A square plate with edge length 9.10 cm and mass 488 g is hinged along one side. If air is blown over the upper surface only, what speed must the air have to hold the plate horizontal? The air has density 1.21 kg/m³.

To solve this problem, we use Bernoulli’s principle and the concept of torque equilibrium.

### Step 1: Understanding the Forces Acting on the Plate
- The square plate has an edge length of **L = 9.10 cm = 0.0910 m**.
- The mass of the plate is **m = 488 g = 0.488 kg**.
- The air is blown over the upper surface only, creating a pressure difference between the top and bottom.
- The density of air is **ρ_air = 1.21 kg/m³**.
- The plate is hinged along one side, so it will stay horizontal when the torque about the hinge due to the weight of the plate is balanced by the torque due to the pressure difference.

### Step 2: Applying Bernoulli’s Principle
The Bernoulli equation for the air above the plate and the stationary air below the plate is:

\[
P_{\text{top}} + \frac{1}{2} \rho v^2 = P_{\text{bottom}}
\]

Since the bottom air is stationary, we assume its velocity is zero. Thus, the pressure difference is:

\[
\Delta P = P_{\text{bottom}} - P_{\text{top}} = \frac{1}{2} \rho v^2
\]

This pressure difference creates an upward force on the plate.

### Step 3: Finding the Lift Force
The force due to the pressure difference is:

\[
F = \Delta P \times A
\]

where the area of the plate is:

\[
A = L^2 = (0.0910)^2 = 0.008281 \text{ m}^2
\]

Thus,

\[
F = \left(\frac{1}{2} \rho v^2\right) \times A
\]

Substituting values,

\[
F = \left(\frac{1}{2} \times 1.21 \times v^2\right) \times 0.008281
\]

\[
F = 0.004999 v^2
\]

### Step 4: Torque Balance
For the plate to remain horizontal, the torque due to the weight about the hinge must equal the torque due to the lift force.

- The weight of the plate acts at its center of mass, which is at a distance **L/2 = 0.0910/2 = 0.0455 m** from the hinge.
- The lift force acts uniformly over the plate, so its effective point of action is also at **L/2 = 0.0455 m**.

Setting up the torque balance equation:

\[
m g \left(\frac{L}{2}\right) = F \left(\frac{L}{2}\right)
\]

Canceling **L/2** from both sides:

\[
m g = F
\]

Substituting values:

\[
(0.488)(9.81) = 0.004999 v^2
\]

\[
4.785 = 0.004999 v^2
\]

Solving for **v**:

\[
v^2 = \frac{4.785}{0.004999}
\]

\[
v^2 = 957.2
\]

\[
v = \sqrt{957.2}
\]

\[
v \approx 30.94 \text{ m/s}
\]

### Final Answer:
The air must have a speed of approximately **30.94 m/s** to hold the plate horizontal.