Two cylinders have equal diameters. are their volumes also equal? is the same true for cones and spheres?

2 answers:

8 0

<span>are their volumes also equal?
</span> Not necessarily, if they have the same height the volume of the two cylinders is the same. However, if they do not have the same height, the volume is different. The volume of a cylinder is:
V = pi * r ^ 2 * h.
the same true for cones and spheres?
For the cone, the same thing happens with the cylinder. It will depend on the height. The volume of the cone is:
V = (1/3) * pi * r ^ 2 * h.
For a sphere YES. the volumes are equal if they have the same diameter since their equation depends only on the radius. The volume of the sphere is:
V = (4/3) * pi * r ^ 3

Paraphin [41]3 years ago

5 0

<span>Although two cylinders may have equal diameters, their volumes are not necessarily equal, as the volume of a cylinder is dependent also on the cylinder's height. The same holds true for cones. Spheres, however, are, by nature, proportional. Therefore, if two spheres have the same diameter, they also have the same volumes.</span>

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Which figure will tessellate the plane? A. regular pentagon B. regular decagon C. regular octagon D. regular hexagon

Harrizon [31]

<h3>Answer: D. regular hexagon</h3>

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In contrast, a regular pentagon has interior angle 108 degrees. This is not a factor of 360, so there is no way to place regular pentagons to have them line up and not be a gap or overlap. This is why regular pentagons do not tessellate the plane. The same can be aside about decagons and octagons as well.

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3 years ago

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anastassius [24]

Answer:

Indefinite integration acts as a tool to solve many physical problems.

There are many type of problems that require an indefinite integral to solve.

Basically indefinite integration is required when we deal with quantities that vary spatially or temporally.

As an example consider the following example:

Suppose that we need to calculate the total force on a object placed in a non- uniform field.

As an example let us consider a rod of length L that posses an charge 'q' per meter length and suppose that we place it in a non uniform electric field which is given by

$E(x)=\frac{E_{o}}{e^{kx}}$

Now in order to find the total force on the rod we cannot use the similar procedure as we can see that the force on the rod varies with the position of the rod.

But if w consider an element 'dx' of the rod at a distance 'x' from the origin the force on this element will be given by

$dF=E(x)\times qdx\\\\dF=\frac{qE_{o}}{e^{kx}}dx$

Now to find the whole force on the rod we need to sum this quantity over the whole length of the rod requiring integration, as shown

$\int dF=\int \frac{qE_{o}}{e^{kx}}dx$

Similarly there are numerous problems considering motion of particles that require applications of indefinite integration.

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The dimensions are 475 m by 237.5 m.

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.. A = x*(950 -x)/2
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