Volume of a Pyramid: 474 ft & 11.0 Acres

  • Thread starter Thread starter r.anthony
  • Start date Start date
  • Tags Tags
    Pyramid Volume
AI Thread Summary
To find the volume of a pyramid with a height of 474 ft and a base area of 11 acres, first convert the base area from acres to square feet, using the conversion of 1 acre = 43,560 ft². The total base area in square feet is calculated as 11 acres multiplied by 43,560 ft². The volume formula for a pyramid, V = 1/3 Bh, requires substituting the base area and height into the equation. It's recommended to convert all measurements to meters for the final volume calculation. The discussion emphasizes understanding the conversion process and applying the volume formula correctly.
r.anthony
Messages
3
Reaction score
0
1. A pyramid has a height of 474 ft and its base covers an area of 11.0 acres. The volume of a pyramid is given by the expression V = 1/3 Bh, where B is the area of the base and h is the height. Find the volume of this pyramid in cubic meters. (1 acre = 43 560 ft2).



Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
Welcome to PF.

One of the rules around here is that we help and don't do it for you. Many people know how to solve your problem but that doesn't help you if they do it for you.

Hence you must make an effort to come up with a solution.

So how would you approach this pyramid?
 
Hey LowlyPion thanks for the welcome. Appreciate the advice. Thinking about it, I guess I would convert the acres of the pyramid into ft^2 first. the height is given so i guess i would multiply that by what 11 acres is in ft^2 then multiply that answer by 1/3. although I'm a bit confused. Since 1 acre is 43560 ft^2, am I right by multiplying 11 by 43560 or do I need to square 43560 first then multiply by 11? then again there's converting ft into m i think
 
Last edited:
Personally, I'd convert everything to meters at the start, since that is the units for the answer.
Google said:
1 acres = 4 046.85642 square meters

You can convert the height straight away.

The rest is ...
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top