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3 years ago

5. If a pilot wants to accelerate, which force

Engineering

1 answer:

Murljashka [212]3 years ago

3 0

Answer:

Thrust

Explanation:

If the pilot wants to accelerate the aircraft, the aircraft needs more power to produce more thrust. The aircraft will go faster when the amount of thrust is greater than the amount of drag.

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Develop a chase plan that matches the forecast and compute the total cost of your plan. (Negative amounts should be indicated by

vladimir2022 [97]

Answer:

The answer for the question :

"Develop a chase plan that matches the forecast and compute the total cost of your plan. (Negative amounts should be indicated by a minus sign. Leave no cells blank - be certain to enter "0" wherever required. Omit the "$" sign in your response.)"

is explained in the attachment.

Explanation:

4 0

3 years ago

Can someone explain it to me!

Darina [25.2K]

Answer:

It will be B

Explanation:

Since resisitors in series are added together, 1 + 1 + 1 would = 3kilo ohms. But with resistors in parallel would be (1/1+1/1)^-1. That would equal 0.5 Now you have two resistors in series for B, and because now that they are in series you add them together, so 0.5 + 1 = 1.5 kilo ohms which is what is needed.

7 0

3 years ago

A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this

tresset_1 [31]

Answer:

a)COP=5.01

b) $W_{in}=2.998$ KW

c)COP=6.01

d) $Q_R=17.99 KW$

Explanation:

Given

$T_L$ = -12°C, $T_H$ =40°C

For refrigeration

We know that Carnot cycle is an ideal cycle that have all reversible process.

So COP of refrigeration is given as follows

$COP=\dfrac{T_L}{T_H-T_L}$ ,T in Kelvin.

$COP=\dfrac{261}{313-261}$

a)COP=5.01

Given that refrigeration effect= 15 KW

We know that $COP=\dfrac{RE}{W_{in}}$

RE is the refrigeration effect

5.01= $\dfrac{15}{W_{in}}$

b) $W_{in}=2.998$ KW

For heat pump

So COP of heat pump is given as follows

$COP=\dfrac{T_h}{T_H-T_L}$ ,T in Kelvin.

$COP=\dfrac{313}{313-261}$

c)COP=6.01

In heat pump

Heat rejection at high temperature=heat absorb at low temperature+work in put

$Q_R=Q_A+W_{in}$

Given that $Q_A=15$ KW

We know that $COP=\dfrac{Q_R}{W_{in}}$

$COP=\dfrac{Q_R}{Q_R-Q_A}$

$6.01=\dfrac{Q_R}{Q_R-15}$

d) $Q_R=17.99 KW$

5 0

3 years ago

Applying the Entropy Balance: Closed Systems Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston–

Mrrafil [7]

Answer:

a) the amount of energy produced in kJ/K is 0.73145 kJ/K

b) the amount of energy produced in kJ/K is 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

Explanation:

Draw the T-s diagram.

$C_p$ = 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280

R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar

Δs = $m[c_p ln(\frac{T_2}{T_1}) - Rln(\frac{P_2}{P_1})]$

Substitute all parameters in the equation

Δs = $5[(0.939) ln(\frac{520}{280}) - (\frac{8.314}{44.01})ln(\frac{20}{2})]$

Δs = 5 kg × 0.14629 kJ/kg.K

= 0.73145 kJ/K

Δs = $m[\frac{s^0(T_2) - s^0(T_1)}{M} - Rln(\frac{P_2}{P_1})]$

Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K

T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K

R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar

Δs = $5[\frac{236.575 - 211.376}{44.01} - (\frac{8.314}{44.01})ln(\frac{20}{2})]$

= 5 kg (0.13795 kJ/kg.K)

= 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

7 0

3 years ago

MITM can present in two forms: eavesdropping and manipulation. Discuss the process involved when an attacker is eavesdropping an

Nikitich [7]

Answer / Explanation:

Eavesdropping attack is also sometimes refereed to as sniffing attack. It is simply the process by which an attacker or hacker tries to penetrate very suddenly into an unaware individuals network or server with the intention to steal information transmitted over the network or server through that computer.

To prevent such attack, there are several mean which include installing network monitoring software to check who else is connected to the network but the most common method of preventing such attack is to encrypt the Hypertext Transfer Protocol (http) and the way to do this is by securing it with a sort of security key.

On installing the security key, the network becomes encrypted and secured such that whatever network transmitted over the network becomes encrypted and unable to read. The protocol then converts to (https).

5 0

3 years ago