutions like this?
Chapter 14 Solutions
Fundamentals of Physics Extended
Additional Science Textbook Solutions
Chemistry: Structure and Properties (2nd Edition)
Microbiology with Diseases by Body System (5th Edition)
Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)
Anatomy & Physiology (6th Edition)
Organic Chemistry (8th Edition)
Human Biology: Concepts and Current Issues (8th Edition)
- An object is placed 24.1 cm to the left of a diverging lens (f = -6.51 cm). A concave mirror (f= 14.8 cm) is placed 30.2 cm to the right of the lens to form an image of the first image formed by the lens. Find the final image distance, measured relative to the mirror. (b) Is the final image real or virtual? (c) Is the final image upright or inverted with respect to the original object?arrow_forwardConcept Simulation 26.4 provides the option of exploring the ray diagram that applies to this problem. The distance between an object and its image formed by a diverging lens is 5.90 cm. The focal length of the lens is -2.60 cm. Find (a) the image distance and (b) the object distance.arrow_forwardPls help ASAParrow_forward
- Physics 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 LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- University Physics Volume 1PhysicsISBN:9781938168277Author:William Moebs, Samuel J. Ling, Jeff SannyPublisher:OpenStax - Rice UniversityPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning





