Compute the number of cubic centimeters of material in the locating saddle shown. Round the answer to the nearest cubic centimeter.

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Answer 1

Textbook Question

Chapter 64, Problem 28A

Compute the number of cubic centimeters of material in the locating saddle shown. Round the answer to the nearest cubic centimeter.
Chapter 64, Problem 28A, Compute the number of cubic centimeters of material in the locating saddle shown. Round the answer

Expert Solution & Answer

To determine

The number of cubic centimetres (Volume) of material in the locating saddle.

Answer to Problem 28A

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31 cm3.

Explanation of Solution

Write the expression of the number of cubic centimetres (Volume) of material in the locating saddle.

V=V1+2V2+V3−V4−V5 ........ (I)

Here, the volume of the part 1 is V1, the volume of the part 2 is V2, the volume of the part 3 is V3, the volume of the part 4 is V4 and the volume of the part 4 is V4.

Write the expression of the volume off the part 1 (Square part).

V1=a2h1 ........ (II)

Here, the side of the part 1 is a and the height of the part 1 is h1.

Write the expression of the volume off the part 2 (Rectangular part).

V2=l2b2h2 ........ (III)

Here, the height of the part 2 is h2, the length of the part 2 is l2 and the width of the part 2 is b2.

Write the expression of the volume off the part 3 (Rectangular part).

V3=l3b3h3 ........ (IV)

Here, the height of the part 3 is h3, the length of the part 3 is l3 and the width of the part 3 is b3.

Write the expression of volume of the part 4 (Semi circular part).

V4=πr2t4 ........ (V)

Here, the radius of the semi circular part is r and the thickness of the semi circular part is t4.

Write the expression of volume of the part 5 (Triangular part).

V5=12b5h5t5 ........ (VI)

Here, the thickness of the triangular part is t5, the height of the part 5 is h5 and the base of the triangular part is b5.

Calculation:

Substitute 6.5 cm for a and 2.5 cm for h1 in Equation (II).

V1=(6.5 cm)2(2.5 cm)=(6.5 cm)(6.5 cm)(2.5 cm)=105.63 cm3

Substitute 6.5 cm for l2, 1 cm for b2 and 3.5 cm for h2 in Equation (III).

V2=(6.5 cm)(1 cm)(3.5 cm)=(6.5 cm)(3.5 cm2)=22.75 cm3

Substitute 6.5 cm for l3, 1.5 cm for b3 and 1 cm for h3 in Equation (IV).

V3=(6.5 cm)(1.5 cm)(1 cm)=(6.5 cm)(1.5 cm2)=9.75 cm3

Substitute 1.5 cm for r and 1 cm for t4 in Equation (V).

V4=π(1.5 cm)2(1 cm)=π(2.25 cm2)(1 cm)=7.07 cm3

Substitute 2.5 cm for b5, 2 cm for h5 and 1 cm for t5 in Equation (VI).

V5=12(2.5 cm)(2 cm)(1 cm)=(0.5)(2.5 cm)(2 cm)(1 cm)=2.5 cm3

Substitute 105.63 cm3 for V1, 2.5 cm3 for V5, 7.07 cm3 for V4, 9.75 cm3 for V3 and 22.75 cm3 for V2 in Equation (I).

V=(105.63 cm3)+2(22.75 cm3)+(9.75 cm3)−(7.07 cm3)−(2.5 cm3)=(105.63 cm3)+(45.5 cm3)+(9.75 cm3)−(7.07 cm3)−(2.5 cm3)=151.31 cm3

Conclusion:

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31 cm3.

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Subject— Applied mathematics

Answer 2

Textbook Question

Chapter 64, Problem 28A

Compute the number of cubic centimeters of material in the locating saddle shown. Round the answer to the nearest cubic centimeter.
Chapter 64, Problem 28A, Compute the number of cubic centimeters of material in the locating saddle shown. Round the answer

Expert Solution & Answer

To determine

The number of cubic centimetres (Volume) of material in the locating saddle.

