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jeka94
4 years ago
8

(Metric) Given the front and right views, create an AutoCAD isometric of the object. Save the drawing as lab5c.dwg. Grid size is

10mm. You may approximate the location of the hidden line as long as it is consistent with the other view
Engineering
1 answer:
Romashka-Z-Leto [24]4 years ago
6 0

Answer:

The explanation is provided below

Explanation:

Consider constructing a diagram using the latest version of Autocad and save the file with the appropriate extension.

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Lawn maintenance is an alternative energy source<br> -true<br> -false
Reika [66]

Answer:

false

Explanation:

4 0
3 years ago
What is the weight of a glider with a mass of 5.3 grams? (Hint: watch your units!)
Vinvika [58]

Answer: 0.053

Explanation:

So, we convert 5.3 grams into kilograms.

5.3g = 0.0053 kg (Since 1kg equals 1000g)

On Earth, gravity is 10 N/kg.

Weight = mass x gravity

Weight = 0.0053kg x 10 N/kg

Weight = 0.053 Newtons (On Earth)

6 0
3 years ago
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 vehicle of 1 200 kg is moving at a speed of 40 km/h on an incline of 1 in 50. The total constant rolling and wind resistance i
Readme [11.4K]

Answer:11.602 KW

Explanation:

mass of vehicle\left ( m\right )=1200 kg

speed=40Km/h

Resistance=600 N

\eta =80%

Gear ratio\left ( G\right )=4:1

D_{effective}=500mm

Net force to overcome by engine is

F=Resistance + sin component of weight

F=600+mgsin\theta

Where tan\theta =[tex]\frac{1}{50}

\theta =1.1457^{\circ}

F=600+1200\times 9.81\times sin\left ( 1.1457\right )

F=600+235.38=835.38 N

power=F.v=835.38\frac{100}{9}

Engine Power=\frac{835.\frac{100}{9}}{\eta }=11.602 KW

4 0
4 years ago
How to pass sharp edged tools to another student in welding
lisov135 [29]

Answer: When walking with a sharp tool, the tool should be carried with the blade down and away from the body. When climbing with a sharp tool, tool belts, or buckets with hand lines should be used so workers can have both hands to grip the ladder.

Explanation:

3 0
3 years ago
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