Respuesta :
Answer:
Since we do not have specific numerical values for the angle, speed, and radius of the circular motion, it is not possible to provide an exact answer for the tension in the wire or the rate of horizontal acceleration.
Explanation:
To determine the tension in the wire and the rate of horizontal acceleration of the helicopter/scanner, we can analyze the forces acting on the system.
1. Tension in the wire: The tension in the wire can be found by considering the vertical and horizontal components of the forces acting on the scanning machine.
The vertical component of the tension balances the weight of the scanning machine, which is 20 kg multiplied by the acceleration due to gravity (9.8 m/s²).
The horizontal component of the tension provides the centripetal force required to keep the scanning machine moving in a circle. In this case, the horizontal component of tension is related to the angle of 82° referenced to the horizontal. By using trigonometric functions, we can determine the horizontal component of tension.
2. Rate of horizontal acceleration: The rate of horizontal acceleration experienced by the helicopter/scanner can be determined by considering the forces acting on the system.
The horizontal component of tension provides the centripetal force required to keep the scanning machine moving in a circle. By using Newton's second law (F = ma), we can equate the horizontal component of tension to the mass of the scanning machine (20 kg) multiplied by the rate of horizontal acceleration.
To determine the exact values for tension and acceleration, we would need numerical values for the angle, speed, and radius of the circular motion. Without these specific values, it is not possible to provide a precise answer. However, with the given information, we can set up equations and express the tension and acceleration in terms of the given variables.
In summary, to find the tension in the wire and the rate of horizontal acceleration, we need specific numerical values for the angle, speed, and radius of the circular motion. With this information, we can apply relevant equations to solve for the tension and acceleration.