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1 year ago

To meet the needs of a client, what is best for an interior designer to do?

Engineering

1 answer:

umka2103 [35]1 year ago

6 0

To meet the needs of a client, what is best for an interior designer to do is research the project requirements .

<h3>What does an interior designer do for a client?</h3>

Interior designers strive to suit the unique requirements of each customer while also creating appealing, practical, and secure workplaces.

Make a life plan. Space planning comes first in the interior design process, according to Nesen. The American Institute of Architects defines space planning as the process of delineating interior establishing circulation patterns, defining spatial areas, and developing plans for furniture and equipment placement.

Be specific about your wants, demands, and objectives. Make a wish list and assist the interior designer in creating an appropriate project brief. Describe the issues you believe need to be resolved. Establish a deadline for doing the assigned work.

To learn more about interior design refer to:

brainly.com/question/27741315

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3 years ago

An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e

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Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in $2.25 * 10^4 BTU$ . It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - $1 BTU = 778.17 lbf*ft$

$2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft$

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:

$Ep = m*g*(h_2 - h_1)\\ W = m*g$

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.

$h_2$ is the final elevation and $h_1$ is the initial elevation.

Replacing W in the Ep equation

$Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\$

Finally, the next step is to replace the variables of the problem.

$h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft$

The elevation at the high point of the road is 12186.5 in ft.

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3 years ago

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Answer:

D) quantity of components required for this type of system

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