Question 1195308
**a) Cross-sectional Area of the Tape**

* **Find the volume of the tape:**
    * Volume = Weight of tape / Density of steel
    * Volume = 2.2 lbs / (1 lb / 3.53 in³) = 7.766 in³

* **Find the cross-sectional area:**
    * Area = Volume / Length 
    * Area = 7.766 in³ / 100 ft * (12 in/ft) = 0.0647 in²

**b) Correction for Increase in Tension per Tape Length**

* **Formula:** 
    * Correction per unit length = (P - P₀) / (A * E) 
        * where:
            * P = Applied tension during measurement (15 lbs)
            * P₀ = Standard tension (12 lbs)
            * A = Cross-sectional area of the tape (0.0647 in²)
            * E = Modulus of elasticity of steel (29 x 10⁶ psi)

* **Calculate:**
    * Correction per unit length = (15 lbs - 12 lbs) / (0.0647 in² * 29 x 10⁶ psi)
    * Correction per unit length = 3 lbs / 1871300 psi 
    * Correction per unit length = 1.604 x 10⁻⁶ in/in

**c) Correction for Increase in Tension - Whole Length**

* **Total Correction:** Correction per unit length * Measured length
    * Total Correction = 1.604 x 10⁻⁶ in/in * 1000 ft * (12 in/ft) 
    * Total Correction = 0.0192 in

**d) Corrected Length for the Effect of Increased Tension**

* **Corrected Length:** Measured length - Correction for tension
    * Corrected Length = 1000.00 ft - (0.0192 in / 12 in/ft) 
    * Corrected Length = 1000.00 ft - 0.0016 ft 
    * Corrected Length = 999.9984 ft

**Therefore:**

* a) Cross-sectional area of the tape: 0.0647 in²
* b) Correction for increase in tension per tape length: 1.604 x 10⁻⁶ in/in
* c) Correction for increase in tension - whole length: 0.0192 in
* d) Corrected length for the effect of increased tension: 999.9984 ft

**Note:**

* This calculation assumes that the tape is perfectly elastic within the range of applied tensions.
* Other factors, such as temperature and sag, can also affect the accuracy of the measurement and may need to be considered for more precise results.