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3 years ago

If you are a government authority what extend will you modify the existing policy

Engineering

1 answer:

mr_godi [17]3 years ago

4 0

Answer:

Explanation:

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How did Atlantis benefit from lessons learned in construction of earlier orbiters?

Alenkasestr [34]

Answer:

Atlantis benefited from lessons learned in the construction and testing of Enterprise, Columbia and Challenger. ... The Experience gained during the Orbiter assembly process also enabled Atlantis to be completed with a 49.5 percent reduction in man hours (compared to Columbia).

Explanation:

8 0

3 years ago

Please answer the questions !

gizmo_the_mogwai [7]

Answer:

120

Explanation:

6 0

3 years ago

Water flows through a nozzle at the end of a fire hose. If the nozzle exit velocity must be 20 m/s and the exit diameter is 40 m

lara31 [8.8K]

Answer:

minimum flow rate provided by pump is 0.02513 m^3/s

Explanation:

Given data:

Exit velocity of nozzle = 20m/s

Exit diameter = 40 mm

We know that flow rate Q is given as

$Q = A \times V$

where A is Area

$A =\frac{\pi}{4} \times (40\times 10^{-3})^2 = 1.256\times 10^{-3} m^2$

$Q = 1.256\times 10^{-3} \times 20 = 0.02513 m^3/s$

minimum flow rate provided by pump is 0.02513 m^3/s

5 0

3 years ago

write down your own definition of Engineering, preferably in 4-5 sentences. Maximum of 150 words for your definition???.

ollegr [7]

Answer:

A charge q1=7.0mc is located at the origin and a second charge q2=-5.0mc is located on the x axis, 0.3m the origin find the electric field at the point p which he's coordinates (0,0.40)m

4 0

3 years ago

Using the Rayleigh criterion, calculate the minimum feature size that can be resolved in a system with a 0.18 NA lens when g-lin

Vladimir79 [104]

Answer:

# for a g line, R = 1.211 μm

# for an I-line, R = 1.013 μm

# for a g line, R = 0.726 μm

# for an I-line, R = 0.243 μm

# for a g line, R = 0.605 μm

# for an I-line, R = 0.608 μm

Explanation:

We know that;

Rayleigh Resolution R = 0.5 × λ/NA

for a g line, λ = 436 nm

for an I-line λ = 365 nm

Now when NA = 0.18

# for a g line, λ = 436 nm

R = 0.5 × 436/0.18 = 1.211 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.18 = 1.013 μm

when NA = 0.30

# for a g line, λ = 436 nm

R = 0.5 × 436/0.30 = 0.726 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 = 0.243 μm

when NA = 0.36

# for a g line, λ = 436 nm

R = 0.5 × 436/0.36 = 0.605 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 = 0.608 μm

6 0

3 years ago