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3 years ago

What is the first step in the Design Process *

Engineering

1 answer:

Gennadij [26K]3 years ago

8 0

Answer:

Define the problem

I hope this helps!

Explanation:

The problem must be know before you can develop and solution, generate concepts, or construct and test a prototype

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Why is GoPro the easiest way to capture and share your experiences?

max2010maxim [7]

Answer:

GoPro cameras have a fixed 170-degree lens. This allows for wide-angle photos and videos. Basically, at 170-degrees, it will capture almost everything in front of the camera. To get absolutely everything, it would need another 10-degrees

Explanation:

3 0

2 years ago

Take water density and kinematic viscosity as p=1000 kg/m3 and v= 1x10^-6 m^2/s. (c) Water flows through an orifice plate with a

guapka [62]

Answer:

$K_v=12.34$

Explanation:

Given;

For orifice, loss coefficient, K₀ = 10

Diameter, D₀ = 45 mm = 0.045 m

loss coefficient of the orifice, Ko = 10

Diameter of the gate valve, Dy = 1.5D₀ = 1.5 × 0.045 m = 0.0675 m

Total head drop, Δhtotal=25 m

Discharge, Q = 10 l/s = 0.01 m³/s

Now,

the velocity of flow through orifice, Vo = Discharge / area of the orifice

Vo = $\frac{0.01}{\frac{\pi}{4}0.045^2}$

Vo = 6.28 m/s

also,

the velocity of flow through gate valve, $V_v$ = Discharge / area of the orifice

$V_v$ = $\frac{0.01}{\frac{\pi}{4}0.0675^2}$

$V_v$ = 2.79 m/s

Now,

the total head drop = head drop at orifice + head drop at gate valve

25 m = $K_o\frac{V_o^2}{2g}+K_v\frac{V_v^2}{2g}$

where,

$K_v$ is the loss coefficient for the gate valve

on substituting the values, we get

25 m = $10\frac{6.28^2}{2\times 9.81}+K_v\frac{2.79^2}{2\times9.81}$

$K_v\frac{2.79^2}{2\times9.81}$ = 4.898

$K_v=12.34$

3 0

4 years ago

10. The repair order is a legal document because

babymother [125]

I think it’s B.) it’s signed by the customer

4 0

4 years ago

Suppose we are managing a consulting team of expert computer hackers, and each week we have to choose a job for them to undertak

lina2011 [118]

Answer:

if number == 1

then

tempSolution= max(l[number],h[number])

else if number == 2 then

tempSolution= max(optimalPlan(1, l, h)+ l[2], h[2])

else

tempSolution= max(optimalPlan(number − 1, l, h) + l[number], optimalPlan(number − 2, l, h) + h[number])

end if

return Value

FindOptimalValue(number, l, h)

for itterator = 1 ! number do

tempSolution[itterator] = 0

end for

for itterator = 1 ! number do

if itterator == 1 then

tempSolution[itterator] max(l[itterator], h[itterator])

else if itterator == 2 then

tempSolution[itterator] max(tempSolution[1] + l[2], h[2])

else

tempSolution[itterator] max(tempSolution[itterator − 1] + l[itterator], tempSolution[itterator − 2] + h[itterator])

end if

end for

return Value[number]

OPtimalPlan(number, l, h, Value)

for itterator = 1 ! number do

WeekVal[itterator]

end for

if tempSolution[number] − l[number] = tempSolution[number − 1] then

WeekVal[number] ”Low stress”

OPtimalPlan(number-1, l, h, Value)

else

WeekVal[number] ”High stress”

OPtimalPlan(number-2, l, h, Value)

end if

return WeekVal

7 0

3 years ago

A closed, 5-m-tall tank is filled with water to a depth of 4 m. The top portion of the tank is filled with air which, as indicat

Brums [2.3K]

Answer:

The pressure that the water exerts on the bottom of the tank is 59.2 kPa

Explanation:

Given;

height of tank, h = 5m

height of water in the tank, $h_w$ = 4m

pressure at the top of the tank, $P_{top}$ = 20 kPa

The pressure exerted by water at the bottom of the tank is the sum of pressure on water surface and pressure due to water column.

$P_{bottom} = \gamma h + P_{top}\\\\P_{bottom} = (9.8*10^3*4 \ \ + \ 20*10^3)Pa\\\\P_{bottom} = 59200 \ Pa\\\\P_{bottom} = 59.2 \ kPa$

Therefore, the pressure that the water exerts on the bottom of the tank is 59.2 kPa

6 0

3 years ago