I need to know how many tanks size is 1.10m diameter and 1.30m in height can I fit in a container with sizes 5.80m lenght, 2.30m height and 2.30m in width

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Posted on Jan 02, 2017

Hi Alex. The 3 measurements you quote are all cubic feet so you cannot ascertain any fixed sizes from that. The figure is arrived at by multiplying the width x height x depth. This means that if the box is 1 foot high and 1 foot wide it would need to be 1.4 feet deep to total 1.4 cubic feet. If it was 6 feet high and 1.4 feet wide and 2 inches deep it would also be 1.4 cubic feet. As you can see the sizes are completely different and the combination of different sizes is just about infinite.

Hope that is of some help, although to be honest I don't see how....

Regards

Steve

Hope that is of some help, although to be honest I don't see how....

Regards

Steve

Jan 21, 2016 | Office Equipment & Supplies

Hi Chris:

There are 10 mm in 1 cm, so the stone is 20 mm.

I'm assuming this is in diameter which is the common way to refer to the size of spherical objects.

20 mm is about .3/4 of an inch, roughly the width of my thumbnail, so the stone would be about the size of a marble that would cover your (or my) thumbnail

There are 10 mm in 1 cm, so the stone is 20 mm.

I'm assuming this is in diameter which is the common way to refer to the size of spherical objects.

20 mm is about .3/4 of an inch, roughly the width of my thumbnail, so the stone would be about the size of a marble that would cover your (or my) thumbnail

May 13, 2015 | Office Equipment & Supplies

About 488 cubic feet or about 3650 US gallons.

Sep 09, 2014 | Office Equipment & Supplies

Here is to get you started. To increase the size of the image do a CTRL Plus (+) in your browser.

You need to calculate the slant height of the pyramid for the formula of the lateral area. You should find a value of** (1/2)*SQRT(203) **or about 7.1239 cm

You need to calculate the altitude (height) of the pyramid from the apex (summit) to the center of the base triangle (center of inscribed circle, barycenter, orthocenter). The hypotenuse of such triangle is the slant height. One leg is the altitude (to be found),**the measure of the second leg is (1/3) the altitude** **of the equilateral triangle** that forms the base. You should find (1/3) m MH= (1/3)* **(11/2)*SQRT(3)**

1. Calculate the area of the base (use a formula for the equilateral triangle or the general formula for a triangle: you have its height MH ).

2. Lateral area = 3 times the area of triangle Triangle ECD (in yellow above).

3. Total area = area of base + lateral area.

4. Volume= (1/3)(Area of base)* (height of pyramid)

If you can see the details on the screen capture below, fine, Press CTRL + in your browser to increase the size.

You need to calculate the slant height of the pyramid for the formula of the lateral area. You should find a value of

You need to calculate the altitude (height) of the pyramid from the apex (summit) to the center of the base triangle (center of inscribed circle, barycenter, orthocenter). The hypotenuse of such triangle is the slant height. One leg is the altitude (to be found),

1. Calculate the area of the base (use a formula for the equilateral triangle or the general formula for a triangle: you have its height MH ).

2. Lateral area = 3 times the area of triangle Triangle ECD (in yellow above).

3. Total area = area of base + lateral area.

4. Volume= (1/3)(Area of base)* (height of pyramid)

If you can see the details on the screen capture below, fine, Press CTRL + in your browser to increase the size.

Mar 29, 2014 | Office Equipment & Supplies

About 1700 cubic feet.

Since you know the circumference, calculate the diameter by dividing it by pi.

Since you know the circumference, calculate the diameter by dividing it by pi.

Feb 17, 2014 | Office Equipment & Supplies

The volume of a right circular cone is given by the formula

V_cone=(Area of base *height)/3

Since cone is circular , its base is a disk with a certain radius r. The formula becomes

I suggest you convert gallons to cubic feet or to cubic meters then put

V=Pi*(r^2)*h/3 where V is the value of volume after conversion.

If you choose the radius, then to isolate h,

If you choose the height, then to solve for r

r^2=(3V)/(Pi*h)

and

Now, it is your turn to carry out the rest of the calculation.

It will consist of the value of the radius, and the value of the height.

If the cone is not right circular, the situation becomes more complicated.

Sep 16, 2013 | Office Equipment & Supplies

If you are looking for just a floor size than you multiply lenght with width to get the square footage.

May 18, 2011 | Office Equipment & Supplies

I am sorry to say it but your units do not make sense: Units of volumes and areas are mixed up.

Let us try to make sense of it.

If you have some container whose base is 3 square meters (not cubic meter sq) and it has a height of 3 inches you can calculate its volume in cubic meters or in cubic inches or cubic feet. To do so convert the height in inches into meters and multiply the area of the base by the height in meters and you will obtain the volume in cubic meters. To get the volume in cubic inches convert the area of the base to square inches and multiply by the height in inches.

However, even after you have expressed the volume in one unit or another (m^3 or in^3) you cannot find the mass of the substance contained (be it in tons, tonnes or kilograms): You need to know the substance and more specifically its density or its specific gravity).

Please, reformulate your question and specify the substance that fills or would fill the volume. Without that information, nothing doing.

Let us try to make sense of it.

If you have some container whose base is 3 square meters (not cubic meter sq) and it has a height of 3 inches you can calculate its volume in cubic meters or in cubic inches or cubic feet. To do so convert the height in inches into meters and multiply the area of the base by the height in meters and you will obtain the volume in cubic meters. To get the volume in cubic inches convert the area of the base to square inches and multiply by the height in inches.

However, even after you have expressed the volume in one unit or another (m^3 or in^3) you cannot find the mass of the substance contained (be it in tons, tonnes or kilograms): You need to know the substance and more specifically its density or its specific gravity).

Please, reformulate your question and specify the substance that fills or would fill the volume. Without that information, nothing doing.

Apr 08, 2011 | Office Equipment & Supplies

We find volume by using V = H*W*L , where H = height , W = width, and L = length. You say H = 2 ft,

V = 80 ft ^3, and W = L - 3 . Put all of these back onto the equation and we will have

80 = 2 * (L - 3) * L . We can condense this to be 80 = (2*L - 2*3 ) * L then becomes 80 = 2*L*L - 2*3*L

simplifying this we get 40 = L*L - 3*L . This is also 40 = L * (L - 3) this is a quadratic equation, meaning that

we will get two answers for L. In this case L = 0 and L = 3. Since 0 ft would mean that the tool box doesn't

have a length we will use L = 3. We can now put this back into our original equation to find out what the

width is. 80 = 2 * W * 3, this is 80/6 = W = 13.3333 . Now we know that W = 13.3333 and L = 3.

V = 80 ft ^3, and W = L - 3 . Put all of these back onto the equation and we will have

80 = 2 * (L - 3) * L . We can condense this to be 80 = (2*L - 2*3 ) * L then becomes 80 = 2*L*L - 2*3*L

simplifying this we get 40 = L*L - 3*L . This is also 40 = L * (L - 3) this is a quadratic equation, meaning that

we will get two answers for L. In this case L = 0 and L = 3. Since 0 ft would mean that the tool box doesn't

have a length we will use L = 3. We can now put this back into our original equation to find out what the

width is. 80 = 2 * W * 3, this is 80/6 = W = 13.3333 . Now we know that W = 13.3333 and L = 3.

Mar 07, 2011 | Office Equipment & Supplies

if container is a cylinder, then formula is pi x radius^2 x height
pi x (18.25)^2 x 61 = 63827.15 cm^3

Feb 27, 2011 | Office Equipment & Supplies

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