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elena55 [62]
3 years ago
13

Select all that apply.

Engineering
1 answer:
garik1379 [7]3 years ago
5 0

uhh , not all girls are like that. many of em these days are more career oriented and thats good i guess , so if anything ur just being judgy a f here.

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As you have learned, not all motor oils are the same. What are two things that make them different?.
saul85 [17]

Answer:

quality ingredients and exceeding industry standards

Explanation:

5 0
2 years ago
Bridge A is the longest suspension bridge in a Country. Bridge B is 5555 feet shortershorter than Bridge A. If the length of Bri
MatroZZZ [7]

Answer:

the length of the bridge B is L = m - 5555 ft

Explanation:

Denoting the length of the bridge B as L we get

length of bridge B = length of bridge A - 5555 ft

since length of bridge A=m

L = m - 5555 ft

4 0
3 years ago
3. When performing overhead work on scaffolding, what protective measures must be taken to prevent objects
hjlf

Answer:

Toeboards, debris nets, or canopies

Explanation:

7 0
3 years ago
Design a PI controller to improve the steady-state error. The system should operate with a damping ratio of 0.8. Compute the ove
blondinia [14]

Answer:

The MATLAB Code for this PI Controller will be:

Kp = 350;

Ki = 300;

Kd = 50;

C = pid(Kp,Ki,Kd)

T = feedback(C*P,1);

t = 0:0.01:2;

step(T,t)

Explanation:

When you are designing a PID controller for a given system, follow the steps shown below to obtain a desired response.

Obtain an open-loop response and determine what needs to be improved

Add a proportional control to improve the rise time

Add a derivative control to reduce the overshoot

Add an integral control to reduce the steady-state error

Adjust each of the gains $K_p$, $K_i$, and $K_d$ until you obtain a desired overall response.

The further explanation is attached in the Word File.

Download docx
5 0
3 years ago
#5 Air undergoes an adiabatic compression in a piston-cylinder assembly from P1= 1 atm and Ti=70 oF to P2= 5 atm. Employing idea
otez555 [7]

Answer:

The work transfer per unit mass is approximately 149.89 kJ

The heat transfer for an adiabatic process = 0

Explanation:

The given information are;

P₁ = 1 atm

T₁ = 70°F = 294.2611 F

P₂ = 5 atm

γ = 1.5

Therefore, we have for adiabatic system under compression

T_{2} = T_{1}\cdot \left (\dfrac{P_{2}}{P_{1}}  \right )^{\dfrac{\gamma -1}{\gamma }}

Therefore, we have;

T_{2} = 294.2611 \times \left (\dfrac{5}{1}  \right )^{\dfrac{1.5 -1}{1.5 }} \approx 503.179 \ K

The p·dV work is given as follows;

p \cdot dV = m \cdot c_v \cdot (T_2 - T_1)

Therefore, we have;

Taking air as a diatomic gas, we have;

C_v = \dfrac{5\times R}{2} = \dfrac{5\times 8.314}{2} = 20.785 \ J/(mol \cdot K)

The molar mass of air = 28.97 g/mol

Therefore, we have

c_v = \dfrac{C_v}{Molar \ mass} = \dfrac{20.785}{28.97} \approx 0.7175 \ kJ/(kg \cdot K)

The work done per unit mass of gas is therefore;

p \cdot dV =W =   1 \times 0.7175 \times (503.179 - 294.2611) \approx 149.89 \ kJ

The work transfer per unit mass ≈ 149.89 kJ

The heat transfer for an adiabatic process = 0.

8 0
3 years ago
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