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Harlamova29_29 [7]
4 years ago
13

A room has a wall with an R value of 18 F sq.ft. hr/BTU. The room is 15 feet long and 11 feet wide with walls that are 9 ft high

. On a particular day the average outside temperature is 40 degrees F. How much heat is transferred through the walls on that day (in BTUs)?
Physics
1 answer:
seropon [69]4 years ago
5 0

Answer:

The heat transferred through the wall that day is  13728 BTUs

Explanation:

Here, we have the area of the wall given as

Area of wall = 2 × Length × Height + 2 × Width × Height

Length = 15 feet

Width = 11 Feet and

Height = 9 feet

Therefore, the area = 2×15×9 + 2×11×9 = 468 ft²

Temperature difference is given by

Average outside temperature - Wall temperature = 40 - 18 = 22 °F

Therefore the heat transferred through the wall that day (24 hours) at 18 sq.ft. hr/BTU is given by;

468 × 22 × 24/18 = 13728 = 13728 BTUs.

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The length of the adult mosquito is typically between 3.0 mm and 6.0 mm. The smallest known mosquitoes are around 2.5 mm, and th
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Unknown:

Length of largest known mosquito in inches = ?

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3 years ago
Consider the three displacement vectors
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Answer:

Explanation:

\overrightarrow{A} = 3\widehat{i}+3\widehat{j}

\overrightarrow{B} = \widehat{i}-4\widehat{j}

\overrightarrow{C} = -2\widehat{i}+5\widehat{j}

(a)

\overrightarrow{D} =\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}

\overrightarrow{D} =\left ( 3+1-2 \right )\widehat{i} +\left ( 3-4+5 \right )\widehat{j}

\overrightarrow{D} =\left 2\widehat{i} +4\widehat{j}

Magnitude of \overrightarrow{D} = \sqrt{2^{2}+4^{2}}

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tan\theta =\frac{4}{2}

θ = 63.44°

(b)

\overrightarrow{E} = - \overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}

\overrightarrow{E} =\left ( - 3- 1 -2 \right )\widehat{i} +\left ( - 3 + 4+5 \right )\widehat{j}

\overrightarrow{E} =- \left 6\widehat{i} +6\widehat{j}

Magnitude of \overrightarrow{E} = \sqrt{6^{2}+6^{2}}

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Let θ be the direction of vector D

tan\theta =\frac{6}{-6}

θ = 135°

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4 years ago
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