diesel engines do not use spark plugs because diesel has more oil, burns slower, and is more difficult to ignite. Therefor spark plugs are insufficient with just spark so instead they use glow plugs witch provides a high amount of heat to the fuel and ignites more quickly and reliably.
<span>The amount of compression pressure in a diesel engine is sufficient to produce enough heat to ignite the fuel. The glow plugs are only used on a cold start.</span>
Answer:
100°heat
Explanation:
since when i calculate this and that, the answer is 100° heat.
sorry if it is inconvenient
0.0179 ohms for copper.
0.0184 ohms for annealed copper
Ď = R (A/l) where
Ď = electrical resistivity
R = electrical resistance of a uniform specimen
A = cross sectional area
l = length
Solve for R by multiplying both sides by l/A
R = Ď(l/A)
The cross section of the wire is pi * 1^2 mm = 3.14159 square mm = 3.14159e-6 square meters.
The length is 3 meters. So l/A = 3/3.14159e-6 = 9.5493e5
Ď for copper is 1.68e-8 so 1.68e-8 * 9.5493e5 = 1.60e-2 ohms at 20 C
But copper has a temperature coefficient (α) of 0.00386 per degree C.
So the resistance value needs to be adjusted based upon how far from 20 C the temperature is.
50 - 20 = 30 C
So 0.00386 * 30 = 0.1158 meaning that the actual resistance at 50 C will be 11.58% higher.
So 1.1158 * 0.016 = 0.0179 ohms.
If you're using annealed copper, the values for Ď and the temperature coefficient change.
Ď = 1.72e-8
α = 0.00393
Doing the math, you get
1.72e-8 * 9.5493e5 * (1 + 30 * 0.00393) = 0.0184 ohms
Answer:
The dimension is 
Explanation:
From the question we are told that
Here ![[J] = \frac{1}{L^2 T}](https://tex.z-dn.net/?f=%5BJ%5D%20%3D%20%5Cfrac%7B1%7D%7BL%5E2%20T%7D)
![[n] =\frac{1}{L^3}](https://tex.z-dn.net/?f=%5Bn%5D%20%3D%5Cfrac%7B1%7D%7BL%5E3%7D)
![[x] = L](https://tex.z-dn.net/?f=%5Bx%5D%20%3D%20L)
So
![\frac{1}{L^2 T} = -D \frac{d(\frac{1}{L^3})}{d[L]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BL%5E2%20T%7D%20%3D%20%20-D%20%5Cfrac%7Bd%28%5Cfrac%7B1%7D%7BL%5E3%7D%29%7D%7Bd%5BL%5D%7D)
Given that the dimension represent the unites of n and x then the differential will not effect on them
So
![\frac{1}{L^2 T} = -D \frac{(\frac{1}{L^3})}{[L]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BL%5E2%20T%7D%20%3D%20%20-D%20%5Cfrac%7B%28%5Cfrac%7B1%7D%7BL%5E3%7D%29%7D%7B%5BL%5D%7D)
=> 
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