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user100 [1]
3 years ago
12

Which of the following allows team members to visualize a design model from a variety of perspectives?

Engineering
2 answers:
julsineya [31]3 years ago
8 0

Answer: from what i know im pretty sure its isometrics or sketches im certain its sketches but not 100%

Explanation: A sketch is a rapidly executed freehand drawing that is not usually intended as a finished work. A sketch may serve a number of purposes: it might record something that the artist sees, it might record

Yuki888 [10]3 years ago
4 0

Answer: Sketches

Explanation:

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Kazeer [188]

Answer:

1,2,3,5

Explanation:

Not positive but i think so

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3 years ago
A projectile is launched horizontally 1m above the ground. If it lands 300m away from the initial launch position, find: a)-the
PIT_PIT [208]

Answer:

(a): The launch velocity is Vx= 666.66 m/s.

(b): The angle wich the projectile contacts the ground is α= 0.38°

Explanation:

h= 1m

g= 9.8 m/s²

h= g*t²/2

t= 0.45 s

Vy= g*t

Vy= 4.42 m/s

d=Vx* t

Vx= 666.66 m/s (a)

α= tg⁻¹ ( Vy/Vx)

α= 0.38° (b)

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3 years ago
Can you identify the major theme illustrated by each of the following examples? If necessary, you may review the themes in Chapt
Bingel [31]
Where are the master plans stored and order given?
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3 years ago
You are traveling upstream on a river at dusk. You see a buoy with the number 5 and a flashing green light . What should you do?
nalin [4]

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Explanation:

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8 0
3 years ago
The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change i
ehidna [41]

Answer:

Change in length = 0.1257 mm

Change in diameter= -0.03771mm

Explanation:

Given

Diameter, d = 15 mm

Length of rod, L = 200mm

F = Force= 300N

d = 0.015m

Ep=2.70 GPa, np=0.4.

First, we have to calculate the normal stress using

σ = F/A where F = Force acting on the Cross-sectional area

A = Area

Area is calculated as πd²/4 where d = 0.015m

A = 22/7 * 0.015²/4

A = 0.000176785714285m²

A = 1.768E-4m²

So, stress. σ = 300N/1.768E-4m²

σ = 1696832.579185520Pa

σ = 1.697MPa

Calculating E(long)

E(long) = σ /Ep

E(long) = 1.697E-3/2.70

E(long) = 0.0006285

At this point, we fan now calculate the change in length of the element;

∆L = E(long) * L

∆L = 0.0006285 * 200mm

∆L = 0.1257mm

Calculating E(lat)

E(lat) = -np * E(long)

E(lat) = -4 * 0.0006285

E(lat) = -0.002514

At this point, we can now calculate the change in diameter of the element;

∆D = E(lat) * D

∆L = -0.002514 * 15mm

∆L = -0.03771mm

8 0
3 years ago
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