Given:
• Initial velocity = 5 m/s
,
• Final velocity = 10 m/s
,
• Initial pressure = 110 kPa
,
• Density = 749 kg/m³
Let's find the value of the final pressure.
Apply Bernoulli's equation:
![(P_1)/(\rho_g)+(v_1^2)/(2g)=(P_2)/(\rho_g)+(v_2^2)/(2g)](https://img.qammunity.org/2023/formulas/physics/college/x6lcs8r0y00rgvahb2nz7hqtc5x3z1c6ai.png)
Now, simplify the equation:
![(P_1-P_2)/(\rho_g)=(v_2^2-v_1^2)/(2)](https://img.qammunity.org/2023/formulas/physics/college/a6tbw5ku7d9qe4st4ganbpqhot38w3zsx9.png)
Where:
P1 is the initial pressure = 110 kPa
P2 is the final pressure
v2 is the final velocity = 10 m/s
v1 is the initial velocity = 5 m/s
ρg is the density = 749 kg/m³
Plug in values into the equation and solve for the final pressure, p2.
We have:
![\begin{gathered} (110-P_2)/(749)=(10^2-5^2)/(2) \\ \\ (110-P_2)/(749)=(100-25)/(2) \\ \\ (110-P_2)/(749)=37.5 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/w2zge2x2q51ijkqzwje8pcpjzm03vp4uho.png)
Soling further:
![\begin{gathered} 110-P_2=37.5*749 \\ \\ 110-P_2=28087.5\text{ Pa} \\ \\ 110-P_2=28.0875\text{ kPa} \\ \\ P_2=110-28.0875 \\ \\ P_2=81.9\approx82\text{ kPa} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/d8h5w77afv4251bcofls8mn5x4jky8drp4.png)
Therefore, the final pressure is 82 kPa
ANSWER:
d. 82 kPa