What is a "correction factor" wrt this problem below?

A surveyor has a steel measuring tape that is calibrated to be 100.000 m long (i.e., accurate to ±1 mm ) at 20 ∘C.

A)  If she measures the distance between two stakes to be 75.175 m on a 3 ∘C day, does she need to add or subtract a correction factor to get the true distance?

So, I realize steel will shrink when it's colder, so she'll have to subtract. But, for part B, I don't even know what a correction factor is, so, what?

B)  How large, in mm, is the correction factor?

2 Answers

Relevance
  • Dixon
    Lv 7
    4 weeks ago
    Favorite Answer

    You need to investigate how much steel contracts between 20 and 3 Celcius. It will be probably given as some factor (ie ratio) per degree C over a reasonable working range of temperature, ie it will be linearised even if it isn't truly linear.

    • Login to reply the answers
  • Anonymous
    4 weeks ago

    If it's colder than the tape contracts so she had to add a correction factor.

    • Login to reply the answers
Still have questions? Get your answers by asking now.