{"id":12127,"date":"2023-02-27T09:17:58","date_gmt":"2023-02-27T09:17:58","guid":{"rendered":"https:\/\/www.greatassignmenthelp.com\/questions\/?p=12127"},"modified":"2023-12-21T05:59:17","modified_gmt":"2023-12-21T05:59:17","slug":"assume-patmos1-00atm-what-is-the-gas-pressure-pgas","status":"publish","type":"post","link":"https:\/\/www.greatassignmenthelp.com\/questions\/assume-patmos1-00atm-what-is-the-gas-pressure-pgas\/","title":{"rendered":"Assume patmos=1.00atm. What is the gas pressure Pgas?"},"content":{"rendered":"<p>Express your answer in pascals to three significant figures. h1=13.5 cm h2=6.00 cm mercury of density= 1.36\u00d710 4 kg\/m 3 Solve the following differential equation: y&#8221; + 4y = sin^3(x) Find answer of form: y(x)=________+C1________+C2________ .<\/p>\n<p><strong>Answer:-\u00a0<\/strong>Well, let&#8217;s start by finding the pressure due to the &#8220;extra&#8221; height of the mercury. p = 1.36e4 kg\/m\u00c2\u0142 \u00c2\u00b7 (0.105m &#8211; 0.05m) \u00c2\u00b7 9.8m\/s\u00c2\u02db = 7330 N\/m\u00c2\u02db = 7330 Pa The pressure at B is clearly p_b = p_atmos = p_gas + 7330 Pa The pressure at A is p_a = p_gas = p_atmos &#8211; 7330 Pa c) 1 atm = 101 325 Pa Then p_gas = 101325 Pa &#8211; 7330 Pa = 93 995 Pa<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Express your answer in pascals to three significant figures. h1=13.5 cm h2=6.00 cm mercury of density= 1.36\u00d710 4 kg\/m 3 Solve the following differential equation: [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[27],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v23.5 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Assume patmos=1.00atm. 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