## Physics: Principles with Applications (7th Edition)

a. $1.29 \times 10^5 N/m^2$ b. $9.7 \times 10^4 N/m^2$
The pressure in the tank is atmospheric pressure, plus the pressure difference indicated by the column of mercury. Use equation 10–3c for the open-tube manometer. Note that 1040 mbar = $1.04 \times 10^5 N/m^2$. a. $P_t=1.04 \times 10^5 N/m^2+(13.6 \times 10^3 kg/m^3)(9.80 m/s^2)(0.185m)=1.29 \times 10^5 N/m^2$ b. $P_t=1.04 \times 10^5 N/m^2+(13.6 \times 10^3 kg/m^3)(9.80 m/s^2)(-0.056m)=9.7 \times 10^4 N/m^2$