Answer:
The correct option is;
Neither Technician A nor B
Explanation:
The evaporative emission monitor or Evaporaive Emission Control System EVAP System monitors enables the Power Control Module of the car to check fuel system leak integrity and the vapor consumption efficiency during engine combustion
It is a requirement of EPA on cars to check the emission of smug forming evaporates from cars
Serious monitor faults can cause the turning on of the check engine lights and the vehicle will not pass OBD II test, but it will not lead to engine shutdown
It runs when the engine is 15 to 85% full and the TP sensor is between 9% and 35%.
Therefore, the correct option is that neither Technician A nor B are correct.
Answer:
metals, composite, ceramics and polymers.
Explanation:
The four categories of engineering materials used in manufacturing are metals, composite, ceramics and polymers.
i) Metals: Metals are solids made up of atoms held by matrix of electrons. They are good conductors of heat and electricity, ductile and strong.
ii) Composite: This is a combination of two or more materials. They have high strength to weight ratio, stiff, low conductivity. E.g are wood, concrete.
iii) Ceramics: They are inorganic, non-metallic crystalline compounds with high hardness and strength as well as poor conductors of electricity and heat.
iv) Polymers: They have low weight and are poor conductors of electricity and heat
Answer:
Explanation:
There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.
We are given the following information:
y₁ = 3 m
y₂ = 3 m − 2.4 m = 0.6 m
t₂ − t₁ = 0.4 s
a = -9.8 m/s²
t₀ = 0 s
v₀ = 0 m/s
We need to find y₀.
Use a constant acceleration equation:
y = y₀ + v₀ t + ½ at²
Evaluated at point 1:
3 = y₀ + (0) t₁ + ½ (-9.8) t₁²
3 = y₀ − 4.9 t₁²
Evaluated at point 2:
0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²
0.6 = y₀ − 4.9 t₂²
Solve for y₀ in the first equation and substitute into the second:
y₀ = 3 + 4.9 t₁²
0.6 = (3 + 4.9 t₁²) − 4.9 t₂²
0 = 2.4 + 4.9 (t₁² − t₂²)
We know t₂ = t₁ + 0.4:
0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)
0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))
0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)
0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)
0 = 2.4 − 3.92 t₁ − 0.784
0 = 1.616 − 3.92 t₁
t₁ = 0.412
Now we can plug this into the original equation and find y₀:
3 = y₀ − 4.9 t₁²
3 = y₀ − 4.9 (0.412)²
3 = y₀ − 0.83
y₀ = 3.83
Rounded to two significant figures, the height of the tree is 3.8 meters.
