Respuesta :
Answer:
Assuming pressure is held constant,the question reduces to a ratio and proportion type of question where;
At constant pressure,
21.0ft³-55.0°F
11.0fy³=11.0ft³/21.0ft³×55°F
The temperature is 28.809°F≈29°F
29°F is the temperature of the helium if the volume of the balloon decreases to 11.0ft³.
What is temperature?
Temperature expresses hotness and coldness or a measure of the average kinetic energy of the atoms or molecules in the system.
Assuming pressure is held constant, the question reduces to a ratio and proportion type of question where;
At constant pressure,
21.0ft³- 55.0°F
11.0fy³ = 11.0ft³ ÷ 21.0ft³ × 55°F
The temperature is 28.809°F ≈ 29°F
Hence, 29°F is the temperature of the helium if the volume of the balloon decreases to 11.0ft³.
Learn more about temperature here:
brainly.com/question/28004147
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