Mathematics: What is the area of the cross section?

What is the area of the cross section?, №15229260, 07.05.2020 21:54

Mathematics

In this problem, we will find the volume of a solid with circular base of radius 2, for which parallel cross-sections perpendicular to the base are squares. To do this, we will assume that the base is the circle x2+y2=4, so that the solid lies between planes parallel to the x-axis at x=2 and x=−2. The cross-sections perpendicular to the x-axis are then squares whose bases run from the semicircle y=√−4−x2 to the semicircle y = √4−x2 Required: a. What is the area of the cross-section at x? b. What is the volume of the solid ?

Step-by-step answer

P Answered by PhD

1

(a) Each cross section is a square whose side length is decided by the distance in the x,y plane between the two curves $y=\sqrt{4-x^2}$ and $y=-\sqrt{4-x^2}$ , which is $2\sqrt{4-x^2}$ . Then each cross section has area

$(2\sqrt{4-x^2})^2=4(4-x^2)=\boxed{16-4x^2}$

(b) The volume of the solid is obtained by integrating the cross-sectional area from x = -2 to x = 2.

$\displaystyle\int_{-2}^2(16-4x^2)\,\mathrm dx=16x-\dfrac43x^3\bigg|_{-2}^2=\boxed{\dfrac{128}3}$

Mathematics

What is the shape of the cross-section formed when a plane containing line ac and line eh intersects this cube? what is the area of the cross-section? in two or more complete sentences, explain how you were able to determine the shape and area of the cross-section. be sure to include the properties of quadrilaterals and the mathematical computations necessary to support your answers

Step-by-step answer

P Answered by Master

1

Given a cube of sides 4 units with vertices: A, B, C, D, E, F, G, H such that the top square is ABCD and the bottom square is EFGH with AC, BD the diagonals of the top square and EG, FH the diagonals of EFGH.

The shape of the cross-section formed when a plane containing line AC and line EH intersects the cube is the rectangle ACEH.

We obtain the value of line AC which is equal to line EH using the pythagoras rule which states that c^2=a^2+b^2

, where c is the length of line AC, a is the length of line AB = 4 and b is the length of line BC = 4.

Thus,
$c^2=4^2+4^2=16+16=32 \\ c= \sqrt{32} =4 \sqrt{2}$

Therefore, the shape of the cross-section formed when a plane containing line AC and line EH intersects the cube is the rectangle ACEH with length of $4 \sqrt{2}$ units and width of 4 units.

Area of a rectangle is given by
Area = length x width
$=4 \sqrt{2}\times4=16 \sqrt{2}$ square units.

Mathematics

What is the shape of the cross-section formed when a plane containing line ac and line eh intersects this cube? what is the area of the cross-section? in two or more complete sentences, explain how you were able to determine the shape and area of the cross-section. be sure to include the properties of quadrilaterals and the mathematical computations necessary to support your answers

Step-by-step answer

P Answered by Specialist

1

Given a cube of sides 4 units with vertices: A, B, C, D, E, F, G, H such that the top square is ABCD and the bottom square is EFGH with AC, BD the diagonals of the top square and EG, FH the diagonals of EFGH.

The shape of the cross-section formed when a plane containing line AC and line EH intersects the cube is the rectangle ACEH.

We obtain the value of line AC which is equal to line EH using the pythagoras rule which states that c^2=a^2+b^2

, where c is the length of line AC, a is the length of line AB = 4 and b is the length of line BC = 4.

Thus,
$c^2=4^2+4^2=16+16=32 \\ c= \sqrt{32} =4 \sqrt{2}$

Therefore, the shape of the cross-section formed when a plane containing line AC and line EH intersects the cube is the rectangle ACEH with length of $4 \sqrt{2}$ units and width of 4 units.

Area of a rectangle is given by
Area = length x width
$=4 \sqrt{2}\times4=16 \sqrt{2}$ square units.

