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1. The total area of fictional Puget Loud, near Seapple, Washington, can be roughly estimated using a rectangular shape that is 110 miles long and 15 miles wide. In Puget Loud, orcas require approximately 5 mi^2 of territory per animal. Based on this information, how many orcas could live in Puget Loud? Show your work.

2. The mass of a ball is 128g. It has a density of 0.5g/cm^3. What is the volume of the ball? Bonus Point: what is the radius of the ball?

 May 22, 2019

Best Answer 

 #1
avatar+195 
+1

1. The total area of Puget Loud is 110 mi x 15 mi = 110*15 = 1650 mi^2. Because every orca needs 5 mi^2 of teritory, the number of orcas that can live in Puget Loud is 1650 mi^2 / 5 mi^2=1650/5 = \(\boxed{330}\)

 

2. \(density = \frac{mass}{volume}\) so \(.5 =\frac{128}{v}\), where v is volume.

\(.5v=128\)

\(v=\boxed{256 cm^3}\)

 

The eaqution for the volume of a sphere is \(v=\frac{4}{3} \pi r^3\) where v is the volume and r is the radius so \(256=\frac{4}{3} \pi r^3\)

Dividing by \(\frac{4}{3} \pi\) yields \(r^3=78842.81\). Taking the cube root of both sides gets \(r = \boxed{42.88}\)

.
 May 22, 2019
 #1
avatar+195 
+1
Best Answer

1. The total area of Puget Loud is 110 mi x 15 mi = 110*15 = 1650 mi^2. Because every orca needs 5 mi^2 of teritory, the number of orcas that can live in Puget Loud is 1650 mi^2 / 5 mi^2=1650/5 = \(\boxed{330}\)

 

2. \(density = \frac{mass}{volume}\) so \(.5 =\frac{128}{v}\), where v is volume.

\(.5v=128\)

\(v=\boxed{256 cm^3}\)

 

The eaqution for the volume of a sphere is \(v=\frac{4}{3} \pi r^3\) where v is the volume and r is the radius so \(256=\frac{4}{3} \pi r^3\)

Dividing by \(\frac{4}{3} \pi\) yields \(r^3=78842.81\). Taking the cube root of both sides gets \(r = \boxed{42.88}\)

power27 May 22, 2019
 #2
avatar+109560 
+1

Nice answers , Power  !!!

 

 

cool cool cool

CPhill  May 22, 2019
 #3
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+2

thank you so much! The explanations helped a lot!

Guest May 22, 2019

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