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Mnenie [13.5K]
2 years ago
13

Transmission cleaners are used: A) Only in conjunction with fuel system cleanersB) Only in the colder monthsC) By themselvesD) I

n conjunction with a transmission fluid exchange
Engineering
2 answers:
valentina_108 [34]2 years ago
5 0

Answer: D

Explanation: When you change your transmission fluid you change your filter. When you do that you clean where the filter was. Then you can put the new filter on and the new fluid.

irinina [24]2 years ago
3 0

Explanation: When you change your transmission fluid you change your filter. When you do that you clean where the filter was. Then you can put the new filter on and the new fluid.

is correct

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Hot water at an average temperature of 88°C and an average velocity of 0.4 m/s is flowing through a 5-m section of a thin-walled
SIZIF [17.4K]

Answer:

a) The rate of heat transfer will be 19.58 Watts.

b) The temperature drop of the hot water will be 0.024 Degree Celcius.

Explanation:

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3 years ago
In a creep test, increasing the temperature will (choose the best answer) A. increase the instantaneous initial deformation B. i
Hitman42 [59]

Answer:

All of the above

Explanation:

firstly, a creep can be explained as the gradual deformation of a material over a time period. This occurs at a fixed load with the temperature the same or more than the recrystallization temperature.

Once the material gets loaded, the instantaneous creep would start off and it is close to electric strain. in the primary creep area, the rate of the strain falls as the material hardens. in the secondary area, a balance between the hardening and recrystallization occurs. The material would get to be fractured hen recrstallization happens.  As temperature is raised the recrystallization gets to be more.

8 0
2 years ago
Which best describes the body in terms of simple machines?
alex41 [277]

Answer:B

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3 years ago
Direction: List down or enumerate the type of outlets you want to install on your dream house. Also indicate the quantity (in pi
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Apartment outlet 4pcs

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6 0
2 years ago
It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
denis23 [38]

The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

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