Generally, frictional losses are more predominant for the machines being not 100% efficient. This friction leads to the loss of energy in the form of heat, into the surroundings. Some of the supplied energy may be utilised to change the entropy (measure of randomness of the particles) of the system.
Answer:
I couldn't find options for your question online, but I can give you an explanation so you can choose the correct option.
Explanation:
A spark knock is a form of unpredictable behavior that occurs in combustion, that is, in the chemical reaction that occurs between oxygen and an oxidizable material. Such combustion is usually manifested by incandescence or flame.
The spark knock is a detonation that occurs when there is a lot of pressure in the fuel.
<u>Some situations in which this can happen are:
</u>
- Engine overloaded.
- Maximum pressure in the cylinders.
- Engine overheated.
- Overheated air.
- Long and excessive engine ignition timing.
- Spark plug at high temperatures.
Answer:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑
= mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑ = mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN
Answer and Explanation:
O decreases linearly with the distance from the generator
Answer:
a. 0.28
Explanation:
Given that
porosity =30%
hydraulic gradient = 0.0014
hydraulic conductivity = 6.9 x 10⁻4 m/s
We know that average linear velocity given as
The velocity in m/d ( 1 m/s =86400 m/d)
v= 0.27 m/d
So the nearest answer is 'a'.
a. 0.28
