If the work required to stretch a spring 3 ft beyond its natural length is 6 ft-lb, how much work is needed to stretch it 9 in. beyond its natural length?

3/8

Step-by-step explanation:

Given that:

Work required to stretch a string:

3 ft = 6ft - lb

The work required to stretch it 9 inches

Given that :

Work(W) = kx

Take the integral of W at 3 fts beyond its natural length;

W = ∫kx dx at 0 to 3

W = k ∫xdx at 0 to 3

W = 6

W = k[x²/2] at x =0 to x = 3

6 = k[3² /2] - 0

6 = k[9/2]

12 = 9k

k = 12/9 = 4/3 = 1.333

Converting inches to feet:

1 inch 0.0833 ft

9 inches = 0.75 ft

W = ∫kx dx at 0.75 to 0

W = 4/3 ∫xdx at 0 to 0.75 = 3/4

W = 4/3[x²/2] at x =0.75 to 0

W = 4/3[(3/4)²/2] at 0.75 to 0

W = 4/3[(9/16) / 2]

W = 4/3 * (9/16 * 1/2)

W = 36/96

W = 6/16 = 3/8 ft - lbs

Accuracy refers to how close the values are to the true result.
Precision refers to how close the values are to each other.

The above example shows that on Sunday, the reading of thermometer was 6 degrees from the actual reading and on the next 4 days the reading was 2 to 5 degrees from the actual reading. Since the measurements are being compared to the actual result and not to each other, the example gives the instant of Accuracy only, not the precision.

So, the answer to this question is option A