The 50.0 g block is 40.0% submerged. A: What is the mass of the submerged portio
ID: 594642 • Letter: T
Question
The 50.0 g block is 40.0% submerged.
A: What is the mass of the submerged portion of the block?
Please show me every step in plain english.
Explanation / Answer
Free-body diagram: ......Fs .......^ .......I ____I____ I..............I I...block...I I________I .....I....^ .....I....I ....V....I ...W...Fb 3 forces are acting on the block: 1. Fs = upward force from scale (up = + Fs) 2. W = downward weight force (down = - W) 3. Fb = upward buoyancy force (up = + Fb) Since the block is at rest, (net force = 0), the force equation is: Fs + Fb - W = 0 ~(eq1) where, Fs = m (shown on scale) x g Fb = ? x V x g W = m (block) x g g = acceleration due to gravity Substituting Fs, Fb and W into ~(eq1): ==> {m (on scale) x g} + {? x V x g} - (m (block) x g} = 0 Dividing both sides of equation by g: m (on scale) + (? x V) - m (block) = 0 ~(eq2) Given: V = volume of water displaced = 40% volume of the block = 0.40 x V (block) Substituting 0.40 x V (block) into ~(eq2): m (on scale) + (? x 0.40 x V (block)) - m (block) = 0 ~(eq3) where, m (on scale) = 5.6 g ? = density of water at 20ºC = 0.99821 g/cm^3 m (block) = 50 g Substituting m (on scale), ? and m (block) into ~(eq3): 5.6 g + (0.99821 g/cm^3 x 0.40 x V (block)) - 50 g = 0 ~(eq4) Simplifying ~(eq4) and solving for V (block): ==> 5.6 g + 0.399284 g/cm^3 x V (block) - 50 g = 0 ==> 0.399284 g/cm^3 x V (block) - 44.4 g = 0 ==> 0.399284 g/cm^3 x V (block) = 44.4 g ==> V (block) = 44.4 g / (0.399284 g/cm^3) ==> V (block) = 111.199 cm^3 ? (block) = density of block = m (block) / V (block) = 50 g / 111.199 cm^3 = 0.449644 g / cm^3 *** Answer: 0.44964 g / cm^3 ***
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