An object with a huge mass also must have a huge
(a) weight.
(b) volume.
(c) size.
(d) surface area.
The correct option for the statement "An object with a huge mass also must have a huge".
Answer to Problem 4RAT
The correct option for the statement "An object with a huge mass also must have a huge" is option (a).
Explanation of Solution
The weight of any substance is a function of mass and gravitational acceleration acts on it. The weight of a body can be calculated by the product of its mass and acceleration due to gravity of the planet. The magnitude of gravitational acceleration is constant at a particular planet.
Thus, it is concluded that if the mass of an object is huge, its weight is also huge. In contrast, the volume, the size and the surface area majorly depend upon the density of the substance. From this, we can conclude that option (a) will be correct in the given options.
Conclusion:
Therefore, the correct option for the statement "An object with a huge mass also must have a huge" is option (a).
Chapter 3 Solutions
Conceptual Physical Science Explorations
Additional Science Textbook Solutions
Campbell Biology in Focus (2nd Edition)
Applications and Investigations in Earth Science (9th Edition)
Organic Chemistry (8th Edition)
Physics for Scientists and Engineers: A Strategic Approach, Vol. 1 (Chs 1-21) (4th Edition)
Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)
Human Physiology: An Integrated Approach (8th Edition)
- What is the resistance (in (2) of a 27.5 m long piece of 17 gauge copper wire having a 1.150 mm diameter? 0.445 ΧΩarrow_forwardFind the ratio of the diameter of silver to iron wire, if they have the same resistance per unit length (as they might in household wiring). d. Ag dFe = 2.47 ×arrow_forwardFind the ratio of the diameter of silver to iron wire, if they have the same resistance per unit length (as they might in household wiring). d Ag = 2.51 dFe ×arrow_forward
- Show that the units 1 v2/Q = 1 W, as implied by the equation P = V²/R. Starting with the equation P = V²/R, we can get an expression for a watt in terms of voltage and resistance. The units for voltage, V, are equivalent to [? v2 v2 A, are equivalent to J/C ✓ X . Therefore, 1 = 1 = 1 A V1 J/s Ω V-A X = 1 W. . The units for resistance, Q, are equivalent to ? The units for current,arrow_forwardPlease solve and answer the question correctly please. Thank you!!arrow_forwardPlease solve and answer the question correctly please. Thank you!!arrow_forward
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningUniversity Physics (14th Edition)PhysicsISBN:9780133969290Author:Hugh D. Young, Roger A. FreedmanPublisher:PEARSONIntroduction To Quantum MechanicsPhysicsISBN:9781107189638Author:Griffiths, David J., Schroeter, Darrell F.Publisher:Cambridge University Press
- Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningLecture- Tutorials for Introductory AstronomyPhysicsISBN:9780321820464Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina BrissendenPublisher:Addison-WesleyCollege Physics: A Strategic Approach (4th Editio...PhysicsISBN:9780134609034Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart FieldPublisher:PEARSON