What is a robot’s work envelope?

Engineering

1 answer:

Dmitry_Shevchenko [17]3 years ago

3 0

Answer:

Explanation:

A robot's work envelope is its range of movement. It is the shape created when a manipulator reaches forward, backward, up and down. These distances are determined by the length of a robot's arm and the design of its axes. ... A robot can only perform within the confines of this work envelope.

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How many times greater is the value of the 2 of the 270413 than the valuce of the 2 in 419427?

nadezda [96]

The value of 2 in 270,413 is 10,000x greater than the value of 2 in 419,427.
Step-by-step explanation:
The value of 2 in 270,413 is 200,000.
The value of 2 in 419,427 is 20
Divide the two numbers together to find your answer:
200,000/20 = 10,000
The value of 2 in 270,413 is 10,000x greater than the value of 2 in 419,427.

Please mark it as brainlest answer:).

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3 years ago

g A 30-m-diameter sedimentation basin has an average water depth of 3.0 m. It is treating 0.3 m3/s wastewater flow. Compute over

stepladder [879]

Answer:

The overflow rate is 4.24×10^-4 m/s.

The detention time is 7069.5 s

Explanation:

Overflow rate is given as volumetric flow rate ÷ area

volumetric flow rate = 0.3 m^3/s

area = πd^2/4 = 3.142×30^2/4 = 706.95 m^2

Overflow rate = 0.3 m^3/s ÷ 706.95 m^2 = 4.24×10^-4 m/s

Detention time = volume ÷ volumetric flow rate

volume = area × depth = 706.95 m^2 × 3 m = 2120.85 m^3

Detention time = 2120.85 m^3 ÷ 0.3 m^3/s = 7069.5 s

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3 years ago

An engine has been diagnosed with blowby.

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B, tech b only, Blow by is describe or defined as air going by the compression rings. Hope it helps

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The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

$Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1$

The Lower Surface Cp is given as

$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$