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larisa86 [58]
2 years ago
11

Technician A says reducing spark advance can cause spark knock. Technician B says excessive carbon deposits can cause spark knoc

k. Who is correct
Engineering
1 answer:
olchik [2.2K]2 years ago
3 0

Answer:

A

Explanation:

Technician A is correct due to its factual statement. ALso I did this...

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A car is about to start but it blows up. what is the problem with the car<br> ?
ratelena [41]

Answer:

because there is a bomb

6 0
3 years ago
Read 2 more answers
How high of a column of sae 30 oil would be required to give the same pressure as 700 mm hg?
Rasek [7]

Hoiu-10,4000 mm.

<h3>Is positive pressure good for PC?</h3>
  • A balanced configuration is the most efficient way to cool your pc although it should tend towards a slight positive pressure if you can help it.
  • Tip: As much as it might seem important, the concept of heat rising doesn't have too much of an effect.

To learn more about it, refer

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7 0
1 year ago
g A pump is required to deliver 100 gpm at a head of 100 ft, but the pump rated capacity is 150 gpm at a head of 100 ft. If the
Thepotemich [5.8K]

Answer: valving the pump discharge to reduce the flow will only result in an increase in the water velocity according to the laws of continuity of flow.

Q = AV = constant.

The pump speed of 150 gpm is, and will remain constant. Valving simply reduces flow area A, which is balanced out by increased velocity V of water through the pipe.

This does not affect the pump speed Q (flow rate) and hence it remains the same.

8 0
3 years ago
What is the resultant force on one side of a 25cm diameter circular plate standing at the bottom of 3m of pool water?
Tom [10]

Answer:

F=1.47 KN

Explanation:

Given that

Diameter of plate = 25 cm

Height of pool h = 3 m

We know that force can be given as

F= P x A

P=ρ x g x h

Now by putting the values

P=1000 x 10 x 3

P= 30 KPa

A=\dfrac{\pi}{4}\times 0.25^2\ m^2

A=0.049\ m^2

F= 30 x 0.049 KN

F=1.47 KN

So the force on the plate will be 1.47 KN.

4 0
2 years ago
A cone penetration test was carried out in normally consolidated sand, for which the results are summarized below: Depth (m) Con
Cerrena [4.2K]

Answer:

hello your question is incomplete attached below is the missing equation related to the question  

answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°

Explanation:

<u>Determine the friction angle at each depth</u>

attached below is the detailed solution

To calculate the vertical stress = depth * unit weight of sand

also inverse of Tan = Tan^-1

also qc is in Mpa while σ0 is in kPa

Friction angle at each depth

2 meters = 40.389°

3.5 meters  = 38.987°

5 meters = 38.022°

6.5 meters = 39.869°

8 meters = 40.265°

6 0
3 years ago
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