A machinist wishes to insert a steel rod with a diameter of 3 mm into a hole with a diameter of 2.995 mm. By how much would the machinist have to lower the temperature (in °C) of the rod to make it fit the hole?

Answers

Answer 1

The machinist would have to lower the temperature of the steel rod by approximately 0.4167 °C to make it fit the hole.

To calculate the temperature change required to make the steel rod fit the hole, we can use the concept of thermal expansion and the coefficient of linear expansion of steel.

Determine the difference in diameters: The difference in diameters between the steel rod and the hole can be calculated as follows:

Difference = Diameter of the rod - Diameter of the hole

Difference = 3 mm - 2.995 mm = 0.005 mm

Convert the difference to meters: To ensure consistent units, we need to convert the difference in diameters to meters:

Difference = 0.005 mm = 0.005 × 10^-3 m = 5 × 10^-6 m

Calculate the linear expansion: The linear expansion of a material can be expressed as the product of its original length, coefficient of linear expansion, and change in temperature. For steel, the coefficient of linear expansion is typically around 12 × 10^-6 per °C.

Rearrange the formula: We can rearrange the formula for linear expansion to solve for the change in temperature:

Change in temperature = Difference / (Coefficient of linear expansion × Original length)

Substitute values and calculate: Assuming the original length of the rod is large enough for a negligible change, we can calculate the change in temperature:

Change in temperature = (5 × 10^-6 m) / (12 × 10^-6 per °C) ≈ 0.4167 °C

Therefore, the machinist would have to lower the temperature of the steel rod by approximately 0.4167 °C (or approximately 199.66°C when rounded) to make it fit the hole.

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Related Questions

Part C
Increase the value of the Applied Force to 150 N. Pause the simulation and observe the magnitudes and directions of the applied force, the friction
force, and the sum of forces. Why do you think the crate moves this time?

Answers

The crate moves this time because the applied force of 150 N is greater than the maximum static friction force that opposes the motion of the crate. Static friction is a force that opposes the relative motion between two objects in contact. The maximum static friction force is determined by the normal force and the coefficient of static friction between the two surfaces in contact. The harder the surfaces are pushed together, the more force is needed to move them. When the applied force exceeds the maximum static friction force, the crate will start to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force.

How can graphs help demonstrate the qualitative relationship that may exist in a set of data to readers?

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Graphs can help to demonstrate the qualitative relationship between a set of data by identifying patterns among these data.

What other information can the data show?Comparison between groups.Proportional relationships.Variation and dispersion.

Graphs can allow readers to identify a lot of information among a set of data. The most common information to be evaluated through the graphs is the existence of patterns between the data.

For example, a line graph can show the change in a variable over time, allowing readers to see if there is a consistent increase or decrease.

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