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2 years ago

When doing perimeter and area do you add or multiply

Physics

2 answers:

pshichka [43]2 years ago

8 0

Answer:

Both

Explanation:

Lisa [10]2 years ago

3 0

Answer:

Perimeter = Add all of the side's lengths

Area = Multiply Length dimension and Width dimension together to get answer (different formulas for different shapes)

Explanation:

When we do perimeter, we're only taking the OUTSIDE line of the shape, we're not taking the inside body of the shape, so, to get the perimeter of a shape that has 4 sides, and the dimensions of 4 in 2 sides and 3 in 2 sides, then, we have to add like this... 4+4+3+3=14

So our perimeter of our <em>example</em> shape is 14

Now, when we are doing area, we're finding the INSIDE part of the shape, and to do that, we have to multiply. Let's take the same dimensions from our last example. If we have 4 in 2 sides and 3 in 2 sides, then we will do 4x3=12 to get our answer.

So our area of our <em>example</em> shape is 12

~Brainly Master - Helping Students~

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If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu

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3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

$\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}$ .

Volume of water:

$V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}$ .

Mass of water:

$m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}$ .

Amount of heat that the 15 L water absorbed:

$Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}$ .

What's the mass of the hot steel tool?

The specific heat of carbon steel is $0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}$ .

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

$Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}$ .

$\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}$ .

$m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}$ .

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Answer:

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Explanation:

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