Question

TV broadcast antennas are the tallest artificial structures on Earth. In 1987, a 72.0-kg physicist placed himself and 400 kg of equipment at the top of one 610-m high antenna to perform gravity experiments. By how much was the antenna compressed, if we consider it to be equivalent to a steel cylinder 0.150 , {m} in radius?

A. 2.63 , {mm}
B. 3.45 , {mm}
C. 4.21 , {mm}
D. 5.12 , {mm}

Answer 1

The antenna was compressed by approximately 2.63 mm.

Step-by-step explanation:

To calculate the compression of the antenna, we can use the formula for the strain of a cylinder: strain = (F * L) / (A * Y), where F is the force applied, L is the original length, A is the cross-sectional area, and Y is the Young's modulus. In this case, the force applied is the weight of the physicist and equipment, which is equal to (72 kg + 400 kg) * 9.8 m/s^2 = 4704 N. The original length is 610 m, the cross-sectional area can be calculated as A = π * r^2 = π * (0.150 m)^2, and the Young's modulus for steel is approximately 2 × 10^11 N/m^2. Plugging these values into the formula, we get strain = (4704 N * 610 m) / (π * (0.150 m)^2 * 2 × 10^11 N/m^2). Solving for strain gives us a value of approximately 4.207 × 10^-6.

To find the compression, we can multiply the strain by the original length of the antenna: compression = strain * L = (4.207 × 10^-6) * 610 m. Plugging in the values, we get a compression of approximately 2.564 mm. Therefore, the correct answer is A. 2.63 mm.