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Unformatted text preview: fnal mass oF liquid water= initial mass o liq- m= 0.5kg - 0.071kg= 0.429kg Problem 2) Car A will have a larger compression on the spring, since both cars have the same speed but dierent masses and mass A> mass B => KE ( car A) > KE ( car B), where as we know KE = 1/2 m v . And considering the system is closed, whichever car has greater change in KE will have the greater change in PE as well, so PE (car A) > PE (car B). With PEs = 1/2 k x , which means that x ( car A) > x (car B). Ice Water Water Eb m (liq) Eth T Eth T Ti= -27C Ti= 10 C mi= 0.5kg T= 0C T= 0 C m= 0.429kg...
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This note was uploaded on 03/27/2011 for the course PHY 7A taught by Professor Pardini during the Winter '08 term at UC Davis.
- Winter '08