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physics 十萬火急! 20分!
A narrow glass tube, sealed at one end, has a column of airenclosed by a thread of mercury120 mm long. When the tube is held with its open end uppermost, the air column is 300 mm long. When the tube is inverted, the air column is 420mm long. Given that the density of mercury 1.36 x 10^4 kg m^-3. Calculatea) the... 顯示更多 A narrow glass tube, sealed at one end, has a column of airenclosed by a thread of mercury120 mm long. When the tube is held with its open end uppermost, the air column is 300 mm long. When the tube is inverted, the air column is 420mm long. Given that the density of mercury 1.36 x 10^4 kg m^-3. Calculate a) the atmospheric pressure, b) the length of air column when the tube is held horizontally. 更新: 我係中5學生 未教mmHg 呢樣 老師說: Pressure of gas inside column x area = mg of mercury + atmospheric pressure ( pressure outside) x area 但我不明白
最佳解答:
(a) Let the atmospheric pressure be Po Pressure of air column when the tube is upright = (Po + 120) mmHg Pressure of sir column when the tube is inverted = (Po - 120) mmHg Using Boyle's Law, (Po + 120) x 300A = (Po-120) x 420A where A is the cross-sectional area of the air column hence, Po = 720 mmHg = (720/1000) x 1.36x10^4 x 9,81 N/m^2 = 9.6 x 10^4 N/m^2 [where 9.81 m/s^2 is the acceleration due to gravity g] (b) When the tube is held horizontally, the pressure of the air in the tube is at atmospheric pressure Po. Using Boyle' Law again, Po x L.A = (Po +120) x 300A where L is the length of air column hence, L = 350 mm
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physics 十萬火急! 20分!
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發問:A narrow glass tube, sealed at one end, has a column of airenclosed by a thread of mercury120 mm long. When the tube is held with its open end uppermost, the air column is 300 mm long. When the tube is inverted, the air column is 420mm long. Given that the density of mercury 1.36 x 10^4 kg m^-3. Calculatea) the... 顯示更多 A narrow glass tube, sealed at one end, has a column of airenclosed by a thread of mercury120 mm long. When the tube is held with its open end uppermost, the air column is 300 mm long. When the tube is inverted, the air column is 420mm long. Given that the density of mercury 1.36 x 10^4 kg m^-3. Calculate a) the atmospheric pressure, b) the length of air column when the tube is held horizontally. 更新: 我係中5學生 未教mmHg 呢樣 老師說: Pressure of gas inside column x area = mg of mercury + atmospheric pressure ( pressure outside) x area 但我不明白
最佳解答:
(a) Let the atmospheric pressure be Po Pressure of air column when the tube is upright = (Po + 120) mmHg Pressure of sir column when the tube is inverted = (Po - 120) mmHg Using Boyle's Law, (Po + 120) x 300A = (Po-120) x 420A where A is the cross-sectional area of the air column hence, Po = 720 mmHg = (720/1000) x 1.36x10^4 x 9,81 N/m^2 = 9.6 x 10^4 N/m^2 [where 9.81 m/s^2 is the acceleration due to gravity g] (b) When the tube is held horizontally, the pressure of the air in the tube is at atmospheric pressure Po. Using Boyle' Law again, Po x L.A = (Po +120) x 300A where L is the length of air column hence, L = 350 mm
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