Q. 15

Expert-verifiedFound in: Page 511

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

A surveyor has a steel measuring tape that is calibrated to be 100.000 m long (i.e., accurate to) at . If she measures the distance between two stakes to be 65.175 m on a day, does she need to add or subtract a correction factor to get the true distance? How large, in mm, is the correction factor?

She needs to subtract a correction factor to get the true distance and the correction factor is mm is 12.18 mm

The process of thermal expansion occurs when objects are heated as well as expanded. In case the temperature of an object changes through , as well as length changes through and the length's fractional changes are effective to the temperature's change

The process of thermal expansion occurs when objects are heated as well as expanded. In case the temperature of an object changes through , as well as length changes through and the length's fractional changes are effective to the temperature's change .

The equation (18.8) in the form is:

(1)

In this case, is the coefficient. Table 18.4 effectively gives the coefficient expression of linear for common materials, for instance

The equation for is

(2)

In this case, the subject of L is the initial length and the temperature effectively changes from

Therefore, the change is temperature is

Thus, plug the value for into equation 2 in order to get

There is a requirement for subtracting the correction factor in terms of getting true distance and the correction factor is same for change in length.

Therefore, correction factor

94% of StudySmarter users get better grades.

Sign up for free