Mathematics

What is the shape of the cross-section formed when a plane containing line AC and line EH intersects this cube? What is the area of the cross-section? In two or more complete sentences, explain how you were able to determine the shape and area of the cross-section. Be sure to include the properties of quadrilaterals and the mathematical computations necessary to support your answers

Step-by-step answer

P Answered by Specialist

(a) Rectangle

$(b)\ Area = 16\sqrt 2$

Step-by-step explanation:

Given

See attachment for cube

Solving (a): Cross-section formed when lines AC and EH meets

Lines AC and EH will meet midway of lines AE and CH

So, the shape that will be formed is quadrilateral and the sides of the quadrilateral is: AE, EH, HC and CA

Where

AE = HC = DG = 4

EH = CA

CA is the diagonal of the top face of cube.

So:

CA^2 = AD^2 + DC^2

So, we have:

CA^2 = 4^2 + 4^2

CA^2 = 16 + 16

CA^2 = 32

Take square roots

$CA = \sqrt{32$

$CA = 4\sqrt{2$

So, we have:

AE = HC =4

$CA = EH = 4\sqrt 2$

The cross-section is a rectangle

Solving (b): The area of the cross-section

Area = Length * Width

$Area = 4 * 4\sqrt 2$

$Area = 16\sqrt 2$

AE = HC = DG = 4

Solving (c): How the cross-section is determined

Copied from (a)

Lines AC and EH will meet midway of lines AE and CH

So, the shape that will be formed is quadrilateral and the sides of the quadrilateral is: AE, EH, HC and CA

What is the shape of the cross-section formed when a plane containing line AC and line EH intersects

Mathematics

In this problem, we will find the volume of a solid with circular base of radius 2, for which parallel cross-sections perpendicular to the base are squares. To do this, we will assume that the base is the circle x2+y2=4, so that the solid lies between planes parallel to the x-axis at x=2 and x=−2. The cross-sections perpendicular to the x-axis are then squares whose bases run from the semicircle y=√−4−x2 to the semicircle y = √4−x2 Required: a. What is the area of the cross-section at x? b. What is the volume of the solid ?

Step-by-step answer

P Answered by PhD

1

(a) Each cross section is a square whose side length is decided by the distance in the x,y plane between the two curves $y=\sqrt{4-x^2}$ and $y=-\sqrt{4-x^2}$ , which is $2\sqrt{4-x^2}$ . Then each cross section has area

$(2\sqrt{4-x^2})^2=4(4-x^2)=\boxed{16-4x^2}$

(b) The volume of the solid is obtained by integrating the cross-sectional area from x = -2 to x = 2.

$\displaystyle\int_{-2}^2(16-4x^2)\,\mathrm dx=16x-\dfrac43x^3\bigg|_{-2}^2=\boxed{\dfrac{128}3}$

Engineering

At steady state, air at 200 kPa, 330 K, and mass flow rate of 0.1 kg/s enters an insulated duct having differing inlet and exit cross-sectional areas. The inlet cross-sectional area is 6 cm2. At the duct exit, the pressure of the air is 100 kPa and the velocity is 250 m/s. Neglecting potential energy effects and modeling air as an ideal gas with constant cp = 1.008 kJ/kg · K, determine: (a) the velocity of the air at the inlet, in m/s. (b) the temperature of the air at the exit, in K. (c) the exit cross-sectional area, in cm2.

Step-by-step answer

P Answered by Master

1

Photos are attached. Step wise solution is given.

1. Velocity of air at inlet is calculated.

2.Temperature of air at exit in Kelvin is also calculated.

3.Exit Cross-Sectional area

Best Regards.

At steady state, air at 200 kPa, 330 K, and mass flow rate of 0.1 kg/s enters an insulated duct havi

Mathematics

Aright rectangular prism has a height of 15 in, and the area of the cross section taken parallel to the base at a level of 5 in above the base is 25 in2. a right triangular prism has a height of 15 in, and the area of the cross section taken parallel to the base at a level of 5 in above the base is 25 in2. do the solids have the same volume? a. yes, because they have the same height and the same cross-sectional area at every level parallel to the bases. b. yes, because the solids are congruent. c. no, because the heights are not equal. d. no, because

Step-by-step answer

P Answered by PhD

28

A. Yes, because they have the same height and the same cross-sectional area at every level parallel to the bases.

A right prism has constant cross-sectional area at every level parallel to the base and the volume is the area of the base times the height.

