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CaHeK987 [17]
3 years ago
7

A roofing estimate uses a specific term known as ________________. (7 letters)

Engineering
2 answers:
erica [24]3 years ago
5 0

Answer: the word shingles is  correct

Explanation:

JulijaS [17]3 years ago
3 0

Answer:

shingles

Explanation:

You might be interested in
Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 280
stiks02 [169]

Answer:

Minimum Allowable diameter for each of the 5 bolts = 0.0394 m = 3.94 cm

Explanation:

Maximum Working Stress = Ultimate Shear Stress/factor of safety

Maximum Working Stress = 280 MPa/3.8 = 73.68 MPa

Working Stress = Applied load/Minimum allowable Area = L/A

Minimum Allowable Area = Applied Load/Maximum Working Stress

A = 450000/73680000 = 0.00611 m²

This area is supplied by 5 bolts, so each bolt supplies A/5 = 0.0061/5 = 0.00122 m²

Cross sectional Area of bolts = πD²/4

0.00122 = πD²/4

D² = 4 × 0.00122/π = 0.00155

D = √0.00155 = 0.0394 m = 3.94 cm

Each of the five bolt can have a minimum diameter of 3.94 cm

Hope this Helps!!!

8 0
3 years ago
A wood pipe having an inner diameter of 3 ft. is bound together using steel hoops having a cross sectional area of 0.2 in^2. The
Minchanka [31]

Answers:

31.7 inches

Explanation:

Given:

Diameter = 3ft

Let D = Diameter

So, D = 3ft. (Convert to inches)

D = 3 * 12in = 36 inches

Coss-sectional area of the steel = 0.2in²

Gauge Pressure (P) = 4psi

Stress in Steel (σ)= 11.4ksi

Force in steel = ½ (Pressure * Projected Area)

Area (A) = 2 * Force/Pressure

Also, Area (A) = Spacing (S) * Wood Pipe Diameter

Area = Area

2*Force/Pressure = Spacing * Diameter

Substitute values I to the above expression

2 * Force / 4psi = S * 36 inches

Also

Force in steel (F) = Stress in steel (σ) × Cross-sectional area of the steel

So, F = 11.4ksi * 0.2in²

F = 11.4 * 10³psi * 0.2in²

F = 2.28 * 10³ psi.in²

So, 2 * Force / 4psi = S * 36 becomes

2 * 2.28 * 10³/4 = S * 36

S = 2 * 2.28 * 10³ / (4 * 36)

S = 4560/144

S = 31.66667 inches

S = 31.7 inches (approximated)

5 0
4 years ago
Think for a moment about the potential problems with big data that the speaker mentioned:
Natalka [10]

Answer: 1. sadly yes, some people are treated unfairly for crimes people yet have commited. 2. no 3. yes

Explanation:

i did this last year

5 0
3 years ago
Read 2 more answers
Find the time-domain sinusoid for the following phasors:_________
sattari [20]

<u>Answer</u>:

a.  r(t) = 6.40 cos (ωt + 38.66°) units

b.  r(t) = 6.40 cos (ωt - 38.66°) units

c.  r(t) = 6.40 cos (ωt - 38.66°) units

d.  r(t) = 6.40 cos (ωt + 38.66°) units

<u>Explanation</u>:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = \sqrt{a^2 + b^2}

Ф = direction = tan⁻¹ (\frac{b}{a})

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{-5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{-5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

3 0
3 years ago
E = 50V, R= 1000 ohms, then I =
Margarita [4]

Explanation:

<em>Current</em><em> </em><em>(</em><em>I)</em><em>=</em><em> </em><em><u>Voltage</u></em><em><u> </u></em><em><u>(</u></em><em><u>V)</u></em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>Resistance </em><em>(Ω)</em><em>. </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>

<em>I=</em><em> </em><em><u>5</u></em><em><u>0</u></em><em><u>V</u></em>

<em> </em><em> </em><em> </em><em> </em><em>1000</em><em>(Ω)</em>

<em>I=</em><em> </em><em><u>5</u></em><em><u>V</u></em>

<em> </em><em> </em><em> </em><em> </em><em>1</em><em>0</em><em>0</em><em>(Ω)</em>

<em>I=</em><em> </em><em>0</em><em>.</em><em>0</em><em>5</em><em> </em><em>(</em><em>Amp)</em><em>. </em><em>Answer</em>

<em>Voltage=</em><em>V</em>

<em>Current</em><em>=</em><em>I</em>

<em>Resistance</em><em>=</em><em>(Ω)</em><em>. </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>(Ω)</em><em>=</em><em>Ohm</em>

<em>And </em><em>SI </em><em>unit</em><em> </em><em>of </em><em>current</em><em> </em><em>is</em><em>. </em><em>Ampere(</em><em>Amp)</em><em> </em>

<em><u>hope</u></em><em><u> this</u></em><em><u> helps</u></em><em><u> you</u></em>

<em><u>have</u></em><em><u> a</u></em><em><u> good</u></em><em><u> day</u></em><em><u>.</u></em>

<em> </em><em> </em><em> </em><em> </em><em> </em>

8 0
3 years ago
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