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connecting buoyancy with displacement | metal | density (g/cm³) | |----…

Question

connecting buoyancy with displacement

metaldensity (g/cm³)
cork0.24
iron7.50
lead11.34
wax0.72

which material will displace a volume of water? iron, lead, and aluminum
which material will displace a volume of water less than its own volume? cork and wax
which material will displace a volume of water equal to its own volume? iron, lead, and aluminum
which material will displace a volume of water greater than its own volume?
options: all of them; cork and wax; iron, lead, and aluminum; none of them

Explanation:

Step1: Recall Archimedes' Principle

Archimedes' principle states that the volume of water displaced by an object is related to whether the object floats or sinks. For an object with density \(
ho_{obj}\) and water with density \(
ho_{water} = 1\space g/cm^3\):

  • If \(

ho_{obj}>
ho_{water}\) (object sinks), it displaces a volume of water equal to its own volume.

  • If \(

ho_{obj}<
ho_{water}\) (object floats), it displaces a volume of water equal to the volume of the submerged part. Since it floats, the submerged volume (and thus water displaced) is less than its total volume only if it's partially submerged, but actually, for floating objects, the mass of displaced water equals the mass of the object. So, the volume of displaced water \(V_{displaced}=\frac{m_{obj}}{
ho_{water}}\), and the object's volume \(V_{obj}=\frac{m_{obj}}{
ho_{obj}}\). Since \(
ho_{obj}<
ho_{water}\), \(V_{displaced}=\frac{
ho_{obj}}{
ho_{water}}V_{obj}<V_{obj}\) (because \(\frac{
ho_{obj}}{
ho_{water}}< 1\)).

  • To displace more than its own volume, we would need \(V_{displaced}>V_{obj}\), which would require \(\frac{

ho_{obj}}{
ho_{water}}V_{obj}>V_{obj}\) or \(
ho_{obj}>
ho_{water}\), but that's not possible. Wait, no—wait, if an object sinks, \(V_{displaced}=V_{obj}\). If it floats, \(V_{displaced}V_{obj}\)? Let's check the densities:

  • Cork: \(

ho = 0.24<1\) (floats), Wax: \(
ho = 0.72<1\) (floats), Aluminum: \(2.64>1\) (sinks), Iron: \(7.50>1\) (sinks), Lead: \(11.34>1\) (sinks).

  • For floating objects (cork, wax), \(V_{displaced}=\frac{

ho_{obj}}{
ho_{water}}V_{obj}\). Since \(
ho_{obj}<1\), \(V_{displaced}ho_{obj}V_{obj}\), mass of displaced water \(m =
ho_{water}V_{displaced}\). So \(V_{displaced}=\frac{
ho_{obj}}{
ho_{water}}V_{obj}\). If \(
ho_{obj}<
ho_{water}\), then \(V_{displaced}ho_{obj}>
ho_{water}\), \(V_{displaced}=V_{obj}\). So when would \(V_{displaced}>V_{obj}\)? That would require \(\frac{
ho_{obj}}{
ho_{water}}V_{obj}>V_{obj}\) or \(
ho_{obj}>
ho_{water}\), but even then, it's equal, not greater. Wait, maybe I made a mistake. Wait, the question is "Which material will displace a volume of water greater than its own volume?". Let's check the options:

  • All of them: No, because sinks have equal, floats have less.
  • Cork and wax: They float, so \(V_{displaced}
  • Iron, lead, aluminum: They sink, so \(V_{displaced}=V_{obj}\), not greater.
  • None of them: Because for any object, if it sinks, \(V_{displaced}=V_{obj}\); if it floats, \(V_{displaced}

Step2: Analyze Each Material

  • Cork: \(

ho = 0.24\space g/cm^3<1\space g/cm^3\) (water). Floats, so \(V_{displaced}=\frac{0.24}{1}V_{obj}=0.24V_{obj}

  • Wax: \(

ho = 0.72\space g/cm^3<1\space g/cm^3\). Floats, \(V_{displaced}=0.72V_{obj}

  • Aluminum, Iron, Lead: Densities > 1, so they sink, \(V_{displaced}=V_{obj}\) (not greater).
  • So no material displaces more than its own volume. So the correct option is "none of them".

Answer:

none of them