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Given a 10 foot length of tree trunk with a radius of 3 feet, what is the weight of the tree trunk section if it has a density of 45 lb/ft3? (to the nearest whole integer)

User Quar
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2 Answers

4 votes
The answer is 12,723 lb

Use the formula for the volume of a cylinder to model the volume of the tree trunk.

V = πr^2h
V = π(3^2)(10)
V = 90π ft^3

therefore,

90π ft^3 × 45 lb/ft^3 = 12723.4502
User Murilo Giacometti
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6.1k points
5 votes

Answer:

Using the formula:


\text{Weight} = \text{Density} * \text{Volume}

As per the statement:

Given a 10 foot length of tree trunk with a radius of 3 feet.

Volume of tree trunk(V) is given by:


V = \pi r^2h

where, r is the radius and h is the length of the tree trunk.

Substitute r = 3 ft and h = 10 ft and use
\pi = 3.14

then;


V = 3.14 \cdot 3^2 \cdot 10 = 282.6 ft^3

It is also given it has a density of 45 lb/ft^3

then;


\text{Weight} = 45 \cdot 282.6 = 12717 Ib

Therefore, the the weight of the tree trunk section is, 12,717 Ib

User Borkweb
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