Answer to Problem 28A

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31 cm3.

Explanation of Solution

Write the expression of the number of cubic centimetres (Volume) of material in the locating saddle.

V=V1+2V2+V3−V4−V5 ........ (I)

Here, the volume of the part 1 is V1, the volume of the part 2 is V2, the volume of the part 3 is V3, the volume of the part 4 is V4 and the volume of the part 4 is V4.

Write the expression of the volume off the part 1 (Square part).

V1=a2h1 ........ (II)

Here, the side of the part 1 is a and the height of the part 1 is h1.

Write the expression of the volume off the part 2 (Rectangular part).

V2=l2b2h2 ........ (III)

Here, the height of the part 2 is h2, the length of the part 2 is l2 and the width of the part 2 is b2.

Write the expression of the volume off the part 3 (Rectangular part).

V3=l3b3h3 ........ (IV)

Here, the height of the part 3 is h3, the length of the part 3 is l3 and the width of the part 3 is b3.

Write the expression of volume of the part 4 (Semi circular part).

V4=πr2t4 ........ (V)

Here, the radius of the semi circular part is r and the thickness of the semi circular part is t4.

Write the expression of volume of the part 5 (Triangular part).

V5=12b5h5t5 ........ (VI)

Here, the thickness of the triangular part is t5, the height of the part 5 is h5 and the base of the triangular part is b5.

Calculation:

Substitute 6.5 cm for a and 2.5 cm for h1 in Equation (II).

V1=(6.5 cm)2(2.5 cm)=(6.5 cm)(6.5 cm)(2.5 cm)=105.63 cm3

Substitute 6.5 cm for l2, 1 cm for b2 and 3.5 cm for h2 in Equation (III).

V2=(6.5 cm)(1 cm)(3.5 cm)=(6.5 cm)(3.5 cm2)=22.75 cm3

Substitute 6.5 cm for l3, 1.5 cm for b3 and 1 cm for h3 in Equation (IV).

V3=(6.5 cm)(1.5 cm)(1 cm)=(6.5 cm)(1.5 cm2)=9.75 cm3

Substitute 1.5 cm for r and 1 cm for t4 in Equation (V).

V4=π(1.5 cm)2(1 cm)=π(2.25 cm2)(1 cm)=7.07 cm3

Substitute 2.5 cm for b5, 2 cm for h5 and 1 cm for t5 in Equation (VI).

V5=12(2.5 cm)(2 cm)(1 cm)=(0.5)(2.5 cm)(2 cm)(1 cm)=2.5 cm3

Substitute 105.63 cm3 for V1, 2.5 cm3 for V5, 7.07 cm3 for V4, 9.75 cm3 for V3 and 22.75 cm3 for V2 in Equation (I).

V=(105.63 cm3)+2(22.75 cm3)+(9.75 cm3)−(7.07 cm3)−(2.5 cm3)=(105.63 cm3)+(45.5 cm3)+(9.75 cm3)−(7.07 cm3)−(2.5 cm3)=151.31 cm3

Conclusion:

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31 cm3.

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Answer 3

Textbook Question

Chapter 64, Problem 28A

Compute the number of cubic centimeters of material in the locating saddle shown. Round the answer to the nearest cubic centimeter.
Chapter 64, Problem 28A, Compute the number of cubic centimeters of material in the locating saddle shown. Round the answer

Expert Solution & Answer

To determine

The number of cubic centimetres (Volume) of material in the locating saddle.

Answer to Problem 28A

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31 cm3.

Explanation of Solution

Write the expression of the number of cubic centimetres (Volume) of material in the locating saddle.

V=V1+2V2+V3−V4−V5 ........ (I)

Here, the volume of the part 1 is V1, the volume of the part 2 is V2, the volume of the part 3 is V3, the volume of the part 4 is V4 and the volume of the part 4 is V4.

Write the expression of the volume off the part 1 (Square part).

V1=a2h1 ........ (II)

Here, the side of the part 1 is a and the height of the part 1 is h1.

Write the expression of the volume off the part 2 (Rectangular part).

