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sasho [114]
3 years ago
11

melinda is using a rectangular brass bar in a sculpture she is creating. the brass bar has a length that is 4 more than 3 times

its width if the area of the rectangular brass bar is 480 inches what are the dimensions answer
Engineering
2 answers:
omeli [17]3 years ago
8 0

Answer:

The dimension of Length = L

The dimension of width = L

The dimension of Area = L²

Explanation:

A dimension is one of the physical properties that are regarded as fundamental measures of a physical quantity, such as mass, length and time.

L - length , W - width , A- Area

A= W × L (Formula for Area)

Given:

L= 4 + 3W ( length that is 4 more than 3 times its width)

Area = 480 inches

A= W × L

Sub. Given into Formula of Area

480= W (4 + 3W)

480= 4W + 3W²

3W² + 4W - 480= 0

a=3 b=4 c= -480

where W = unknown using the quadratic equation

W= -b ± √ (b² - 4ac) / 2a

W= -4 ± √ (4² - 4.3.-480)/ 2.3

W= -4 ± √ (16 + 5760)/ 6

W= -4 ± √ (5776)/ 6

W = (-4 ± 76)/6

W = (-4 + 76)/6 or (-4-76)/6

W = 72/6 or -80/6

W = 12 or - 13.6

We choose w=12 inches

Sub. W into L

L= 4 + 3W

L= 4 + 3×12

L= 40 inches

Area= W × L

= 40 inches × 12 inches

Area = 480 inches ²

Dimension wise

The dimension of Length = L

The dimension of width = L

The dimension of Area = L²

lawyer [7]3 years ago
6 0

Answer:

180 x 60 inches

Width = 60 inches

Length = 180 inches

Explanation:

Given

Let L = Length

W = Width

P = Perimeter

Length = 3 * Width

L = 3W

Perimeter of Brass = 480 inches

P = 480

Perimeter is given as 2(L + W);

So, 2 (L + W) = 480

L + W = 480/2

L + W = 240

Substitute 3W for L; so,

3W + W = 240

4W = 240

W = 240/4

W = 60 inches

L = 3W

L = 3 * 60

L = 180 inches

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GrogVix [38]

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Program that removes all spaces from the given input

Explanation:

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5 0
3 years ago
Read 2 more answers
A sheet of steel 3-mm thick has nitrogen atomospheres on both sides at 900 C and is permitted to achieve a steady-state di usion
kati45 [8]

Answer:

X_B = 1.8 \times 10^{-3} m = 1.8 mm

Explanation:

Given data:

Diffusion constant for nitrogen is = 1.85\times 10^{-10} m^2/s

Diffusion flux = 1.0\times 10^{-7} kg/m^2-s

concentration of nitrogen at high presuure = 2 kg/m^3

location on which nitrogen  concentration is 0.5 kg/m^3   ......?

from fick's first law

J = D \frac{C_A C_B}{X_A X_B}

Take C_A as point  on which nitrogen concentration is 2 kg/m^3

x_B = X_A + D\frac{C_A -C_B}{J}

Assume X_A is zero at the surface

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4 0
3 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during
astra-53 [7]

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

\sigma_c = tensile stress = 200 Mpa

K = plane strain fracture toughness= 55 Mpa\sqrt{m}

\lambda= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for \lambda

Multiplying both sides of equation (1) by Y\sqrt{\pi\lambda}

\sigma_c Y\sqrt{\pi\lambda}=K   (2)

Sequaring both sides of equation (2):

(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2  

\sigma^2_c Y^2 \pi\lambda=K^2   (3)

Dividing both sides by \sigma^2_c Y^2 \pi we got:

\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2   (4)

Replacing the values into equation (4) we got:

\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

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