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Deffense [45]
3 years ago
6

____ emotions can influence your driving. A. Only some B. All of your C. Only negative D. Only positive

Engineering
2 answers:
lutik1710 [3]3 years ago
8 0

Answer:

It is A all of your

Zigmanuir [339]3 years ago
7 0
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Given the system of equations−3x2+7x3=2,x1+2x2–x3=3,5x1−2x2=2(a). Compute the determinant. (b). Use Cramer’s rule to solve for t
34kurt

Answer:

(x1,x2,x3)=(0.985507,1.463768,0.913043)

Explanation:

<em>note: </em>

<em>solution is provided in word document please find the attached document.</em>

Download docx
5 0
3 years ago
Describe and compare the characteristics of (a) proportional control, (b) proportional plus integral control, (c) proportional p
stira [4]

Answer:

The answer is below

Explanation:

1. Proportional Control is a form of control engineering in which an output is directly proportional to the error signal.

Characteristics of proportional control are:

* It is utilized when the deviation between the input and output is small

* It is also utilized when the deviation is not sudden.

* It reduces steady-state error

* It speeds up the response of the overdamped system

2. Proportional plus Integral Control is a form of control engineering in which a collective proportion and integral control of the output is equivalent to the combined proportion and integral of the error signal.

Characteristics of proportional plus integral control are:

* it can revert the controlled variable to the original set point

* It decreases steady-state error

* It quickens up the reaction of the overdamped system

3. Proportional plus integral plus derivative control is mostly applicable in operating the process elements such as temperature, pressure, speed, etc. It is recommended for industrial use.

Characteristics of Proportional plus integral plus derivative control are:

* It enhances the temporary reaction of the system.

* It also lessens steady-state error

* It accelerates the response of the overdamped system

4 0
3 years ago
Assume that the voltage applied to a load is V = 208-30° V and the current flowing through the load is I = 515° A. (a) Calcula
RoseWind [281]
Idk I just need point an you probably already solved this by now
3 0
3 years ago
an electric circuit includes a voltage source and two resistances (50 and 75) in parallel. determine the voltage source required
ASHA 777 [7]

Answer:

The voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.

Explanation:

Given;

Resistance, R₁ = 50Ω

Resistance, R₂ = 75Ω

Total resistance, R = (R₁R₂)/(R₁ + R₂)

Total resistance, R = (50 x 75)/(125)

Total resistance, R = 30 Ω

According to ohms law, sum of current in a parallel circuit is given as

I = I₁ + I₂

I = \frac{V}{R_1} + \frac{V}{R_2}

Voltage across each resistor is the same

1.6 = \frac{V}{R_2}  

V = 1.6 x R₂

V = 1.6 x 75

V = 120 V

Therefore, the voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.

This voltage is also the same for 50 ohms resistance but the current will be 2.4 A.

3 0
3 years ago
Read 2 more answers
An AC power generator produces 50 A (rms) at 3600 V. The voltage is stepped up to 100 000 V by an ideal transformer and the ener
RSB [31]

Given:

I_{rms} = 50 A

voltage, V = 3600V

step-up voltage, V' = 100000 V

Resistance of line, R = 100\ohm

Solution:

To calculate % heat loss in long distance power line:

Power produced by AC generator, P = 50\times 3600 W

P = 180000 W = 180 kW

At step-up voltage, V = 100000V or 100 kV

current, I = \frac{P}{V'}

I = \frac{1800000}{100000}

I = 1.8 A

Power line voltage drop is given by:

V_{drop} = I\times R

V_{drop} = 1.8\times 100

V_{drop} = 180 V

Power dissipated in long transmission line P_{dissipated} = V_{drop}\times I

Power dissipated in long transmission line P_{dissipated} = 180\times 1.8 = 324 W

% Heat loss in power line, P_{loss} = \frac{P_{dissipated}}{P}\times 100

% Heat loss in power line, P_{loss} = \frac{324}{180000}\times 100

P_{loss} = 0.18%

 

5 0
3 years ago
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