Akka sif hin galin beckuun mataan isaa beckumsa! 9. ne constant of oroportionality in Hooke's aw is known as A orce corstant 8 Gravity constan: 0 Medulus of Elasticity D Pressure cunstant 10 for a reid body to be in translationa equilior um, the net external force acting on it must be A. Maximum B. Constant C Zero D. Equal to its weight 11 For a rigid body to be in rotational equilibrium the net external torque acting on it must be: A. Maximum B. Constant C. Minimum D. Zero 12. The first condition of equilibrium states that the vector sum of all external forces must be zero IIF 0). This condition ensures: A. Rotational equilibrium B Stable equilibrium C. Translational equilibrium D. Neutral equilibrium 13. The second condition of equilibrium states that the vector sum of all external torques about any axis must be zero (sum tau =0) This condition ensures A. Translational equilibrium B. Unstable equilibrium C Rotational equilibrium D. Linear motion 14. The point where the entire weight of an object can be considered to act is called the A. Center of mass (center of gravity) B Pivot point C. Equilibrium point D. Fulcrum 15. Which type of equilibrium is experienced by a cone resting on its base? A Unstable equilibrium B Neutral equilibrium C. Stable equilibrium D. None of the above III. Fill in the gaps 21 __ is the boundary between elasticity and plasticity. 22 __ is the state of balance at the opposing force are add up to zero 23. You have a rock with a volume of 15cm3 and a mass of 45 E. What is its density? 24. A circular rod of cross-sectional area 100mm2 has a tensile force of 100 kN applied to it.Calculate the value for the stress in the rod IV. Follow all necessary steps to solve the next problems. 25. A wire is stretched 3 mm by a force of 150 N. Assuming the elastic limit is not exceeded, calculate the force that will stretch the wire 5 mm Hebin yaaluun furmaata!

Multiple Choice **9. The constant of proportionality in Hooke's law is known as:** - **A. Force constant** (Also called the spring constant, denoted by ( k ).) **10. For a rigid body to be in translational equilibrium, the net external force acting on it must be:** - **C. Zero** **11. For a rigid body to be in rotational equilibrium, the net external torque acting on it must be:** - **D. Zero** **12. The first condition of equilibrium states that the vector sum of all external forces must be zero (\(\sum F = 0\)). This condition ensures:** - **C. Translational equilibrium** **13. The second condition of equilibrium states that the vector sum of all external torques about any axis must be zero (\(\sum \tau = 0\)). This condition ensures:** - **C. Rotational equilibrium** **14. The point where the entire weight of an object can be considered to act is called the:** - **A. Center of mass (center of gravity)** **15. Which type of equilibrium is experienced by a cone resting on its base?** - **C. Stable equilibrium** --- ### Fill in the Gaps **21.** **Elastic limit** is the boundary between elasticity and plasticity. **22.** **Equilibrium** is the state of balance at which the opposing forces add up to zero. **23.** You have a rock with a volume of \(15\,cm^3\) and a mass of 45 g. What is its density? \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{45\,g}{15\,cm^3} = 3\,g/cm^3 \] **24.** A circular rod of cross-sectional area \(100\,mm^2\) has a tensile force of 100 kN applied to it. Calculate the value for the stress in the rod. First, convert units: - Area: \(100\,mm^2 = 100 \times 10^{-6}\,m^2 = 1 \times 10^{-4}\,m^2\) - Force: ( \(100\,kN = 100,000\,N\) ) \[ \text{Stress} = \frac{\text{Force}}{\text{Area}} = \frac{100,000\,N}{1 \times 10^{-4}\,m^2} = 1,000,000,000\,N/m^2 = 1\,GPa \] --- ### Problem Solving **25.** A wire is stretched 3 mm by a force of 150 N. Assuming the elastic limit is not exceeded, calculate the force that will stretch the wire 5 mm. Hooke’s Law: \(F \propto x\), so \(\frac{F_1}{x_1} = \frac{F_2}{x_2}\) Given: - \(F_1 = 150\,N\) - \(x_1 = 3\,mm\) - \(x_2 = 5\,mm\) - \(F_2 = ?\) \[ \frac{150}{3} = \frac{F_2}{5} \] \[ 50 = \frac{F_2}{5} \] \[ F_2 = 50 \times 5 = 250\,N \] **Final Answer:** The force required to stretch the wire 5 mm is **250 N**. --- If you need further explanations or clarifications, feel free to ask!