Mathematics

Aright rectangular prism and an oblique triangular prism are both 12 centimeters tall and have the same volume. what statement must be true about the two solids? a. the area of the cross-sections of the prisms are multiples of each other. b. the vertical cross-sections of the prisms at the same width must have the same area. c. the cross-sections of the prisms are the same shape. d. the horizontal cross-sections of the prisms at the same height must have the same area.

Step-by-step answer

P Answered by Master

19

The volume of a prism is the product obtained when multiplying the area of its base to its height. If the two given solid figures have the same volume and the same height then, the areas of their bases are also equal. Thus, the answer is letter D.

Engineering

At steady state, air at 200 kPa, 325 K, and mass flow rate of 0.5 kg/s enters an insulated duct having differing inlet and exit cross-sectional areas. The inlet cross-sectional area is 6 cm2. At the duct exit, the pressure of the air is 100 kPa and the velocity is 250 m/s. Neglecting potential energy effects and modeling air as an ideal gas with constant cp = 1.008 kJ/kg.K, determine (a) the velocity of the air at the inlet, in m/s. (b) the temperature of the air at the exit, in K. (c) the exit cross-sectional area, in cm2.

Step-by-step answer

P Answered by Master

velocity of the air at the inlet = 388.64 m/s

temperature = 368.92 K

cross-sectional area = 21.176 cm²

Explanation:

given data

pressure p1 = 200 kPa

temperature t1 = 325 K

mass flow rate n = 0.5 kg/s

inlet cross-sectional area = 6 cm²

pressure of the air = 100 kPa

velocity = 250 m/s.

cp = 1.008 kJ/kg.K

solution

as we know that

mass = $\rho$ A V ..............1

and

pV = nRT ..........2

so we can say

$\frac{n}{V} = \frac{p}{RT}$ ........3

so $\rho = \frac{p}{RT}$

n = $\frac{p}{RT} (AV)$ ..........4

put her value we get

0.5 = $\frac{200\times 10^3}{287\times 325} (6\times 10^{-4}\times V )$

solve we get

V = 388.64 m/s

and

we apply now here steady flow energy that is

$n (h1 + \frac{v1^2}{2000} +\frac{gz1}{1000} ) +Q = n (h2 + \frac{v2^2}{2000} +\frac{gz2}{1000} ) + W$ ..........5

here no heat and work

so Q = 0 and W = 0

we get here h1 - h2 that is

h1 - h2 = $\frac{1}{2000} (v2^2-v1^2)$ ..............6

Cp(T1-T2) = $\frac{1}{2000} (v2^2-v1^2)$

put here value and we get

0.008 ( 325 - T2) = $\frac{1}{2000} (250^2-388.64^2)$

T2 = 368.92 K

and

for Area2 we put value in equation 4

0.5 = $\frac{100\times 10^3}{287\times 368.92} (A2\times 250 )$

solve it we get

A2 = 21.176 cm²

Mathematics

Aright rectangular prism and an oblique triangular prism are both 12 centimeters tall and have the same volume. what statement must be true about the two solids? a. the area of the cross-sections of the prisms are multiples of each other. b. the vertical cross-sections of the prisms at the same width must have the same area. c. the cross-sections of the prisms are the same shape. d. the horizontal cross-sections of the prisms at the same height must have the same area.

Step-by-step answer

P Answered by Specialist

19

The volume of a prism is the product obtained when multiplying the area of its base to its height. If the two given solid figures have the same volume and the same height then, the areas of their bases are also equal. Thus, the answer is letter D.

What is the area of the cross section?

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