13. A 0.200-kg aluminum plate, initially at 20°C, slides down a 13.0-m-lon to th

Question

13. A 0.200-kg aluminum plate, initially at 20°C, slides down a 13.0-m-lon to the horizontal. The force of kinetic friction exactly balances the component of gravity down the plane so that the plate once started, glides down at constant velocity. If 90% of the mechanical energy of the system is absorbed by the aluminum is 900 J/kg)°C) g surface, inclined at a 25.0° angle aluminum, what is its temperature increase at the bottom of the incline? (Specific heat for a. 0.115 C b. 0.0538 C c. 0.269 Co d. 0.78 Co e. 48.5 C ANSWER: B A waterfall is 128 m high. What is the increase in water temperature at the bottom of the falls if all the initial potential energy goes into heating the water? -9.8 m's'.ew J/kg°C) a, 0.15°C b. 0.30°C c. 0.90°C d. 0.37 C e. 0.60°C NSWER: b 15. What is the temperature increase of 3.00 kg of water when heated by a 900-W immersion heater for 9.00 min? (cw 4 1 86 J/kg-PC) a. 77.4°C b. 48.4°C c. 38.7 d. 19.4 e. 0.645 ANSWER: A slice of bread contains about 125 kcal. If specific heat of a person were 1.00 kcal/kg C, by hovw many °C would the temperature of a 65.0-kg person increase if all the energy in the bread were converted to heat? 16. a. 3.85°C b. 0.96°C c. 1.92 d. 2.40°C e. 1.44 ANSWER: When heating 2 kg of ice from-40°C,2 kg of water from 20°C to 50°C, and 2 kg of steam from 100°C to 125°C, which

Explanation / Answer

13) height of object

Sin theta = perpendicula / Hypotenous

Sin (25) = height / hypo.

0.422618262 = height / 13

height = 5.494 m

Energy = mgh = .200 x 9.8 x 5.494 = 10.76 J

90 % of energy = 10.76 x 90 / 100 = 9.684 J [= Q]

Q = mC del T

del T = Q / mC

del T = 9.684 / (.200 x 900)

del T = 0.0538 degree (Option B)

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