V2=l2b2h2 ........ (III)

Here, the height of the part 2 is h2, the length of the part 2 is l2 and the width of the part 2 is b2.

Write the expression of the volume off the part 3 (Rectangular part).

V3=l3b3h3 ........ (IV)

Here, the height of the part 3 is h3, the length of the part 3 is l3 and the width of the part 3 is b3.

Write the expression of volume of the part 4 (Semi circular part).

V4=πr2t4 ........ (V)

Here, the radius of the semi circular part is r and the thickness of the semi circular part is t4.

Write the expression of volume of the part 5 (Triangular part).

V5=12b5h5t5 ........ (VI)

Here, the thickness of the triangular part is t5, the height of the part 5 is h5 and the base of the triangular part is b5.

Calculation:

Substitute 6.5 cm for a and 2.5 cm for h1 in Equation (II).

V1=(6.5 cm)2(2.5 cm)=(6.5 cm)(6.5 cm)(2.5 cm)=105.63 cm3

Substitute 6.5 cm for l2, 1 cm for b2 and 3.5 cm for h2 in Equation (III).

V2=(6.5 cm)(1 cm)(3.5 cm)=(6.5 cm)(3.5 cm2)=22.75 cm3

Substitute 6.5 cm for l3, 1.5 cm for b3 and 1 cm for h3 in Equation (IV).

V3=(6.5 cm)(1.5 cm)(1 cm)=(6.5 cm)(1.5 cm2)=9.75 cm3

Substitute 1.5 cm for r and 1 cm for t4 in Equation (V).

V4=π(1.5 cm)2(1 cm)=π(2.25 cm2)(1 cm)=7.07 cm3

Substitute 2.5 cm for b5, 2 cm for h5 and 1 cm for t5 in Equation (VI).

V5=12(2.5 cm)(2 cm)(1 cm)=(0.5)(2.5 cm)(2 cm)(1 cm)=2.5 cm3

Substitute 105.63 cm3 for V1, 2.5 cm3 for V5, 7.07 cm3 for V4, 9.75 cm3 for V3 and 22.75 cm3 for V2 in Equation (I).

V=(105.63 cm3)+2(22.75 cm3)+(9.75 cm3)−(7.07 cm3)−(2.5 cm3)=(105.63 cm3)+(45.5 cm3)+(9.75 cm3)−(7.07 cm3)−(2.5 cm3)=151.31 cm3

Conclusion:

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31 cm3.

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Answer 4

Textbook Question

Chapter 64, Problem 28A

Compute the number of cubic centimeters of material in the locating saddle shown. Round the answer to the nearest cubic centimeter.
Chapter 64, Problem 28A, Compute the number of cubic centimeters of material in the locating saddle shown. Round the answer

Expert Solution & Answer

To determine

The number of cubic centimetres (Volume) of material in the locating saddle.

Answer to Problem 28A

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31 cm3.

Explanation of Solution

Write the expression of the number of cubic centimetres (Volume) of material in the locating saddle.

V=V1+2V2+V3−V4−V5 ........ (I)

Here, the volume of the part 1 is V1, the volume of the part 2 is V2, the volume of the part 3 is V3, the volume of the part 4 is V4 and the volume of the part 4 is V4.

Write the expression of the volume off the part 1 (Square part).

V1=a2h1 ........ (II)

Here, the side of the part 1 is a and the height of the part 1 is h1.

Write the expression of the volume off the part 2 (Rectangular part).

V2=l2b2h2 ........ (III)

Here, the height of the part 2 is h2, the length of the part 2 is l2 and the width of the part 2 is b2.

Write the expression of the volume off the part 3 (Rectangular part).

V3=l3b3h3 ........ (IV)

Here, the height of the part 3 is h3, the length of the part 3 is l3 and the width of the part 3 is b3.

Write the expression of volume of the part 4 (Semi circular part).

V4=πr2t4 ........ (V)

Here, the radius of the semi circular part is r and the thickness of the semi circular part is t4.

