To reduce the mass of an aluminum bulkhead for a spacecraft, a machinist drills

Question

To reduce the mass of an aluminum bulkhead for a spacecraft, a machinist drills an array of holes in the bulkhead. The bulkhead is a triangular-shaped plate with a base and height of 13.0 ft and 8.00 ft, respectively, and a thickness of 1.0* 10^1 mm. For the density of aluminum, use rho = 2.80 times 10^3 kg/m^3. How many 5.0 cm diameter holes must be drilled clear through the bulkhead to reduce its mass by 6.50 kg? Show your conversion factors. Use the correct number of significant figures in your final answer.

Explanation / Answer

ANSWER

here we have to remove 6.5 kg of the mass by drilling holes which are having 5cm diameter (radius r =1/2 diameter) and with thikness t= 10 mm

and the density is d= 2800 kg /m3

as the density is in MKS system convert the radiusand thikness in to meters

    r = 1/2(5cm ) = 2.5 cm = 2.5/100   m = 0.025 m

    t = 10mm =10*10 -3 m = 0.01m

if we know the volume of the each hole then multiplying thisvolume with density we get the mass of the each hole

so the volume V= r2 t =1.96*10-5 m3

mass m = dV =0.05488 kg

so the number of holes =total mass / mass of each hole

                                  = 6.5 kg / 0.05488

                                 =118.44 holes

Regards!!!

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