Compare the Young's modulus E∗ of two open-cell foams. Which Young's modulus is higher? Recall that C1=1.

1. polystyrene: (Es∼3.5GPa), ρ∗/ρs=0.3
2. Al 7075-T651: (Es∼72GPa), ρ∗/ρs=0.1

To compare the Young's modulus (\(E^*\)) of the two open-cell foams, we need to use the given equations and information.

Young's modulus is defined as the ratio of stress to strain within the elastic limit of a material. In this case, we have two foams: polystyrene and Al 7075-T651. We are given the values of Young's modulus (\(E_s\)) for each material.

Let's compare the two foams using the given information:

1. Polystyrene:
\(E_s \approx 3.5 \, \text{GPa}\)
\( \frac{\rho^*}{\rho_s} = 0.3\)

2. Al 7075-T651:
\(E_s \approx 72 \, \text{GPa}\)
\( \frac{\rho^*}{\rho_s} = 0.1\)

To compare the Young's moduli, we need to find the ratio of \(\frac{E^*}{E_s}\) for both foams.

For polystyrene:
\(\frac{E^*}{E_s} = C_1 \times \left( \frac{\rho^*}{\rho_s} \right) \times \left( \frac{E^*}{E_s} \right)\)
As C1 = 1, the equation simplifies to:
\(\frac{E^*}{E_s} = \frac{\rho^*}{\rho_s}\)

We can substitute the given values and solve for \(\frac{E^*}{E_s}\) for polystyrene:
\(\frac{E^*}{3.5 \times 10^9 \, \text{Pa}} = 0.3\)

Simplifying, we get:
\(E^* = 0.3 \times 3.5 \times 10^9 \, \text{Pa}\)
\(E^* = 1.05 \times 10^9 \, \text{Pa}\)

Similarly, for Al 7075-T651 foam:
\(\frac{E^*}{72 \times 10^9 \, \text{Pa}} = 0.1\)

Simplifying, we get:
\(E^* = 0.1 \times 72 \times 10^9 \, \text{Pa}\)
\(E^* = 7.2 \times 10^9 \, \text{Pa}\)

Comparing the values of \(E^*\) for both foams, we can see that the Young's modulus for Al 7075-T651 foam (7.2 GPa) is higher than that of polystyrene foam (1.05 GPa).

Therefore, the higher Young's modulus is found in Al 7075-T651 foam.