Write the expression of volume of the part 5 (Triangular part).

V5=12b5h5t5 ........ (VI)

Here, the thickness of the triangular part is t5, the height of the part 5 is h5 and the base of the triangular part is b5.

Calculation:

Substitute 6.5 cm for a and 2.5 cm for h1 in Equation (II).

V1=(6.5 cm)2(2.5 cm)=(6.5 cm)(6.5 cm)(2.5 cm)=105.63 cm3

Substitute 6.5 cm for l2, 1 cm for b2 and 3.5 cm for h2 in Equation (III).

V2=(6.5 cm)(1 cm)(3.5 cm)=(6.5 cm)(3.5 cm2)=22.75 cm3

Substitute 6.5 cm for l3, 1.5 cm for b3 and 1 cm for h3 in Equation (IV).

V3=(6.5 cm)(1.5 cm)(1 cm)=(6.5 cm)(1.5 cm2)=9.75 cm3

Substitute 1.5 cm for r and 1 cm for t4 in Equation (V).

V4=π(1.5 cm)2(1 cm)=π(2.25 cm2)(1 cm)=7.07 cm3

Substitute 2.5 cm for b5, 2 cm for h5 and 1 cm for t5 in Equation (VI).

V5=12(2.5 cm)(2 cm)(1 cm)=(0.5)(2.5 cm)(2 cm)(1 cm)=2.5 cm3

Substitute 105.63 cm3 for V1, 2.5 cm3 for V5, 7.07 cm3 for V4, 9.75 cm3 for V3 and 22.75 cm3 for V2 in Equation (I).

V=(105.63 cm3)+2(22.75 cm3)+(9.75 cm3)−(7.07 cm3)−(2.5 cm3)=(105.63 cm3)+(45.5 cm3)+(9.75 cm3)−(7.07 cm3)−(2.5 cm3)=151.31 cm3

Conclusion:

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31 cm3.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The number of cubic centimetres (Volume) of material in the locating saddle.

Answer to Problem 28A

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31 cm3.

Explanation of Solution

Write the expression of the number of cubic centimetres (Volume) of material in the locating saddle.

V=V1+2V2+V3−V4−V5 ........ (I)

Here, the volume of the part 1 is V1, the volume of the part 2 is V2, the volume of the part 3 is V3, the volume of the part 4 is V4 and the volume of the part 4 is V4.

Write the expression of the volume off the part 1 (Square part).

V1=a2h1 ........ (II)

Here, the side of the part 1 is a and the height of the part 1 is h1.

Write the expression of the volume off the part 2 (Rectangular part).

V2=l2b2h2 ........ (III)

Here, the height of the part 2 is h2, the length of the part 2 is l2 and the width of the part 2 is b2.

Write the expression of the volume off the part 3 (Rectangular part).

V3=l3b3h3 ........ (IV)

Here, the height of the part 3 is h3, the length of the part 3 is l3 and the width of the part 3 is b3.

Write the expression of volume of the part 4 (Semi circular part).

V4=πr2t4 ........ (V)

Here, the radius of the semi circular part is r and the thickness of the semi circular part is t4.

Write the expression of volume of the part 5 (Triangular part).

V5=12b5h5t5 ........ (VI)

Here, the thickness of the triangular part is t5, the height of the part 5 is h5 and the base of the triangular part is b5.

Calculation:

Substitute 6.5 cm for a and 2.5 cm for h1 in Equation (II).

V1=(6.5 cm)2(2.5 cm)=(6.5 cm)(6.5 cm)(2.5 cm)=105.63 cm3

Substitute 6.5 cm for l2, 1 cm for b2 and 3.5 cm for h2 in Equation (III).

V2=(6.5 cm)(1 cm)(3.5 cm)=(6.5 cm)(3.5 cm2)=22.75 cm3

Substitute 6.5 cm for l3, 1.5 cm for b3 and 1 cm for h3 in Equation (IV).

