Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
a 1.40 g sample of gold having a density of 19.32 g/cm^3 is hammered into a very thin sheet. if the area of the sheet is 35.0ft^2, what is the average thickness of the sheet expressed in centimeters?
 one year ago
 one year ago
a 1.40 g sample of gold having a density of 19.32 g/cm^3 is hammered into a very thin sheet. if the area of the sheet is 35.0ft^2, what is the average thickness of the sheet expressed in centimeters?
 one year ago
 one year ago

This Question is Closed

.Sam.Best ResponseYou've already chosen the best response.1
Using back the formula, \[\rho= \frac{m}{V}\] \[V= \frac{m}{\rho}\] \[V= \frac{1.40 g }{19.32 g/cm^3}\] \[V= 0.072cm^3\]  Changing 35.0ft^2, \[35.0ft^2=\frac{929cm^2}{1ft^2}=32515cm^2\]  We assume that the metal sheet is either cubed or cuboid, \[V=Length \times Base \times Height\] \[V=Aree \times Height\] \[Height=\frac{Volume}{Area}\] \[Height=\frac{0.072cm^3}{32515cm^2}\] \[Height=2.21 \times 10^{6}cm\]
 one year ago

dddanBest ResponseYou've already chosen the best response.0
thanks can you explain how you get from v=basexhightxlength to v=areaxheight?
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.