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Question Number 123336 by Lordose last updated on 25/Nov/20

$$$$$$\mathrm{A}\mathrm{mercury}\mathrm{based}\mathrm{barometer}\mathrm{reads}$$$$760\mathrm{mmhg}\mathrm{at}\mathrm{the}\mathrm{base}\mathrm{of}\mathrm{a}\mathrm{mountain}\mathrm{and}$$$$704.0\mathrm{mmhg}\mathrm{at}\mathrm{the}\mathrm{top}.\mathrm{Find}\mathrm{the}\mathrm{height}$$$$\mathrm{of}\mathrm{the}\mathrm{mountain}\mathrm{if}\mathrm{the}\mathrm{density}\mathrm{of}$$$$\mathrm{mercury}\mathrm{and}\mathrm{air}\mathrm{are}\mathrm{respectively}$$$${13600\mathrm{kgm}}^{-3}\mathrm{and}1.{25\mathrm{kgm}}^{-3}(\mathrm{g}={10\mathrm{ms}}^{-2})$$

Answered by help last updated on 25/Nov/20

$${\left\{pgh\right\}}_1={\left\{pgh\right\}}_2$$$$13.6\{76-70.4\}=0.00125\times h$$$$h=??$$

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