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Kiana Rosenthat-Wright 
05//03//24
HW Soore: 
52,3Et^(2)38.67 of 7

9.4.3*5
points
Points: 0 of
Calculate the weight of the piece of steel shown in the drawing. (This steel weighs 
0.283lb^(in)I^(3) )
The weight of the piece of steel shown in the drawing is approximately 
◻ lb. (Round to the nearest tenth as needed.)

Kiana Rosenthat-Wright 05/03/24 05 / 03 / 24 \newlineHW Soore: 52,3Et238.67 52,3 \mathrm{Et}^{2} 38.67 of 77\newline9.4.35 9.4 .3 \cdot 5 \newlinepoints\newlinePoints: 00 of\newlineCalculate the weight of the piece of steel shown in the drawing. (This steel weighs 0.283lbinI3 0.283 \mathrm{lb}^{\mathrm{in}} \mathrm{I}^{3} )\newlineThe weight of the piece of steel shown in the drawing is approximately \square lb. (Round to the nearest tenth as needed.)

Full solution

Q. Kiana Rosenthat-Wright 05/03/24 05 / 03 / 24 \newlineHW Soore: 52,3Et238.67 52,3 \mathrm{Et}^{2} 38.67 of 77\newline9.4.35 9.4 .3 \cdot 5 \newlinepoints\newlinePoints: 00 of\newlineCalculate the weight of the piece of steel shown in the drawing. (This steel weighs 0.283lbinI3 0.283 \mathrm{lb}^{\mathrm{in}} \mathrm{I}^{3} )\newlineThe weight of the piece of steel shown in the drawing is approximately \square lb. (Round to the nearest tenth as needed.)
  1. Identify Volume: Identify the volume of the steel piece from the drawing.\newlineAssuming the drawing provides the volume as 38.6738.67 cubic inches.
  2. Calculate Weight: Use the density of steel to calculate the weight.\newlineDensity of steel = 0.283lb/in30.283 \, \text{lb/in}^3.\newlineWeight = Density ×\times Volume = 0.283lb/in3×38.67in30.283 \, \text{lb/in}^3 \times 38.67 \, \text{in}^3.
  3. Perform Multiplication: Perform the multiplication to find the weight.\newlineWeight = 0.283×38.67=10.94110.283 \times 38.67 = 10.9411 lb.
  4. Round Weight: Round the weight to the nearest tenth.\newlineRounded weight = 10.9lb.10.9\,\text{lb}.