A rectangular steel plate that measures 4.43 in. by 8.81 in. by 3 in. thick, with a 4 diameter hole in the center, is being drawn. (a) Find the volume of steel left after the hole is cut. (b) If this steel weighs 30 oz/in ³, what weight should be given on the drawing? Follow the rules for working with measurements. A) The volume is 26.5 in ³, the weight is 2,380 oz. B) The volume is 79.4 in ³, the weight is 7,940 oz. C) The volume is 105 in ³, the weight is 2,380 oz. D) The volume is 79.4 in ³, the weight is 2,380 oz. E) The volume is 26.5 in ³, the weight is 7,940 oz.

Question

Quizplus · Accepted Answer

First, calculate the volume of the rectangular steel plate without the hole: 4.43 in. * 8.81 in. * 3 in. = 117.3 in^3. Then, calculate the volume of the hole (cylinder) using the formula πr^2h, where r = diameter/2 = 2 in., and h = 3 in.: π * 2^2 * 3 ≈ 37.7 in^3. Subtract the volume of the hole from the total volume to get the volume of steel left: 117.3 in^3 - 37.7 in^3 = 79.6 in^3 (rounded to 79.4 in^3 due to significant figures). For the weight, multiply the volume of steel left by the weight per unit volume: 79.6 in^3 * 30 oz/in^3 = 2,388 oz, which rounds to 2,380 oz due to significant figures and measurement rules.

A Rectangular Steel Plate That Measures 4