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Jobisdone [24]
2 years ago
15

Write down the tracking error such that the adaptive cruise control objective is satisfied.

Engineering
1 answer:
bija089 [108]2 years ago
5 0

Answer:
The most common reason a cruise control stops working is due to a blown fuse or a defective brake pedal switch. It can also be caused by issues with the throttle control system or the ABS. In older cruise control systems it can be caused by a broken vacuum line.

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Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cool
MariettaO [177]

Answer: hello your question is incomplete below is the missing part

question :Determine the temperature of the cooled water exiting the cooling tower

answer : T  = 43.477° C

Explanation:

Temp of water at exit = 45°C

mass flow rate of cooling tower = 15,000 kg/s

Temp of makeup water = 20°C

Assuming an atmospheric pressure of = 101.3 kPa

<u>Determine temperature of the cooled water exiting the cooling tower</u>

Water entering cooling tower at 45°C

Given that Latent heat of water at 45°C = 43.13 KJ/mol

Cp(wet air) = 1.005+ 1.884(y1)

where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273

Hence : Cp(wet air) = 29.145 +  (0.01273) (33.94) = 29.577 KJ/kmol°C

<u>First step : calculate the value of Q </u>

Q = m*Cp*(ΔT) + W(latent heat)

Q = 321.6968 (29.577) (40-30) +  43.13 (18.26089)

Q =  95935.8547 KJ/s

Given that mass rate of water = 15000 kg/s

<u>Hence the temperature of the cooled water can be calculated using the equation below</u>

Q = m*Cp*∆T

Cp(water) = 4.2 KJ/Kg°C

95935.8547 = (15000)*(4.2)*(45 - T)

( 45 - T ) = 95935.8547/ 63000.    ∴ T  = 43.477° C

5 0
3 years ago
A sample of sand weighs 490 g in stock and 475 in Oven Dry (OD) condition, respectively. If absorption capability of the sand is
Ivahew [28]

The weight of the specimen in SSD condition is 373.3 cc

<u>Explanation</u>:

a) Apparent specific gravity = \frac{A}{A-C}

Where,

A = mass of oven dried test sample in air = 1034 g

B = saturated surface test sample in air = 1048.9 g

C = apparent mass of saturated test sample in water = 975.6 g

apparent specific gravity = \frac{A}{A-C}

                                         = \frac{1034}{1034-675 \cdot 6}

Apparent specific gravity = 2.88

b) Bulk specific gravity G_{B}^{O D}=\frac{A}{B-C}

G_{B}^{O D}=\frac{1034}{1048.9-675 \cdot 6}

       =  2.76

c) Bulk specific gravity (SSD):

G_{B}^{S S D}=\frac{B}{B-C}

=\frac{1048 \cdot 9}{1048 \cdot 9-675 \cdot 6}

G_{B}^{S S D} = 2.80

d) Absorption% :

=\frac{B-A}{A} \times 100 \%

=\frac{1048 \cdot 9-1034}{1034} \times 100

Absorption = 1.44 %

e) Bulk Volume :

v_{b}=\frac{\text { weight of dispaced water }}{P \omega t}

=\frac{1048 \cdot 9-675 \cdot 6}{1}

= 373.3 cc

5 0
3 years ago
Maggie discovered that a pipe in her basement has sprung a leak. She calls a plumber but in the meantime she grabs a roll of duc
dem82 [27]

Answer:

The answer is "Option a".

Explanation:

Myelination was the myelinization mechanism of a neuron axon. The endothelium is enveloped all around the axon and isolates the axon that inhibits the neuronal message from leaking with the other neuronal axons. Inside this example, therefore, its tubes tape worked similarly to those of myelin sheath, which stops brain transmission.

3 0
3 years ago
What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin
DochEvi [55]

Answer:

m_{LP}=0.45\,kg

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]

Q_{water} = 3599.435\,kJ

The heat liberated by the LP gas is:

Q_{LP} = \frac{3599.435\,kJ}{0.16}

Q_{LP} = 22496.469\,kJ

A kilogram of LP gas has a minimum combustion power of 50028\,kJ. Then, the required mass is:  

m_{LP} = \frac{22496.469\,kJ}{50028\,\frac{kJ}{kg} }

m_{LP}=0.45\,kg

6 0
3 years ago
Removing Shingles from a roof is called a
torisob [31]
D remodeling roofing
7 0
3 years ago
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