A block of wood has a mass of 3.50 kg and a density of 587 kg/m3.It is to be loa
ID: 1728869 • Letter: A
Question
A block of wood has a mass of 3.50 kg and a density of 587 kg/m3.It is to be loaded with lead so that it will float in water with0.93 of its volume immersed. The density of lead is 1.13 X 10^4 kg/m^3. (a) What mass of lead is needed if the lead is on top of the wood?(in kg) (b) What mass of lead is needed if the lead is attached below the wood? (in kg) This is what I solved out for (a) For some reason I am still getting the wrong answer... could someone correct what I have done for(a) and help me figure out(b)? Thank you for any help ^^Explanation / Answer
Volume of block = mass / density = 3.50 / 587 = 0.0059625 m3 . Volume of water displaced is 0.93 times this,or 0.93 * 0.0059625 = 0.0055451m3 . weight of water displaced = v g = 1000 * 0.0055451 * 9.80 = 54.342 Newtons. . The weight of water displaced is equal to the upward buoyancyforce, which must equal the downward force of the weight of theblock and the lead. So.. . B = weight of block + weight of lead . 54.342 = mg + mg 54.342 =3.50 * 9.80 + m * 9.80 . m = 2.045 kg of lead isneeded (part a) (Ithink you have your 0.93 in the wrongplace) . For part b... . Now the buoyant force will be equal to the weight of waterdisplaced, which will equal that displaced by the wood and by thelead! So... . B = weight of wood + weight oflead . water displaced by wood + waterdisplaced by lead = weight of wood + weightof lead . 54.342 + VLg = mg + mL g . Divide by g . 5.5451 + 1000 *VL = 3.5 + mL . 2.0451 + 1000 *mL / 11300 = mL . 2.0451 + 0.0885mL = mL . mL = 2.0451 / (1-0.0885) = 2.244 kg of lead . mL = 2.0451 / (1-0.0885) = 2.244 kg of lead .Related Questions
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