A metal rod and a metal ball more submerged in H2O

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SUMMARY

The discussion focuses on calculating the acceleration of a metal ball submerged in water when the temperature of the water is increased from 30 K to 65 K. Given the specific heat capacities of the materials involved, the volume and mass of the metal rod and ball, and the thermal expansion coefficient, the acceleration of the ball is determined to be 1.8 x 10^10 m/s². The initial length of the rod is 6 m, with the ball having a volume of 6 cm³ and a mass of 60 g, while the rod has a volume of 9 cm³ and a mass of 607 g.

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  • Understanding of hydrostatics and buoyancy principles
  • Knowledge of specific heat capacity and thermal expansion
  • Familiarity with basic physics equations for acceleration
  • Ability to perform unit conversions and calculations involving mass and volume
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  • Learn about specific heat capacity calculations in thermal physics
  • Explore the effects of temperature changes on material properties
  • Investigate the relationship between thermal expansion and acceleration in fluids
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of submerged objects in varying temperatures, particularly in hydrostatic contexts.

Megaevelie
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A metal rod and a metal ball more submerged in h2o, such that one end of the rod is adjacent to the ball. if the h2o will be heated from 30 k to 65 k, what is the acceleration of the ball? the initial length of the rod is 6 m.

Vball= 6 cm3
Vrod= 9 cm3
Crod= 13 cal/g.c degrees
Cball= 8 cal/g.c degrees
Ch2o= 1.0 cal/g.c degrees
Mball= 60 g
Mrod= 607 g
α= 0.005/c degrees

Topic is about Hydrostatic

I was absent during the assigning of this homework. I really need help! Thank you! :)
 
Last edited:
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Answer is 1.8 x 10^10
 

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