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Question Number 183767 by Tawa11 last updated on 29/Dec/22

Answered by MikeH last updated on 30/Dec/22

$$P_0=1.3\times {10}^5\mathrm{N}{\mathrm{m}}^{-2}$$$${\mathrm{P}}_{\mathrm{gauge}}=\rho \mathrm{g}h+P_0$$$$\Rightarrow P_{\mathrm{gauge}}=[\left(1000\right)(9.8)\left(10\right)+1.3\times {10}^5]\mathrm{N}{\mathrm{m}}^{-2}$$$$\Rightarrow P_{\mathrm{gauge}}=2.28\times {10}^5\mathrm{N}{\mathrm{m}}^{-2}$$

Commented by Tawa11 last updated on 30/Dec/22

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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