Answer:
Technician B is correct
Explanation:
An oxygen sensor will generate about 1.0 volts when the fuel mixture is rich and there is little unburned oxygen in the exhaust. When the mixture is lean, the sensor's output voltage will drop down to about 0.1 volts.
An o2 sensor cannot accurately measure how rich or how well an exhaust system is.
Therefore, Technician B is correct.
Answer:
a) F₁₂₀ = 1.34 pa A , b) F₂₀ = 0.746 pa A
Explanation:
Part. A
. The definition of pressure is
P = F / A
As the air can approach an ideal gas we can use the ideal gas equation
P V = n R T
Let's write this equation for two temperatures
P₁ V = n R T₁
P₂2 V = n R T₂
P₁ / P₂ = T₁ / T₂
point 1 has a pressure of P₁ = pa and a temperature of (20 + 273) K, point 2 is at (120 + 273) K, we calculate the pressure P₂
P₂ = P₁ T₂ / T₁
P₂ = pa 393/293
P₂ = 1.34 pa
We calculate the strength
P₂ = F₁₂₀ / A
F₁₂₀ = 1.34 pa A
Part B
In this case the data is
Point 1 has a temperature of 393K and an atmospheric pressure (P₁ = pa), point 2 has a temperature of 293K, let's calculate its pressure
P₁ / P₂ = T₁ / T₂
P₂ = P₁ T₂ / T₁
P₂ = pa 293/393
P₂ = 0.746 pa
Let's calculate the force (F20), from this point
F₂₀ / A = 0.746 pa
F₂₀ = 0.746 pa A
Answer:
-27.3 m/s
Explanation:
Given:
y₀ = 38 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2a (y − y₀)
v² = (0 m/s)² + 2 (-9.8 m/s²) (0 m − 38 m)
v = -27.3 m/s
Or, you can solve with energy.
PE = KE
mgh = ½ mv²
v² = 2gh
v = -27.3 m/s
The answer is circle graph because it shows how a part relates to the whole number.= )