Welcome Guest Search | Active Topics |

Tag as favorite
EK Study Guide Lecture 2 #51 (pg.73)
Options
Previous Topic Next Topic
Janelle_5719
#1 Posted : Tuesday, June 02, 2020 11:18:22 PM
Rank: Newbie

Groups: Registered
Joined: 5/18/2020
Posts: 8

Thanks: 0 times
Was thanked: 0 time(s) in 0 post(s)
I don’t understand how they even began to solve this problem. I originally attempted to find the density of the block and of the brick then convert it to volume but because the styrofoam has no mass I couldn't solve it that way. Could you explain how to properly solve this? I looked at the solution and it made no sense to me.
Back to top
Report?
INSTR_Kailey_109
#2 Posted : Thursday, June 04, 2020 10:01:58 PM
Rank: Advanced Member

Groups: Registered
Joined: 6/4/2020
Posts: 31

Thanks: 0 times
Was thanked: 0 time(s) in 0 post(s)
Hello! This is definitely a tricky question.

Recall that for a floating object, the mass of water displaced = mass of brick+styrofoam (in this case mass of styrofoam is 0). So m_water = m_brick. Also recall that density, rho=m/v or m = rho x v. if m_water = m_brick then rho_water x v_water = rho_brick x v_brick. we know that rho_brick/rho_water = 1.4. If we rearrage our previous formula, rho_brick/rho_water= v_water/v_brick = 1.4. We also know that the volume of water dispaced = 1/2 styrofoam volume, so v_water = 1/2 v_styro.

so: v_water = 1/2v_styro = 1.4v_brick. (1.4~3/2) —> to rearrange this: v_styro ~ (3/2)*(2)v_brick ~ 3v_brick giving you the answer C.
Back to top
Report?
Users browsing this topic
Guest
Tag as favorite
You cannot post new topics in this forum.
You cannot reply to topics in this forum.
You cannot delete your posts in this forum.
You cannot edit your posts in this forum.
You cannot create polls in this forum.
You cannot vote in polls in this forum.

Clean Slate theme by Jaben Cargman (Tiny Gecko)
Powered by YAF | YAF © 2003-2009, Yet Another Forum.NET
This page was generated in 0.042 seconds.