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2 years ago

7

Why would an aerospace engineer limit the maximum angle of deflection of the control surfaces?

1 answer:

WITCHER [35]2 years ago

4 0

Answer:

You can create high drag which allows a steeper angle without increasing your air speed on landing. you can reduce the length of landing role. Flaps are also used to increase the drag they are retracted when they are not needed. it is adviseable to down he flaps during the time of take off.

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The art of manipulating, influencing, or deceiving you into taking some action that isn’t in your best interest or in the best i

Mrac [35]

Social engineering is the answer

7 0

2 years ago

If the efficiency of the boiler is 91.2 % , the overall efficiency of the turbine, which includes the Carnot efficiency and its

Tju [1.3M]

Answer:

Net efficiency of generating unit = 42.08 - 5 = 37.08 %

Explanation:

We have given that efficiency of the boiler = 91.2 % = 0.912

Carnot efficiency = 46.9 % = 0.469

Efficiency of generator = 98.4% =0.984

We have to find the efficiency of total generating unit

For finding the efficiency of total generating unit we have to multiply all the efficiencies

So efficiency of generating unit = 0.912×0.469×0.984 = 0.4208 = 42.08 %

For plant losses we have to subtract 5%

So net efficiency of generating unit = 42.08 - 5 = 37.08 %

4 0

3 years ago

A 100-ampere resistor bank is connected to a controller with conductor insulation rated 75°C. The resistors are not used in conj

Naily [24]

Answer:

answer

Explanation:

6 0

3 years ago

A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T

anyanavicka [17]

Answer:

R min = 28.173 ohm

R max = 1.55 × $10^{4}$ ohm

Explanation:

given data

capacitor = 0.227 μF

charged to 5.03 V

potential difference across the plates = 0.833 V

handled effectively = 11.5 μs to 6.33 ms

solution

we know that resistance range of the resistor is express as

V(t) = $V_o \times e^{t\RC}$ ...........1

so R will be

R = $\frac{t}{C\times ln(\frac{V_o}{V})}$ ....................2

put here value

so for t min 11.5 μs

R = $\frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}$

R min = 28.173 ohm

and

for t max 6.33 ms

R max = $\frac{6.33}{11.5} \times 28.173$

R max = 1.55 × $10^{4}$ ohm

4 0

2 years ago

A centrifugal pump is required to pump water to an open water link situated 4 km away from the location of the pump through a pi

11111nata11111 [884]

Answer:

P= 5.5 bar

Explanation:

Given that

L= 4000 m

d= 0.2 m

Friction factor(F) = 0.01

speed V= 2 m/s

Head = 5 m

Head loss due to friction

$h_f=\dfrac{FLV^2}{2gd}$

$h_f=\dfrac{0.01\times 4000\times 2^2}{2\times 9.81\times 0.2}$

$h_f=40.77m$

So the total head(H) = 5 + 40.77 + 10.3 =56.07

Where 10.3 m is the atmospheric head.

We know that

P=ρ g H

So total Pressure

P= 1000 x 9.81 x 56.07 Pa

$P=5.5\times 10^5\ Pa$

P= 5.5 bar

5 0

3 years ago