What does the Angle X =
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The way we approach a question like this, is by looking at what portion of the circle we actually have to calculate the area of. We know that a circle is a shape that goes 360 degrees around a point. Therefore we know that, for example, a semicircle, or half a circle, would cover half the amount of degrees. This is important, because as well as covering half the number of degrees, it also covers half the area. For the figure in your picture, we are told that the portion missing from the circle is 90 degrees (shown by the right angle symbol in the center).
this means that 90 degrees worth of the circle is missing, so if we do 360÷90, we find that the answer equals
. This means that we are dealing with of a circle. So, we simply find the area of the circle normally, and then multiply by <span>:
The formula for area of a circle is </span>
, so in this case, it can be said to be 
, which is equal to 314.15
Now we simply take that area and multiply it by the fraction we established earlier:
314.15 x = 235.61
Therefore the area of the figure is 235.61
I hope this helped and please remember to try and not only understand the answer, but the maths that worked it out too :))
this means that 90 degrees worth of the circle is missing, so if we do 360÷90, we find that the answer equals
The formula for area of a circle is </span>
Now we simply take that area and multiply it by the fraction we established earlier:
314.15 x
Therefore the area of the figure is 235.61
I hope this helped and please remember to try and not only understand the answer, but the maths that worked it out too :))
Answer:
1.314 MJ
Step-by-step explanation:
As water is removed from the tank, decreasing amounts are raised increasing distances. The total work done is the integral of the work done to raise an incremental volume to the required height.
There are a couple of ways this can be figured. The "easy way" involves prior knowledge of the location of the center of mass of a cone. Effectively, the work required is that necessary to raise the mass from the height of its center to the height of the discharge pipe.
The "hard way" is to write an expression for the work done to raise an incremental volume, then integrate that over the entire volume. Perhaps this is the method expected in a Calculus class.
<h3>Mass of water</h3>The mass of the water being raised is the product of the volume of the cone and the density of water.
The cone volume is ...
V = 1/3πr²h . . . . . . for radius 2 m and height 8 m
V = 1/3π(2 m)²(8 m) = 32π/3 m³
The mass of water in the cone is then ...
M = density × volume
M = (1000 kg/m³)(32π/3 m³) ≈ 3.3510×10^4 kg
<h3>Center of mass</h3>The center of mass of a cone is 1/4 of the distance from the base to the point. In this cone, it is (1/4)(8 m) = 2 m from the base.
<h3>Easy Way</h3>The discharge pipe is 2 m above the base of the cone, so is 4 m above the center of mass. The work required to lift the mass from its center to a height of 4 m above its center is ...
W = Fd = (9.8 m/s²)(3.3510×10^4 kg)(4 m) = 1.3136×10^6 J
<h3>Hard Way</h3>
As the water level in the conical tank decreases, the remaining volume occupies a space that is similar to the entire cone. The scale factor is the ratio of water depth to the height of the tank: (y/8). The remaining volume is the total volume multiplied by the cube of the scale factor.
V(y) = (32π/3)(y/8)³
The differential volume at height y is the derivative of this:
dV = π/16y²
The work done to raise this volume of water to a height of 10 m is ...
(9.8 m/s²)(1000 kg/m³)(dV)((10 -y) m) = 612.5π(y²)(10 -y) J
The total work done is the integral over all heights:
It takes about 1.31 MJ of work to empty the tank.