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The mass of a certain wire varies jointly as its length and the square of its diameter. If 750 m of wire with a diameter of 2.5 mm has a mass of 45 kg, what will be the mass of 1.5 km of the wire with a diameter of 4.5 mm?

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Read the excerpt from Mark Twain's "The Adventures of Tom Sawyer":
[tex]"Tom appeared on the sidewalk with a bucket of whitewash and a long-handled brush. He surveyed the fence, and all gladness left him and a deep melancholy settled down upon his spirit. Thirty yards of board fence nine feet high. Life to him seemed hollow, and existence but a burden."[/tex]
Which best describes Tom’s attitude toward the task of whitewashing the fence?
O
O
O
O Resigned
Enthusiastic
Unconcerned
Reluctant
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Response:
Read the excerpt from Mark Twain's "The Adventures of Tom Sawyer":

"Tom appeared on the sidewalk with a bucket of whitewash and a long-handled brush. He surveyed the fence, and all gladness left him and a deep melancholy settled down upon his spirit. Thirty yards of board fence nine feet high. Life to him seemed hollow, and existence but a burden."

Which best describes Tom’s attitude toward the task of whitewashing the fence?

A. Resigned
B. Enthusiastic
C. Unconcerned
D. Reluctant



Answer :

GinnyAnswer
To solve this problem of determining the mass of a wire based on its length and diameter, we employ the principle of joint variation. According to the problem, the mass of the wire varies jointly with its length and the square of its diameter. Mathematically, this relationship can be expressed as:

[tex]\[ \text{mass} = k \times \text{length} \times (\text{diameter})^2 \][/tex]

where [tex]\( k \)[/tex] is the constant of proportionality.

### Step 1: Determine the constant of proportionality [tex]\( k \)[/tex]

We are given that when the length ([tex]\( L_1 \)[/tex]) is 750 meters, the diameter ([tex]\( D_1 \)[/tex]) is 2.5 millimeters, and the mass ([tex]\( M_1 \)[/tex]) is 45 kilograms.

Using the joint variation formula, we have:
[tex]\[ 45 = k \times 750 \times (2.5)^2 \][/tex]

First, calculate the square of the diameter:
[tex]\[ (2.5)^2 = 6.25 \][/tex]

Substitute this value back into the equation:
[tex]\[ 45 = k \times 750 \times 6.25 \][/tex]

Next, solve for [tex]\( k \)[/tex] by isolating it:
[tex]\[ 45 = k \times 4687.5 \][/tex]
[tex]\[ k = \frac{45}{4687.5} \][/tex]
[tex]\[ k = 0.0096 \][/tex]

Thus, the constant of proportionality [tex]\( k \)[/tex] is 0.0096.

### Step 2: Calculate the mass for the second wire

Next, we use the same relationship to find the mass ([tex]\( M_2 \)[/tex]) of a second wire with a length [tex]\( L_2 \)[/tex] of 1.5 kilometers (which is 1500 meters) and a diameter [tex]\( D_2 \)[/tex] of 4.5 millimeters.

Again, the formula is:
[tex]\[ \text{mass} = k \times \text{length} \times (\text{diameter})^2 \][/tex]

Substitute in the known values:
[tex]\[ M_2 = 0.0096 \times 1500 \times (4.5)^2 \][/tex]

Calculate the square of the diameter:
[tex]\[ (4.5)^2 = 20.25 \][/tex]

Now, substitute this value back into the equation:
[tex]\[ M_2 = 0.0096 \times 1500 \times 20.25 \][/tex]

Finally, perform the calculations:
[tex]\[ M_2 = 0.0096 \times 1500 \times 20.25 \][/tex]
[tex]\[ M_2 = 0.0096 \times 30375 \][/tex]
[tex]\[ M_2 = 291.6 \][/tex]

Thus, the mass of the second wire, which is 1.5 km long with a diameter of 4.5 mm, is approximately 291.6 kilograms.

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