V3=(6.5 cm)(1.5 cm)(1 cm)=(6.5 cm)(1.5 cm2)=9.75 cm3

Substitute 1.5 cm for r and 1 cm for t4 in Equation (V).

V4=π(1.5 cm)2(1 cm)=π(2.25 cm2)(1 cm)=7.07 cm3

Substitute 2.5 cm for b5, 2 cm for h5 and 1 cm for t5 in Equation (VI).

V5=12(2.5 cm)(2 cm)(1 cm)=(0.5)(2.5 cm)(2 cm)(1 cm)=2.5 cm3

Substitute 105.63 cm3 for V1, 2.5 cm3 for V5, 7.07 cm3 for V4, 9.75 cm3 for V3 and 22.75 cm3 for V2 in Equation (I).

V=(105.63 cm3)+2(22.75 cm3)+(9.75 cm3)−(7.07 cm3)−(2.5 cm3)=(105.63 cm3)+(45.5 cm3)+(9.75 cm3)−(7.07 cm3)−(2.5 cm3)=151.31 cm3

Conclusion:

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31 cm3.

Answer 8

Expert Solution & Answer

To determine

The number of cubic centimetres (Volume) of material in the locating saddle.

Answer to Problem 28A

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31 cm3.

Explanation of Solution

Write the expression of the number of cubic centimetres (Volume) of material in the locating saddle.

V=V1+2V2+V3−V4−V5 ........ (I)

Here, the volume of the part 1 is V1, the volume of the part 2 is V2, the volume of the part 3 is V3, the volume of the part 4 is V4 and the volume of the part 4 is V4.

Write the expression of the volume off the part 1 (Square part).

V1=a2h1 ........ (II)

Here, the side of the part 1 is a and the height of the part 1 is h1.

Write the expression of the volume off the part 2 (Rectangular part).

V2=l2b2h2 ........ (III)

Here, the height of the part 2 is h2, the length of the part 2 is l2 and the width of the part 2 is b2.

Write the expression of the volume off the part 3 (Rectangular part).

V3=l3b3h3 ........ (IV)

Here, the height of the part 3 is h3, the length of the part 3 is l3 and the width of the part 3 is b3.

Write the expression of volume of the part 4 (Semi circular part).

V4=πr2t4 ........ (V)

Here, the radius of the semi circular part is r and the thickness of the semi circular part is t4.

Write the expression of volume of the part 5 (Triangular part).

V5=12b5h5t5 ........ (VI)

Here, the thickness of the triangular part is t5, the height of the part 5 is h5 and the base of the triangular part is b5.

Calculation:

Substitute 6.5 cm for a and 2.5 cm for h1 in Equation (II).

V1=(6.5 cm)2(2.5 cm)=(6.5 cm)(6.5 cm)(2.5 cm)=105.63 cm3

Substitute 6.5 cm for l2, 1 cm for b2 and 3.5 cm for h2 in Equation (III).

V2=(6.5 cm)(1 cm)(3.5 cm)=(6.5 cm)(3.5 cm2)=22.75 cm3

Substitute 6.5 cm for l3, 1.5 cm for b3 and 1 cm for h3 in Equation (IV).

V3=(6.5 cm)(1.5 cm)(1 cm)=(6.5 cm)(1.5 cm2)=9.75 cm3

Substitute 1.5 cm for r and 1 cm for t4 in Equation (V).

V4=π(1.5 cm)2(1 cm)=π(2.25 cm2)(1 cm)=7.07 cm3

Substitute 2.5 cm for b5, 2 cm for h5 and 1 cm for t5 in Equation (VI).

V5=12(2.5 cm)(2 cm)(1 cm)=(0.5)(2.5 cm)(2 cm)(1 cm)=2.5 cm3

Substitute 105.63 cm3 for V1, 2.5 cm3 for V5, 7.07 cm3 for V4, 9.75 cm3 for V3 and 22.75 cm3 for V2 in Equation (I).

