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Marina86 [1]
3 years ago
15

Which of the following can minimize engine effort in save fuel

Engineering
2 answers:
pentagon [3]3 years ago
7 0

Answer:

hmm

Explanation:

How to Save Fuel

Keep your vehicle's engine in good condition.

Maintain proper air pressure in your tires.

The faster you travel, the greater your fuel consumption is.

Try to speed up gradually when you stop for a traffic light.

Try to drive smoothly.

Do not allow your engine to idle.

hope this helps

Anton [14]3 years ago
4 0

Answer:

accelerating gradually and changing gears smoothly

Explanation:

anticipating stops avoiding applying brakes for sudden stops.

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Write a Nios II assembly program that reads binary data from the Slider Switches, SW11-0, on the DE2-115 Simulator, and display
Citrus2011 [14]

Algorithm of the Nios II assembly program.

  • Attain data for simulation from the  SW11-0, on the DE2-115 Simulator
  • The data will be  read from the switches in loop.
  • The decimal output is displayed using the seven-segment displays and done using the loop.
  • The program is ended by the user operating the SW1 switch

and

The decimal equivalent on the seven-segment displays HEX3-0 is

  • DE2-115
  • DE2-115_SW11
  • DE2-115_HEX3
  • DE2-115_HEX4
  • DE2-115_HEX5
  • DE2-115_HEX6
  • DE2-115_HEX7

<h3>The  Algorithm and decimal equivalent on the seven-segment displays HEX3-0</h3>

Generally,  the program will be written using a  cpulator simulator in order to attain best result.

We are to

  • Attain data for simulation from the  SW11-0, on the DE2-115 Simulator
  • The data will be  read from the switches in loop.
  • The decimal output is displayed using the seven-segment displays and done using the loop.
  • The program is ended by the user operating the SW1 switch

This will be the Algorithm of the Nios II assembly program .

Hence, the decimal equivalent on the seven-segment displays HEX3-0 is

  • DE2-115
  • DE2-115_SW11
  • DE2-115_HEX3
  • DE2-115_HEX4
  • DE2-115_HEX5
  • DE2-115_HEX6
  • DE2-115_HEX7

For more information on Algorithm

brainly.com/question/11623795

6 0
2 years ago
Which of the following is NOT true about hydraulic valves? A. Directional control valves determine the path of a fluid in a give
Lelechka [254]

Answer: Option D is not true of hydraulic valves. A hydraulic valve is a device that can change the opening degree of liquid flow path

Explanation:

The pilot check valve allows flow of liquid in one direction and blocks flow in the opposite direction

5 0
3 years ago
An asphalt concrete mixture includes 94% aggregate by weight. The specific gravities of aggregate and asphalt are 2.65 and 1.0,
garik1379 [7]

Answer:

2.0%

Explanation:

Percentage of aggregate = 94%

Specific gravity = 2.65

Specific gravity of asphalt = 1.9

Density of mix = 147pcf = 147lb/ft³

Total weight of mix: (volume = 1ft³)

= (147lb/ft³)(1ft³)

= 147lb

Percentage weight of asphalt in<u> mix:</u>

100% - 94%

= 6%

Weight of asphalt binders

= 6% x 147lb

= 8.82lb

Weight of aggregate in mix:

= 94% x 147

= 138.18lb

Specific weight of asphalt binder:

(Gab)(Yw)

Yw = specific Weight of water

= 62.4lb

Gab = specific gravity of asphalt binder

= 1.0

(62.4lb)(1.0)

= 62.4 lb/ft³

Volume of asphalt in binder:

8.82/62.4

= 0.14ft³

Specific weight of binder in mix:

2.65 x 62.4lb/ft³

= 165.36 lb/ft³

Volume of aggregate:

= 138.18/165.36

= 0.84ft³

Volume of void in the mix:

1ft³ - 0.84ft³ - 0.14ft³

= 0.02ft³

<u>The percentage of void in total mix:</u>

VTM = (0.02ft³/1ft³)100

= 2.0%

8 0
3 years ago
Steam flowing through a long, thin walled pipe maintains the pipe wall at a uniform temperature of 500 K. The pipe is covered wi
hjlf

Answer: I would love to learn this

Explanation:

3 0
3 years ago
Express (118)10 and (-49)10 in 8-bit binary one’s complement form and then add the numbers. What would be the representation (-0
stealth61 [152]

Answer:

118_{10}= 0110111 2) -49_{10}=110001_{2} 3) 0_{10}=0:16 \Rightarrow 0_{10}=0_{16}

Explanation:

1) Expressing the Division as the summation of the quotient and the remainder

for

118, knowing it is originally a decimal form:

118:2=59  +(0), 59/2 =29 + 1, 29/2=14+1, 14/2=7+0, 7/2=3+1, 3/2=1+1, 1/2=0+1

118_{10}= 0110111

2) -49_{10}

Similarly, we'll start the process with the absolute value of -49 since we want the positive value of it. Then let's start the successive divisions till zero.

|-49|=49

49:2=24+1, 24:2=12+0,12:2=6+0,6:2=3+0,3:2=1+1,1:2=0+1

100011

-49_{10}=110001_{2}

3) (-0)_{10}

The first step on that is dividing by 16, and then dividing their quotient again by 16, so on and adding their remainders. Simply put:

(-0)_{10}=0:16=0 \Rightarrow (0)_{10}=0_{16} \:or\\(0)_{16}=0000000000000000

5 0
3 years ago
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