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Alexxandr [17]
3 years ago
14

the diameter of a quarter is 24.26 millimeters and the width is 1.75 millimeters. What is the approximate volume of the roll of

quarters, to the nearest tenth?

Mathematics
1 answer:
eimsori [14]3 years ago
6 0

Answer:

volume of the roll ≈ 32340.7 mm³

Step-by-step explanation:

The diameter of a quarter is given as 24.26 millimetres. The width is given as 1.75 millimetres. The width is the thickness of the quarter which is the height.

The roll or stack of quarters will conveniently a solid shape similar to a cylinder. The volume of the roll of the quarters is the same as the volume of a cylinder. The picture below is an example of a roll of quarters.

volume of a cylinder =  πr²h

r = radius = 24.26/2 = 12.13 millimetres

h = height = 1.75 × number of quarters

volume of a cylinder =  πr²h

Usually the number of quarters in a role =  40

h = 1.75 × 40 = 70 millimetres.

volume of a cylinder =  πr²h

volume of the roll =  3.14 × 12.13² × 70

volume of the roll =  3.14 × 147.1369 × 70

volume of the roll =  3.14 × 10299.583

volume of the roll =  32340.690620

volume of the roll ≈ 32340.7 mm³

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Answer:

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Step-by-step explanation:

Let´s analyze each movement and find its perpendicular components. Then, we are going to add the components along the x and y axes to get the resultant and finally to calculate its magnitude and direction.

To determine the perpendicular components, we will use right triangles trigonometrics ratios.

Please see images for each displacement in attached file.

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Ax= AcosФ = 2.50cos45° = 1.77 km (neg. direction)

Ay= AsenФ = 2.50sen45° = 1.77 km (pos. direction)

  • <u>Displacement B: 4.70 km and 60° south of east</u>

Bx= BcosФ = 4.70cos60° = 2,35 km (pos. direction)

By= BsenФ = 4.70sen60° = 4,07 km (neg. direction)

  • <u>Displacement C: 1.30 km and 25° south of west </u>

Cx= CcosФ = 1.30cos25° = 1,18 km (neg. direction)

Cy= CsenФ = 1.30sen25° = 0.55 km (neg. direction)

  • <u>Displacement D: 5.10 km due east (0°)</u>

Dx= DcosФ = 5.10 km (pos. direction)

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Ey= EcosФ = 1.70cos5° = 1.69 km (pos. direction)

  • <u>Displacement F: 7.20 km and 55° south of west</u>

Fx= FcosФ = 7.20cos55° = 4.13 km (neg. direction)

Fy= FsenФ = 7.20sen55° = 5.90 km (neg. direction)

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Gx= GcosФ = 2.80cos10° = 2.76 km (pos. direction)

Gy= GsenФ = 2.80sen10° = 0.49 km (pos. direction)

Now, we add the components along the x- and y- axis to find the components of the resultant (R):

Rx = Ax + Bx + Cx + Dx + Ex + Fx + Gx

Ry = Ay + By + Cy + Dy + Ey + Fy + Gy

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Ry= (1.77)+(-4.07)+(-0.55)+(0)+(1.69)+(-5.90)+(0.49)

<em>Note: minus (-) symbol to negative directions</em>

Rx = 3.28 km

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Let´s use the Theorem of Pythagoras to find the magnitud ot the resultant:

R = \sqrt{Rx^{2} +Ry^{2} }

R =\sqrt{(3.28)^{2}+(-6.57)^{2} }

R = 7,34 km

To find the direction of the resultant:

tanФ = \frac{Ry}{Rx}

tanФ = \frac{6.57}{3.28}

tanФ = 2.00

Ф = tan-1 (2.00)

Ф = 63,43° north of east

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The correct option is:

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