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viktelen [127]
3 years ago
7

Tech A says that low compression on a single cylinder will cause an engine not to start. Tech B says that low compression on a s

ingle cylinder will affect the engine cranking sound. Who is correct
Engineering
1 answer:
Klio2033 [76]3 years ago
4 0

Answer:

Technician B

Explanation:

A vehicle can start and drive with low compression on one cylinder, but it will act like it's misfiring. In a clear flood mode, if the vehicle is started with low compression in one cylinder, it will have an abnormal cranking sound.

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Pedro holds a heavy science book over his head for 10 minutes. Petro is doing work during that time. True or False
algol [13]

Answer:

True because he is working his arms to lift and hold the weight

Explanation:

4 0
3 years ago
On the pavement indicate that the adjacent lane is traveling in the same direction and passing is permitted
Mkey [24]

A broken yellow line on the pavement tells that the adjacent lane is traveling in the opposite direction and passing is permitted.

A broken white line on the pavement show that the adjacent lane is traveling in the same direction and passing is permitted.

<h3>What does pavement markings show?</h3>

Pavement markings are known to be tools that are used to pass infor or messages to roadway users.

Note  that they tell the part of the road that one need to use, give information about conditions ahead, and others

Note that A broken yellow line on the pavement tells that the adjacent lane is traveling in the opposite direction and passing is permitted.

Learn more about pavement markings from

brainly.com/question/10179521

#SPJ1

6 0
2 years ago
The purification of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of h
gtnhenbr [62]

Answer: 5.36×10-3kg/h

Where 10-3 is 10 exponential 3 or 10 raised to the power of -3.

Explanation:using the formula

M =JAt = -DAt×Dc/Dx

Where D is change in the respective variables. Insulting the values we get,

=5.1 × 10-8 × 0.13 × 3600 × 2.9 × 0.31 / 4×10-3.

=5.36×10-3kg/h

6 0
3 years ago
An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

3 0
3 years ago
Subject: Electronics
Maslowich

Answer:

U just believe in yourself ..........

Explanation:

<em>If </em><em>there </em><em>a</em><em>r</em><em>e </em><em>more </em><em>electrons </em><em>than </em><em>protons </em><em>in </em><em>a </em><em>piece </em><em>of </em><em>matter </em><em>it </em><em>will </em><em>have </em><em>a </em><em>negative</em><em> </em><em>charge </em><em>.</em><em> </em><em>i</em><em>f</em><em> </em><em>there </em><em>are </em><em>fever </em><em>it </em><em>will </em><em>have </em><em>positive</em><em> </em><em>charge </em><em>and </em><em>if </em><em>there </em><em>are </em><em>e</em><em>qual </em><em>numbers </em><em>it </em><em>will </em><em>have </em><em>neutral</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

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hope it was helpful to you.....

6 0
2 years ago
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