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1 answer:
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Explanation:
Draw a free body diagram for each disc.
Disc A has three forces acting on it: 86.5 N up, T₁ down, and Wa down.
∑F = ma
86.5 N − T₁ − Wa = 0
Wa = 86.5 N − T₁
ma × 9.8 m/s² = 86.5 N − 55.6 N
ma = 3.2 kg
Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.
∑F = ma
T₁ − T₂ − Wb = 0
Wb = T₁ − T₂
mb × 9.8 m/s² = 55.6 N − 36.5 N
mb = 1.9 kg
Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.
∑F = ma
T₂ − T₃ − Wc = 0
Wc = T₂ − T₃
mc × 9.8 m/s² = 36.5 N − 9.6 N
mc = 2.7 kg
Disc D has two forces acting on it: T₃ up and Wd down.
∑F = ma
T₃ − Wd = 0
Wd = T₃
md × 9.8 m/s² = 9.6 N
md = 0.98 kg
Answer:
0.273 liters are needed to accomplish this task without boiling.
Explanation:
The minimum boiling point of kerosene is . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.
If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:
(1)
Where:
- Density of kerosene, measured in kilograms per cubic meter.
- Volume of kerosene, measured in cubic meters.
,
- Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.
,
- Initial temperatures of kerosene and tin, measured in degrees Celsius.
- Final temperatures of the kerosene-tin system, measured in degrees Celsius.
Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.
From (1), we clear the volume of kerosene:
If we know that ,
,
,
,
,
and
, then the volume of the liquid needed to accomplish this task without boiling is:
0.273 liters are needed to accomplish this task without boiling.