V=(105.63 cm3)+2(22.75 cm3)+(9.75 cm3)−(7.07 cm3)−(2.5 cm3)=(105.63 cm3)+(45.5 cm3)+(9.75 cm3)−(7.07 cm3)−(2.5 cm3)=151.31 cm3

Conclusion:

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31 cm3.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The number of cubic centimetres (Volume) of material in the locating saddle.

Answer 12

Answer to Problem 28A

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31 cm3.

Answer 13

Explanation of Solution

Write the expression of the number of cubic centimetres (Volume) of material in the locating saddle.

V=V1+2V2+V3−V4−V5 ........ (I)

Here, the volume of the part 1 is V1, the volume of the part 2 is V2, the volume of the part 3 is V3, the volume of the part 4 is V4 and the volume of the part 4 is V4.

Write the expression of the volume off the part 1 (Square part).

V1=a2h1 ........ (II)

Here, the side of the part 1 is a and the height of the part 1 is h1.

Write the expression of the volume off the part 2 (Rectangular part).

V2=l2b2h2 ........ (III)

Here, the height of the part 2 is h2, the length of the part 2 is l2 and the width of the part 2 is b2.

Write the expression of the volume off the part 3 (Rectangular part).

V3=l3b3h3 ........ (IV)

Here, the height of the part 3 is h3, the length of the part 3 is l3 and the width of the part 3 is b3.

Write the expression of volume of the part 4 (Semi circular part).

V4=πr2t4 ........ (V)

Here, the radius of the semi circular part is r and the thickness of the semi circular part is t4.

Write the expression of volume of the part 5 (Triangular part).

V5=12b5h5t5 ........ (VI)

Here, the thickness of the triangular part is t5, the height of the part 5 is h5 and the base of the triangular part is b5.

Calculation:

Substitute 6.5 cm for a and 2.5 cm for h1 in Equation (II).

V1=(6.5 cm)2(2.5 cm)=(6.5 cm)(6.5 cm)(2.5 cm)=105.63 cm3

Substitute 6.5 cm for l2, 1 cm for b2 and 3.5 cm for h2 in Equation (III).

V2=(6.5 cm)(1 cm)(3.5 cm)=(6.5 cm)(3.5 cm2)=22.75 cm3

Substitute 6.5 cm for l3, 1.5 cm for b3 and 1 cm for h3 in Equation (IV).

V3=(6.5 cm)(1.5 cm)(1 cm)=(6.5 cm)(1.5 cm2)=9.75 cm3

Substitute 1.5 cm for r and 1 cm for t4 in Equation (V).

V4=π(1.5 cm)2(1 cm)=π(2.25 cm2)(1 cm)=7.07 cm3

Substitute 2.5 cm for b5, 2 cm for h5 and 1 cm for t5 in Equation (VI).

V5=12(2.5 cm)(2 cm)(1 cm)=(0.5)(2.5 cm)(2 cm)(1 cm)=2.5 cm3

Substitute 105.63 cm3 for V1, 2.5 cm3 for V5, 7.07 cm3 for V4, 9.75 cm3 for V3 and 22.75 cm3 for V2 in Equation (I).

V=(105.63 cm3)+2(22.75 cm3)+(9.75 cm3)−(7.07 cm3)−(2.5 cm3)=(105.63 cm3)+(45.5 cm3)+(9.75 cm3)−(7.07 cm3)−(2.5 cm3)=151.31 cm3

Conclusion:

The number of cubic centimetres (Volume) of material in the locating saddle is 151.31 cm3.

Answer 14

Ch. 64 - In order to make a conical duct a circular sheet...Ch. 64 - Prob. 2A Ch. 64 - Prob. 3A Ch. 64 - Convert 12.02 sq in. to square centimeters.Ch. 64 - Prob. 5A Ch. 64 - Prob. 6A Ch. 64 - Prob. 7A Ch. 64 - Compute the volume of each sphere 7 through 14....Ch. 64 - Prob. 9A Ch. 64 - Compute the volume of each sphere 7 through 14....

Compute the number of cubic centimeters of material in the locating saddle shown. Round the answer to the nearest cubic centimeter.

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Answer to Problem 28A

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