### Educational Text: Analysis of Shear Strain and Deformation Angle**Problem Statement:**A brass nameplate with dimensions \(2.00 \, \text{cm} \times 10.0 \, \text{cm} \times 30.0 \, \text{cm}\) is subjected to a force \( F \) of \(2.00 \times 10^5 \, \text{N}\) acting on the upper left side and the bottom right side, as depicted in the figure.---**Diagram Explanation:**1. **Object:** The diagram shows a blue rectangular nameplate labeled "Anthony."2. **Forces:** Orange arrows labeled \( F \) indicate the direction of the applied forces on opposite sides of the nameplate.3. **Deformation:** - The top edge of the nameplate is displaced horizontally, denoted by \( x \). - The vertical height of the nameplate is labeled as \( h \). - The angle of deformation between the original and displaced states is denoted as \( \phi \).---**Tasks:**1. **Determine the Shear Strain \(\frac{x}{h}\):** The shear strain is the ratio of the horizontal deformation \( x \) of the top edge to the vertical height \( h \) of the nameplate. \[ \frac{x}{h} = \_\_ \]2. **Determine the Angle \(\phi\) of Deformation:** Evaluate the angle indicating the deformation from the original position to the displaced position. \[ \phi = \_\_ \]---In summary, this task involves calculating the shear strain and the angle of deformation due to the applied forces, which can help illustrate the material response under shear stress. A sphere of copper has a radius of 8.00 cm and is compressed uniformly by a force of \(5.00 \times 10^8 \, \text{N}\). The bulk modulus for copper is \(140 \times 10^9 \, \text{N/m}^2\).Calculate the sphere's change in volume \(\Delta V\) after compression.\[\Delta V = \, \boxed{} \, \text{m}^3\]Calculate the sphere’s final radius \(R\).\[R = \, \boxed{} \, \text{m}\]

Given that: The dimension of the brass plate is 2 cm×10 cm ×30 cm The force applied is 2×105 N. The shear modulus of brass is σ=30×109 Nm2 it is required to find the shear strain and the angle of deformation. Since you have asked multiple questions, we will solve the first question for you. If you want any specific question to be solved then please specify the question number or post only that question The area of the plate will be as: A=length ×breadthA=30×2 cm2A=60×10-4 m2 Now the stress can be calculated as: σ=FAσ=2×10560×10-4σ=0.033×109 N/m2 Now we know the relation, σ=σbrr=xh=σσbxh=0.033×10930×109xh=1.1×10-3 The angle of deformation can be calculated as: θ=tan-1xhθ=tan-11.1×10-3θ=0.063° The final answer: The shear stress is 1.1×10-3 . The angle of deformation will be as θ=0.063°.