3. a cube made of an unknown material has a dimension of 6 cm on all edges. if the upper face is displaced…

3. a cube made of an unknown material has a dimension of 6 cm on all edges. if the upper face is displaced laterally by 0.1 mm by a force of 2 x 10^6 dynes acting parallel to the surface, calculate the shear modulus of the material.\n4. if the density of seawater is 1.03 g/cm^3 at the surface, what is its density at a depth where the pressure is 10^9 dynes/cm^2?

3. a cube made of an unknown material has a dimension of 6 cm on all edges. if the upper face is displaced laterally by 0.1 mm by a force of 2 x 10^6 dynes acting parallel to the surface, calculate the shear modulus of the material.\n4. if the density of seawater is 1.03 g/cm^3 at the surface, what is its density at a depth where the pressure is 10^9 dynes/cm^2?

Answer

Answer:

  1. (2\times10^{10}\ dyne/cm^{2})
  2. Cannot be determined with the given information

Explanation:

Step1: Calculate shear strain

Shear strain (\gamma=\frac{\Delta x}{h}), where (\Delta x = 0.1\ mm=0.01\ cm) and (h = 6\ cm). So (\gamma=\frac{0.01}{6}).

Step2: Calculate shear stress

Shear stress (\tau=\frac{F}{A}), (F = 2\times 10^{6}\ dyne), (A=6\times6\ cm^{2}= 36\ cm^{2}), so (\tau=\frac{2\times 10^{6}}{36}\ dyne/cm^{2}).

Step3: Calculate shear modulus

Shear modulus (G=\frac{\tau}{\gamma}), substituting (\tau) and (\gamma) values: (G=\frac{\frac{2\times 10^{6}}{36}}{\frac{0.01}{6}}=\frac{2\times 10^{6}\times6}{36\times0.01}= 2\times10^{10}\ dyne/cm^{2}).

Step4: Analyze density - depth problem

To find density at a depth given surface - density and pressure, we need the bulk - modulus of the material. Since the bulk - modulus is not given, we cannot calculate the density at the depth with the information provided.