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zalisa [80]
2 years ago
13

Users of E85 see a drop in gas mileage because ethanol can't produce the same

Engineering
1 answer:
liraira [26]2 years ago
7 0

Answer:

A

Explanation:

Due to ethanol's lower energy content, FFVs operating on E85 get roughly 15% to 27% fewer miles per gallon than when operating on regular gasoline, depending on the ethanol content.

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Suppose you have two boxes in front of you. One box contains a Thevenin Equivalent (voltage source in series with a resistor) an
fomenos

Answer:

1. Measure the temperature of the boxes and leave them unconnected.

2. Norton reduces his circuit down to a single resistance in parallel with a constant current source. A real-life Norton equivalent circuit would be continuously wasting power (as heat) as the current source dumps energy into the resistor, even when externally unconnected, while a Thevenin equivalent circuit would sit there doing nothing.

3. The Norton equivalent box would get warm and eventually run out of power. The Thevenin equivalent box would stay at ambient temperature.

8 0
3 years ago
What is the primary function of NCEES?
joja [24]
National Council of Examiners for engineering and surveying a nonprofit organization
7 0
3 years ago
What is the average distance (in terms of R) between the mobile on the fringe of the serving cell and the second and third tier
kramer

Answer:

do you need all work shown for this?

4 0
2 years ago
You have a Switch LED Lightbulb with the following information: It is equivalent to a 40 Watt incandescent lightbulb. It uses 8
DochEvi [55]

The lightning efficiency based on the scenario depicted will be C. 56 lumens/Watt, more efficient.

<h3>How to calculate the lightning efficiency</h3>

The efficiency of the incandescent bulb will be:

= 450/40 = 11.25 lumens per watt.

The efficiency of the LED bulb will be:

= 450/8 = 56 lumens per watt.

In this case, the LED bulb is more efficient than the incandescent bulb.

Therefore, the lighting efficiency will be 56 lumens/Watt, more efficient

Learn more lightning efficiency on:

brainly.com/question/25927632

4 0
2 years ago
two pints of ethyl ether evaporate over a period of 1,5 hours. what is the air flow necessary to remain at 10% or less of ethyl
lesya692 [45]

The air flow necessary to remain at the lower explosive level is 4515. 04cfm

<h3>How to solve for the rate of air flow</h3>

First we have to find the rate of emission. This is solved as

2pints/1.5 x 1min

= 2/1.5x60

We have the following details

SG = 0.71

LEL = 1.9%

B = 10% = 0.1 a constant

The molecular weight is given as 74.12

Then we would have Q as

403*100*0.2222 / 74.12 * 0.71 * 0.1

= Q = 4515. 04

Hence we can conclude that the air flow necessary to remain at the lower explosive level is 4515. 04cfm

Read more on the rate of air flow on brainly.com/question/13289839

#SPJ1

7 0
1 year ago
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