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inysia [295]
3 years ago
5

Please choose a specific type of stability or control surface (e.g., a canard) and explain how it is used, what it is used for,

and the pros and cons of the device or system.
Engineering
1 answer:
Thepotemich [5.8K]3 years ago
3 0

Answer:

small forewing

<u>pro</u> : Can be used in place of tail plane configuration

<u>con</u> : Can be very complex and difficult to use

Explanation:

A canard is generally used to provide some form of stability to an unstable or semi stable system.

An example of a Canard is a small forewing placed in an aircraft that will help stabilize the aircraft when in motion( in air ). because an airplane is generally an unstable system on its own

<u>pro </u>: Can be used in place of tail plane configuration

<u>con</u> : Can be very complex and difficult to use

You might be interested in
Technician A says that the neutral safety switch must be bad and should be replaced. Technician B says that the neutral safety s
lukranit [14]

Answer:

Technician A is correct

Explanation:

The neutral safety switch when bad must be replaced and not adjusted as suggested by technician B because if the neutral safety switch is bad the Engine might not crank when put in neutral but it will crank when put in park and this is very bad for the life of the Engine it is better to replace it. A test for a bad/faulty neutral safety switch will be required to ascertain the level of damage it might cause to the Engine and prompt replacement is essential as well.  because the neutral helps to prevent the car from starting when it is already engaged in a gear position therefore protecting the car from sudden collisions

3 0
2 years ago
Read 2 more answers
Which of the following are TRUE concerning rectifier circuits? Select all that apply.
Dovator [93]

Explanation:

a. The output of an ideal full wave rectifier is zero volts only when the input is zero volts.

True

The output of an ideal full wave rectifier is zero volts only when the input is zero since 1st diode is forward biased during one half cycle of the input and 2nd diode is forward biased during the other half cycle of the input therefore, it fully utilizes both the input cycles so the output voltage is only zero when the input is zero.

b. Half-wave rectifier circuits need a minimum of 4 diodes to operate.

False

A Half-wave rectifier circuits need a minimum of 1 diode to operate, whereas a full-wave bridge rectifier need minimum of 4 diodes to operate.

c. In an ideal full wave bridge rectifier, half of the diodes are in the ON state and half of the diodes are in the OFF state at any given time the input voltage is not zero.

True

A full-wave bridge rectifier consists of 4 diodes, where 2 diodes are functional in half of the cycle(so the other 2 are off) and other 2 diodes are functional in the other half cycle( so the other 2 are off).

d. The output of an ideal half wave rectifier is zero volts only when the input is zero volts.

False

The output of an ideal half wave rectifier is zero during half of the cycle when the diode is reversed biased and doesn't conduct even though input voltage is not zero volts at this point.

e. A turn-on voltage of a diode (y,) greater than zero can cause the output of a full wave rectifier to be zero volts even when the input is not zero volts.

False

A turn-on voltage of a diode (y,) greater than zero cannot cause the output of a full wave rectifier to be zero rather there will be a little voltage drop across the output of full wave rectifier due to this turn-on voltage of diode which is usually 0.7 volts for silicon based diodes.

f. An advantage of the half wave rectifier is that is can use a smoothing capacitor, while a full wave rectifier cannot.

False

Smoothing capacitor can be used in both half wave rectifier as well as full wave rectifier.

8 0
3 years ago
One kilogram of a contaminant is spilled at a point in a 4m deep reservoir and is instantaneously mixed over the entire depth. (
rosijanka [135]

Answer:

0.028 kg per Seconds; 8.4

Explanation:

N-S, 4 x 100 = 400 m^2, concentration 5/400= 0.0125 kg/s

E-W 4 x 100 =400 m^2 , => 10/400 =0.025 kg/s

(0.0125^2 + 0.025^2)^(1/2) =0.028 kg/s

in 5 minutes = 5 x 60 x 0.028 =8.4

8 0
3 years ago
Substance Specific Heat (cal/g°C) Specific Heat (J/g°C) water 1.00 4.18 steam 0.96 4.02 alcohol 0.59 2.47 ice 0.50 2.09 wood 0.4
Jobisdone [24]

Answer:

C) 43,2°C

Explanation:

<em>Sensible heat</em> is the amount of thermal energy that is required to change the temperature of an object, the equation for calculating the heat change is  given by:

Q=msΔT

where:

  • Q, heat that has been absorbed or realeased by the substance [J]
  • m, mass of the substance [g]
  • s, specific heat capacity [J/g°C]
  • ΔT, changes in the substance temperature [°C]

To solve the problem, we clear ΔT of the equation and then replace our data:

Q=890 [J],

m=16,6 [g],

s=2,74 [J/g°C]

Q=msΔT.......................ΔT=Q/ms

ΔT=\frac{890 J}{16,6g*2,47\frac{J}{gC}}=21,7°C

As:

ΔT=Tfinal-Tinitial

Tfinal=ΔT+Tinitial

Tfinal=21,7+21,5=43,2°C

The final temperature of the ethanol is 43,2°C.

4 0
3 years ago
In this problem, we will explore how deepening the pipeline affects performance in two ways: faster clock cycle and increased st
Nataliya [291]

Answer:

a) the speed up is 1.45

b) the speed up is 1.42

Explanation:

Given

5 stage pipeline = 1 ns clock cycle

12 stage pipeline = 0.6 ns clock cycle

a) The speed up is

speed-upx_{pipeline} =\frac{CPI_{unpipelined} }{CPI_{pipelined} } (\frac{cycle-time_{unpipelined} }{cycle-time_{pipelined} } ) (eq. 1)

The CPIpipelined is

CPI_{pipelined} =CPI_{ideal} +average-stall-cycle/instructions (eq. 2)

The execution time is

execution-time=instruction*CPI_{pipelined} *cycle-time_{unpipelined} (eq. 3)

The CPI for 5-stage pipeline is

CPI = 6/5

cycle time = 1 s

The CPI for 12-stage pipeline is

CPI = 11/8

cycle time = 0.6 s

Replacing values in equation 1

speed-up_{pipeline} =\frac{I*\frac{6}{5}*1 }{I*\frac{11}{8}*0.6 } =\frac{1.2}{1.375*0.6} =1.45

b) The equation for CPI instruction in 5-stage is

CPI_{5} =CPI_{time} +instructions-of-5-stage*cycletime*number-of-cycles=\frac{6}{5} +(\frac{20}{100} )*0.05*2=1.22

For CPI instruction in 12-stage is

CPI_{12} =\frac{11}{8} +(\frac{20}{100} )*0.05*5=1.43

The speed up is using the equation 1

speed-up=\frac{1*1.22*1}{1*1.43*0.6} =1.42

3 0
2 years ago
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