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dimulka [17.4K]

3 years ago

9

Sam, a carpenter, is asked to identify the abilities he has that are important to his work. What are the top abilities he might

list? (Select all that apply.)
inductive reasoning

problem sensitivity

visualization

manual dexterity

fluency of ideas

2 answers:

Sergeeva-Olga [200]3 years ago

6 0

Answer

visualization, problem sensitivity, and manual dexterity

Explanation

rodikova [14]3 years ago

6 0

Answer:

it is visualization, problem sensitivity, and manual dexterity

Explanation:

I just did the assignment on edge and got it right

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What term describes a guardrail system

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Answer:

The ( Fall Protection in Construction ) is the term which relates to these mechanisms , i.e , guardrail system, a positioning system and a personal fall arrest system.

Explanation:

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A scale on a blue print drawing of a house shows that 666 centimeters represents 333 meters.

Dennis_Churaev [7]

Answer:

545454cm

Explanation:

The blue print drawing of the house shows 666 centimeters, but the real picture of the house is 333 meters. So let the number of cm on the blueprint that represent the distance of 272727 meters be x. Firstly convert the meters to centimeters 666cm = 333m, x=272727m ; then cross multiply, 666cm=33300cm x=27272700cm ; x =(666cm×272727cm)/33300cm =545454cm.

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3 years ago

A vertical plate has a sharp-edged orifice at its center. A water jet of speed V strikes the plate concentrically. Obtain an exp

Vesnalui [34]

Answer:

Force = 455672061 N

Explanation:

given data

V = 515 ft/s

D = 54 in = 4.5 ft

d = 51 in = 4.25 ft

to find out

Evaluate the force

solution

we know that force is express as

force F = ρ ΔA v² ..................1

put here value we get

force F = 1000 × $\frac{\pi}{4}$ [ D² - d² ] × v²

F = 1000 × $\frac{\pi}{4}$ [ 4.5² - 4.25² ] × 515²

Force = 455672061 N

8 0

4 years ago

An engineer must design a rectangular box that has a volume of 9 m3 and that has a bottom whose length is twice its width. What

Jet001 [13]

Answer:

$Length =3$ $Height = 2$ and $Width = \frac{3}{2}$

Explanation:

Given

$Volume = 9m^3$

Represent the height as h, the length as l and the width as w.

From the question:

$Length = 2 * Width$

$l = 2w$

Volume of a box is calculated as:

$V = l*w*h$

This gives:

$V = 2w *w*h$

$V = 2w^2h$

Substitute 9 for V

$9 = 2w^2h$

Make h the subject:

$h = \frac{9}{2w^2}$

The surface area is calculated as:

$A = 2(lw + lh + hw)$

Recall that: $l = 2w$

$A = 2(2w*w + 2w*h + hw)$

$A = 2(2w^2 + 2wh + hw)$

$A = 2(2w^2 + 3wh)$

$A = 4w^2 + 6wh$

Recall that: $h = \frac{9}{2w^2}$

So:

$A = 4w^2 + 6w * \frac{9}{2w^2}$

$A = 4w^2 + 6* \frac{9}{2w}$

$A = 4w^2 + \frac{6* 9}{2w}$

$A = 4w^2 + \frac{3* 9}{w}$

$A = 4w^2 + \frac{27}{w}$

To minimize the surface area, we have to differentiate with respect to w

$A' = 8w - 27w^{-2}$

Set A' to 0

$0 = 8w - 27w^{-2}$

Add $27w^{-2}$ to both sides

$27w^{-2} = 8w$

Multiply both sides by $w^2$

$27w^{-2}*w^2 = 8w*w^2$

$27 = 8w^3$

Make $w^3$ the subject

$w^3 = \frac{27}{8}$

Solve for w

$w = \sqrt[3]{\frac{27}{8}}$

$w = \frac{3}{2}$

Recall that : $h = \frac{9}{2w^2}$ and $l = 2w$

$h = \frac{9}{2 * \frac{3}{2}^2}$

$h = \frac{9}{2 * \frac{9}{4}}$

$h = \frac{9}{\frac{9}{2}}$

$h = 9/\frac{9}{2}$

$h = 9*\frac{2}{9}$

$h= 2$

$l = 2w$

$l = 2 * \frac{3}{2}$

$l = 3$

Hence, the dimension that minimizes the surface area is:

$Length =3$ $Height = 2$ and $Width = \frac{3}{2}$

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Answer:

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