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3 years ago

14

While discussing electronic gasoline injection systems, Technician A states that most multiport EFI systems use two injectors fo

r each engine cylinder. Technician B says such a system uses only one injector per engine cylinder. Who is right?

1 answer:

lutik1710 [3]3 years ago

8 0

Answer:Technician B is right.

Explanation: A multi port system is an advanced type of Electronic Fuel Inection system that uses one injection per cylinder. The microcontroller which is a part in the automated system of the Multi port system is responsible for monitoring and controlling the fuel injection process leading to fuel efficiency.

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Communicating results allows scientists to learn from each other's results.

sveta [45]

True, scientists often talk to each other to figure out if their results were similar and what they could have done better.

Although, talking to other scientists does have risks, other scientists could copy your work and further better it.

So, your final answer is TRUE, sorry for the long answer, I needed to have a word count about 20 characters and then I got carried away! lol

3 0

3 years ago

Read 2 more answers

Estimate how much solar energy reaches the earth per year (in Joule).

Alexxandr [17]

Each hour 430 quintillion Joules of energy from the sun hits the Earth.

In a year it is very hard to determine because of the night and different light levels.

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3 years ago

A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

Elden [556K]

Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

$J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}$

$B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\$

$B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T$

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

$B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T$

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0

3 years ago

find the potential energy of an aircraft weighing 10000 bs at 5000 ft true altitude and 125 kts true air speed

Cloud [144]

Answer:

$U=5*10^7ft.Ib$

Explanation:

From the question we are told that

Weight $W= 10000bs$

Altitude $H=5000ft$

Speed $V=125kts\\1kts=0.514m/s\\V=125*0.514=>64.25m/s$

Generally the equation for Potential energy ids mathematically given as

$Potential\ Energy\ U=mgh$

$U=Wh$

$U=10000*5000$

$U=5*10^7ft.Ib$

6 0

3 years ago

A 500 N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area

Ghella [55]

Answer:

d. 10000N

Explanation:

When a force ( $F_1$ ) is exerted on the smaller area piston ( $A_1$ ), the pressure that originates therein is transmitted to the larger area piston( $A_2$ ). According to Pascal's principle the pressure on the smallest piston ( $P_1=\frac{F_1}{A_1}$ ) will be equal to the pressure on the largest piston ( $P_2=\frac{F_2}{A_2}$ ):

$\frac{F_1}{A_1}=\frac{F_2}{A_2}\\F_2=F_1\frac{A_2}{A_1}\\F_2=500N\frac{40cm}{2cm}\\F_2=10000N$

7 0

3 years ago