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f.3 physics問題一條!!!!急!
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The change in resistance for decrease in temperature is (50 - 20) / (80 - 10) units/℃ = 3/7 unit/℃ Therefore, the resistance of the thermistor at 0℃ (melting ice) is 50 units + 10 * 3/7 units = 54.3 units
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f.3 physics問題一條!!!!急!
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When a thermistor is immersed in water at 10 °C, its resistance is 50 units. When it is immersed in water at 80 °C, its resistance is 20 units. What is its resistance if it is immersed in pure melting ice? Assume that the resistance of the thermistor varies linearly with temperature.ans: 54.3... 顯示更多 When a thermistor is immersed in water at 10 °C, its resistance is 50 units. When it is immersed in water at 80 °C, its resistance is 20 units. What is its resistance if it is immersed in pure melting ice? Assume that the resistance of the thermistor varies linearly with temperature. ans: 54.3 units 我想要睇個step 唔該哂!最佳解答:
The change in resistance for decrease in temperature is (50 - 20) / (80 - 10) units/℃ = 3/7 unit/℃ Therefore, the resistance of the thermistor at 0℃ (melting ice) is 50 units + 10 * 3/7 units = 54.3 units
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