Final answer:
The force exerted by a gymnast upon landing relates to their deceleration and gravity. Using Newton's second law, a 40.0-kg gymnast decelerating at 7*g would exert a significant force. However, the weight of a 46.2 kg gymnast is simply 46.2 kg * 9.81 m/s^2, representing their weight due to Earth's gravitational pull.
Step-by-step explanation:
The question is asking how much force a gymnast weighs while standing stationary on a scale after landing from a height, which is a classic physics problem related to Newton's laws of motion and the concept of deceleration due to gravity. To calculate the force that a gymnast experiences upon landing, we can apply Newton's second law, which states that the force is equal to mass times acceleration (F = m * a). In this case, the acceleration includes both the acceleration due to gravity and the additional deceleration caused by landing on the mat.
To calculate the force exerted by a 40.0-kg gymnast who decelerates at 7 times the acceleration due to gravity (where g = 9.81 m/s2), we use the equation F = m * a, where a is 7*g. Thus, we have F = 40 kg * (7 * 9.81 m/s2). This would give us a force that the gymnast must exert to decelerate safely. However, the weight of the gymnast is asked in the initial question. The weight of any object is the force due to gravity, which is m * g. Therefore, a 46.2 kg gymnast would weigh 46.2 kg * 9.81 m/s2 on Earth.