Respuesta :
Short simple answer: The mountain climber stretches her rope by approximately 0.23 cm.
Explanation: To find the amount the rope stretches, we can use Hooke's Law, which states that the amount of stretching or compression of a material is directly proportional to the force applied to it. The formula for Hooke's Law is:[ F = k \cdot \Delta L ]Where:( F ) is the force applied (weight of the climber),( k ) is the spring constant (Young's modulus for the rope),( \Delta L ) is the change in length.We first need to find the force applied by the climber, which is equal to her weight. The weight can be calculated using the formula:[ F = m \cdot g ]Where:( m ) is the mass of the climber (62 kg),( g ) is the acceleration due to gravity (9.8 m/s²).[ F = 62 , \text{kg} \times 9.8 , \text{m/s}^2 = 607.6 , \text{N} ]Now, we can rearrange Hooke's Law to solve for ( \Delta L ):[ \Delta L = \frac{F}{k} ][ \Delta L = \frac{607.6 , \text{N}}{5 \times 10^9 , \text{N/m}^2} ][ \Delta L \approx 0.0012152 , \text{m} ]Converting this to centimeters:[ \Delta L \approx 0.12152 , \text{cm} ]So, the rope stretches by approximately 0.23 cm when the climber hangs 35 m below the rock outcropping.
Explanation: To find the amount the rope stretches, we can use Hooke's Law, which states that the amount of stretching or compression of a material is directly proportional to the force applied to it. The formula for Hooke's Law is:[ F = k \cdot \Delta L ]Where:( F ) is the force applied (weight of the climber),( k ) is the spring constant (Young's modulus for the rope),( \Delta L ) is the change in length.We first need to find the force applied by the climber, which is equal to her weight. The weight can be calculated using the formula:[ F = m \cdot g ]Where:( m ) is the mass of the climber (62 kg),( g ) is the acceleration due to gravity (9.8 m/s²).[ F = 62 , \text{kg} \times 9.8 , \text{m/s}^2 = 607.6 , \text{N} ]Now, we can rearrange Hooke's Law to solve for ( \Delta L ):[ \Delta L = \frac{F}{k} ][ \Delta L = \frac{607.6 , \text{N}}{5 \times 10^9 , \text{N/m}^2} ][ \Delta L \approx 0.0012152 , \text{m} ]Converting this to centimeters:[ \Delta L \approx 0.12152 , \text{cm} ]So, the rope stretches by approximately 0.23 cm when the climber hangs 35 m below the rock outcropping.
