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3 years ago

How does a turbo charger work

Engineering

2 answers:

yKpoI14uk [10]3 years ago

8 0

Answer:

The turbocharger on a car applies a very similar principle to a piston engine. It uses the exhaust gas to drive a turbine. This spins an air compressor that pushes extra air (and oxygen) into the cylinders, allowing them to burn more fuel each second.

Explanation:

Romashka-Z-Leto [24]3 years ago

7 0

Answer:

Explanation:

You might be interested in

In the following code, determine the values of the symbols here and there. Write the object code in hexadecimal. (Do not predict

allsm [11]

Answer:

Answer explained below

Explanation:

The value of here is 9

The value of there is hexadecimal value of DECO here, d = 0x39 aaaa (aaaa is the memory address of here )

We have the object code :-

let's take there address is 0x0007

0x0005 BR there :- 0x120020

0x0007 here: .WORD 9

310003 there: DECO here,d - 0x390007

310005 STOP

.END

4 0

3 years ago

The two types of outlets that are found in an electrical system are:______

Dmitry [639]

Answer:

a. lighting and receptacle outlets

Explanation:

The two types of outlets that are found in an electrical system are

a. lighting and receptacle outlets

Outlets allow electrical equipment to connect to the electrical grid. The electrical grid provides alternating current to the outlet.

3 0

3 years ago

John has just graduated from State University. He owes $35,000 in college loans, but he does not have a job yet. The college loa

IRISSAK [1]

Answer:

The correct response is "821.88". A further explanation is given below.

Explanation:

According to the question,

The largest amount unresolved after five years would have been:

= $35000\times (\frac{F}{P}, 4 \ percent,5 )$

= $35000\times 1.216 7$

= $42584.50$

Now,

time (t) will be:

= $5\times 12$

= $60 \ monthly \ payments$

So, monthly payment will be:

= $42582.85\times (\frac{A}{P}, 0.5 \ percent,60 )$

= $42584.50\times 0.0193$

= $821.88$

6 0

3 years ago

Discuss the organizational system that you believe would be the most effective for the safety officer in a medium-sized (100-200

marin [14]

Answer:

A safety manager is a person who designs and maintains the safety elements at workplace. A balance should be required for production and the job in providing work environment. As a safety officer in a medium sized manufacturing facility the following organizational system can be designed and maintained:

Maintaining a workplace as per the guidelines by Occupational safety and health association. The rules and regulation should be such that maintains the manufacturing facilities.

For warning to workers proper labelling, floor mapping, signs, posters should be used.

Procurement and usage of safe tools.

A guideline that describes safety standard and precautionary measures should be available to the workers. They should be aware about all the steps that needs to be taken in crisis.

Ensuring that the workers have enough training safety and health or accident prevention.

Identify and eliminate the hazardous elements from the workplace.

A strict action should be taken against the worker in case of violation of rules and not adhering with guidelines.

3 0

3 years ago

If you are interested only in the temperature range of 20° to 40°C and the ADC has a 0 to 3V input range, design a signal condit

mario62 [17]

Explanation:

Temperature range → 0 to 80'c

respective voltage output → 0.2v to 0.5v

required temperature range 20'c to 40'c

Where T = 20'c respective voltage

$\begin{aligned}v_{20} &=0.2+\frac{0.5-0.8}{80} \times 20 \\&=0.2+\frac{0.3}{80} \times 20 \\V_{20} &=0.275 v\end{aligned}$

$\begin{aligned}\text { when } T=40^{\circ} C & \text { . } \\v_{40} &=0.2+\frac{0.5-0.2}{80} \times 40 \\&=0.35 V\end{aligned}$

Therefore, Sensor output changes from 0.275v to 0.35volts for the ADC the required i/p should cover the dynamic range of ADC (ie - 0v to 3v)

so we have to design a circuit which transfers input voltage 0.275volts - 0.35v to 0 - 3v

Therefore, the formula for the circuit will be

$\begin{array}{l}v_{0}=\left(v_{i n}-0.275\right) G \\\sigma=\ldots \frac{3-0}{0.35-0.275}=3 / 0.075=40 \\v_{0}=\left(v_{i n}-0.275\right) 40\end{array}$

The simplest circuit will be a op-amp

NOTE: Refer the figure attached

Vs is sensor output

Vr is the reference volt, Vr = 0.275v

$\begin{aligned}v_{0}=& v_{s}-v_{v}\left(1+\frac{R_{2}}{R_{1}}\right) \\\Rightarrow & \frac{1+\frac{R_{2}}{R_{1}}}{2}=40 \\& \frac{R_{2}}{R_{1}}=39 \quad \Rightarrow\end{aligned}$

choose R2, R1 such that it will maintain required ratio

The output Vo can be connected to voltage buffer if you required better isolation.

3 0

3 years ago