What is the coeffecient of drag for the shape that has diameter 7.32[cm]? .
Use three sig. figs. or N/A if there is not enough information.
2. You have a large wind tunnel of cross-sectional area 0.95[m2]. You have a set of shapes that are all the same shape except they each have a different cross-sectional area. The wind speed is held constant at 3.2[ms-1]. Assume air density is 1 [kg m-3]. The slope of Drag versus Area for this data is 0.45. [Slope units left out purposefully by me for sake of next question.]
What is the coeffecient of drag for the shape? .
Use three sig. figs. or N/A if there is not enough information.
3. The slope from the plot of Drag versus Area in the question above has units of
[N]
[m^2]
[kgm^-3]
No units
[Nm^-2]
All three have me stumped
1. To determine the coefficient of drag for a shape with a diameter of 7.32 cm, you need more information. The coefficient of drag depends on factors such as the shape of the object, the Reynolds number (which accounts for the object's size, velocity, and density), and the roughness of the surface. Without additional information, it is not possible to calculate the coefficient of drag. Therefore, the answer is N/A (not applicable).
2. In this scenario, you have a wind tunnel with a cross-sectional area of 0.95 m2. You are testing different shapes with different cross-sectional areas and measuring the drag at a constant wind speed of 3.2 m/s. Assuming the air density is 1 kg/m3, and given that the slope of the Drag versus Area plot is 0.45, you can calculate the coefficient of drag for the shape using the equation:
Coefficient of drag = slope / (density * wind speed^2)
Coefficient of drag = 0.45 / (1 * 3.2^2)
Coefficient of drag ≈ 0.045
Therefore, the coefficient of drag for the shape is approximately 0.045.
3. The slope from the plot of Drag versus Area in the above question represents the change in drag divided by the change in area. To determine the units of the slope, we need to analyze the units of drag and area.
Drag is typically measured in units of force, such as Newtons (N).
Area is measured in units of length squared, such as square meters (m^2).
Since the slope represents the change in drag (N) divided by the change in area (m^2), the slope has units of Newtons per square meter (N/m^2) or Pascals (Pa), which is the unit of pressure.
Therefore, the correct units for the slope from the plot of Drag versus Area are [Nm^-2].