Question

A cord of mass 0.75 kgkg is stretched between two supports 6.0 mm apart.

Answer 1

The question deals with high school Physics, involving concepts of tension, forces, and material deformation. Utilizing a standard formula for Young's modulus, one can determine the modulus by having the total force applied, the wire's cross-sectional area, the extension due to this force, and the wire's original length.

Step-by-step explanation:

The question appears to be related to the subject of Physics, specifically to the topic of tension, forces, and material deformation. Students are often tasked with calculating tension, elongation, and stress-strain relationships, all of which are foundational concepts in the study of mechanics within physics.

These topics are generally covered in high school physics courses and can also extend into introductory college-level courses.

Example Calculation

Let's take a specific example from the list:

60. Young's modulus calculation. A wire is stretched by an additional force, and the extension is measured. We're given an initial and final force (100 N each addition), which leads to a total of 200 N acting on the wire.

The extension caused by the second 100-N weight is 3.0 mm.

The formula for Young's modulus (E) is E = (F/A) / (ΔL/L0), where F is the force applied, A is the cross-sectional area, ΔL is the change in length, and L0 is the original length.

To find Young's modulus, we need:

The force causing the extension (which is 200 N because it's the sum of the two weights).

The cross-sectional area of the wire, which can be calculated as π * (D/2)², where D is the diameter.

The extension ΔL, which is 3.0 mm.

The original length of the wire L0, which is 2.0 m.

By inserting these values into the formula, the value of Young's modulus for the wire can be calculated.

Answer 2

To find the tensions in the cables and the mass of the equipment, we use the principle of equilibrium and set up equations based on the forces acting on the scaffold. Solving these equations, we find that the tension in the right cable is 392 N, the tension in the left cable is 784 N, and the mass of the equipment is 80 kg.

Step-by-step explanation:

In order to find the tensions in the cables and the mass of the equipment, we can start by determining the forces acting on the scaffold. The weight of the scaffold, painter, and equipment create a downward force of (40.0 kg + 80.0 kg) × 9.8 m/s² = 1176 N.

Since the tension in the left cable is twice that in the right cable, we can let the tension in the right cable be T and the tension in the left cable be 2T. Using the principle of equilibrium, we can write the following equations:

ΣFx = T - 2T = -T = 0 (since the scaffold is in equilibrium horizontally)

ΣFy = 1176 N - T - 2T - 80.0 kg × 9.8 m/s² = 0 (since the scaffold is in equilibrium vertically)

Simplifying the second equation, we find:

T = 392 N

Therefore, the tension in the left cable is 2T = 784 N. And the mass of the equipment is 80 kg.