60000 Pa + 1000 kg/m3 /2 ( (2.547 m/s) ^2 - (10.20 m/s) ^2)

Let's break this down and solve the expression step by step.

The given expression is:

\[
60000 \, \text{Pa} + \frac{1000 \, \text{kg/m}^3}{2} \left( (2.547 \, \text{m/s})^2 - (10.20 \, \text{m/s})^2 \right)
\]

### Step 1: Square the velocities
First, we square the velocities \(2.547 \, \text{m/s}\) and \(10.20 \, \text{m/s}\):

\[
(2.547)^2 = 6.485809 \, \text{m}^2/\text{s}^2
\]
\[
(10.20)^2 = 104.04 \, \text{m}^2/\text{s}^2
\]

### Step 2: Subtract the squared values
Now, subtract the results:

\[
6.485809 - 104.04 = -97.554191 \, \text{m}^2/\text{s}^2
\]

### Step 3: Multiply by \( \frac{1000 \, \text{kg/m}^3}{2} \)
Next, multiply by the factor \( \frac{1000 \, \text{kg/m}^3}{2} \):

\[
\frac{1000}{2} = 500 \, \text{kg/m}^3
\]
\[
500 \times (-97.554191) = -48777.0955 \, \text{Pa}
\]

### Step 4: Add 60000 Pa
Finally, add \(60000 \, \text{Pa}\) to the result:

\[
60000 + (-48777.0955) = 11222.9045 \, \text{Pa}
\]

### Final Answer:
\[
\boxed{11222.90 \, \text{Pa}}
\]

Would you like more details or have any questions on the steps?

Here are some related questions to explore:

1. How does changing the velocity affect the pressure in this equation?
2. What does the term \( \frac{1}{2} \rho v^2 \) represent in fluid dynamics?
3. If the velocity \( v_2 \) was unknown, how could you solve for it?
4. What units must all terms have to maintain dimensional consistency in this type of equation?
5. How would the result change if the density \( \rho \) were increased?

**Tip:** When dealing with fluid dynamics, terms involving \( v^2 \) often come from energy or pressure equations like Bernoulli's principle.

Explore a detailed solution to the problem involving pressure and velocity changes using Bernoulli's principle. This problem covers key fluid dynamics concepts, such as kinetic energy and pressure, with step-by-step calculations. Suitable for high school and college-level students.

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