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3 years ago

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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A Michelson interferometer operating at a 500 nm wavelength has a 3.73-cm-long glass cell in one arm. To begin, the air is pumpe

LekaFEV [45]

Answer:

The number of bright-dark fringe is 42

Solution:

As per the question:

Wavelength of light, $\lambda = 500\ nm = 500\times 10^{- 9}\ m$

Length of the glass cell, x = 3.73 cm = 0.0373 m

Refractive index, $\mu = 1.00028$

Now,

To calculate the bright-dark fringe shifts, we use the formula given below:

$d_{m} = \frac{2x}{\lambda }\times (\mu - 1)$

Now, substituting the appropriate values in the above formula:

$d_{m} = \frac{2\times 0.0373}{500\times 10^{- 9}}\times (1.00028 - 1)$

$d_{m} = 41.77$ ≈ 42

6 0

3 years ago

A smoking lounge is to accommodate 19 heavy smokers. The minimum fresh air requirement for smoking lounges is specified to be 30

igor_vitrenko [27]

Answer 1: minimum required flow rate of fresh air is 0.57 m^3/ses

Explanation: since the minimum requirement per person is 30 L/sec

Converting to m^3 it becomes

30/1000 = 0.03 m^3/sec

For 19 heavy smoker we will require

19 * 0.03 = 0.57m^3/sec

Answer 2: diameter of the duct will be 0.3m

Explanation: since flow rate is

Q =0.57m^3/sec

Also

Q = AV (continuity equation)

Where A is the duct area and V is the velocity of air flow in m/sec

0.57m^3/sec = A * 8m/sec

A = 0.57/8 = 0.071m^2

Area of the duct is that of a circle

A = 3.142 *(d^2 ÷4)

d^2 = (0.017 * 4)/3.142 = 0.09

d is square root of 0.09

d = 0.3m

7 0

3 years ago

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2.5×10-4

krok68 [10]

Answer:

1788.9 MPa

Explanation:

The magnitude of the maximum stress (σ) can be calculated usign the following equation:

$\sigma = 2\sigma_{0} \sqrt{\frac{a}{\rho}}$

Where:

ρ: is the radius of curvature = 2.5x10⁻⁴ mm (0.9843x10⁻⁵ in)

σ₀: is the tensile stress = 100x10⁶ Pa (14500 psi)

2a: is the crack length = 4x10⁻² mm (1.575x10⁻³ in)

Hence, the maximum stress (σ) is:

$\sigma = 2*100\cdot 10^{6} Pa \sqrt{\frac{(4 \cdot 10^{-2} mm)/2}{2.5 \cdot 10^{-4} mm}} = 1.79 \cdot 10^{6} Pa = 1788.9 MPa$

Therefore, the magnitude of the maximum stress is 1788.9 MPa.

I hope it helps you!

5 0

4 years ago

Due at 11:59pm please help

sergeinik [125]

I believe it’s c table

7 0

3 years ago

A computer has a two-level cache. Suppose that 60% of the memory references hit on the first level cache, 35% hit on the second

Andreyy89

Answer:

t=14ns

Explanation:

We make the relation between the specific access time and the memory percentage in each level, so

$60\% \Rightarrow 60/100 = 0.60\\35\% \Rightarrow 35/100 = 0.35\\05\% \Rightarrow 05/100 = 0.05$

$t= 0.6(5) + 0.35(5+15) + 0.05(5+15+60)\\t= 0.6(5) + 0.35(20) + 0.05(80)\\t= 3 + 7 + 4\\t= 14 ns$

Average Access Time is 14 nsec.

4 0

3 years ago