Answer:
C. Multipoint fuel injection
Explanation:
A fuel injection system can be defined as a system found in the engine of most automobile cars, used for the supply of a precise amount of fuel or fuel-air mixture to the cylinders in an internal combustion engine through the use of an injector.
There are different types of fuel injection system and these includes;
I. Central-point injection.
II. Throttle (single point) body injection.
III. Gasoline direct injection.
IV. Multipoint (port) fuel injection.
Multipoint fuel injection is a type of fuel injection system that operates with fuel injectors located only in the intake manifold near each intake valve and sprays fuel toward the valve. As a result, it allows for the supply of a precise amount of fuel and as such creating a better air-fuel ratio for automobile cars.
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Admission to the Engineering course at Cambridge is highly competitive, both in terms of the numbers and quality of applicants. In considering applicants, Colleges look for evidence both of academic ability and of motivation towards Engineering. There are no absolute standards required of A Level achievement, but it should be noted that the average entrant to the Department has three A* grades. You need to get top marks in Maths and Physics.All Colleges strongly prefer applicants for Engineering to be taking a third subject that is relevant to Engineering.
Hope that helps and good luck if you are applying. Can you please mark this as brainliest and press the thank you button and if you have any further questions please let me know!!
Answer:
a) λ = 2 meters
b) λ = 1.76 meters
c)In free space: length of phasing section = 0.5 meters and in dielectric medium: length of phasing section =0.44 meter
Explanation:
Given that:
velocity factor = 0.88
operating frequency = 150 MHz = 150 × 10⁶ Hz
phasing section is a quarter-wavelength = λ/4
a) Wavelength of free space:
λ = c/f
where c is the speed of light = 3×10⁸
λ = (3×10⁸)/150 × 10⁶
λ = 2 meters.
b) Wavelength of dielectric:
Velocity of dielectric = speed of light (c) × velocity factor
Velocity of dielectric = 3×10⁸ × 0.88 = 2.64 × 10⁸ m/s
wavelength dielectric = Velocity of dielectric /f = (2.64 × 10⁸)/150 × 10⁶ =1.76
wavelength dielectric = 1.76 meters
c) length of the phasing section in meters:
l = λ/4
In free space:
l = 2/4 = 0.5 meter
In dielectric
l = 1.76/4 = 0.44 meter