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2 years ago

12

During delivery of a 2023 ariya equipped with propilot assist 2. 0, what should you point out to your customers about the turn s

ignal stalk?.

1 answer:

marin [14]2 years ago

4 0

You should point out to your customers that the turn signal stalk should be turned ON when they are about 100 feet away from an intersection.

<h3>What is a turn signal stalk?</h3>

A turn signal stalk can be defined as a slender lever which is usually mounted on or near the steering wheel of an automobile vehicle, and it's used by a driver to control a signal or perform a function.

In conclusion, it is very important that you should point out to your customers that the turn signal stalk should be turned ON when they are about 100 feet away from an intersection.

Read more on an intersection here: brainly.com/question/22580476

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You might be interested in

If it took 2 hours to clean 4 houses how many houses could be cleaned in 3 hours

Lena [83]

Answer:

6 houses

Explanation:

because

2hrs=4 houses which means you are cleaning 2houses in one hour

so in 3 hours you will houses because you will clean 2 houses in one hour

I hope this helped you sorry if I am wrong

6 0

3 years ago

An inventor claims to have devised a cyclical power engine that operates with a fuel whose temperature is 750 °C and radiates wa

Phantasy [73]

Answer:

Yes

Explanation:

Given Data

Temprature of source=750°c=1023k

Temprature of sink =0°c=273k

Work produced=3.3KW

Heat Rejected=4.4KW

Efficiency of heat engine(η)= $\frac{Work produced}{Heat supplied}$

and

Heat Supplied ${\left (Q_s\right)}=Work Produced(W)+Heat rejected\left ( Q_r \right )$

${Q_s}=3.3+4.4=7.7KW$

η= $\frac{3.3}{7.7}$

η=42.85%

Also the maximum efficiency of a heat engine operating between two different Tempratures i.e. Source & Sink

η=1- $\frac{T_ {sink}}{T_ {source}}$

η=1- $\frac{273}{1023}$

η=73.31%

Therefore our Engine Efficiency is less than the maximum efficiency hence the given claim is valid.

5 0

3 years ago

Can someone pls give me the answer to this?

olganol [36]

I think option c 12 is currect

7 0

3 years ago

Think of the differences between circuit-switching and packet-switching paradigms in the Internet core design. Assume an Interne

dem82 [27]

Answer:

0.264 ; 0.079

Explanation:

Given that:

Sample size, n = 100

Probability of being active, p = 1% = 1/100 = 0.01

Using the binomial probability relation :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

Probability that more than 1 user will be active

P(x > 1) = 1 - [p(x=0) + p(x = 1)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x > 1) = 1 - [0.366 + 0.370]

P(x > 1) = 0.264

2.)

Probability that more than 2 user will be active

P(x > 2) = 1 - [p(x=0) + p(x = 1) + p(x = 2)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x = 2) = 100C2 * 0.01^2 * 0.99^98 = 0.185

P(x > 1) = 1 - [0.366 + 0.370 + 0.185]

P(x > 1) = 0.079

7 0

3 years ago

An industrial plant consists of several 60 Hz single-phase motors with low power factor. The plant absorbs 600 kW with a power f

Gelneren [198K]

Answer:

(a) $Q=332$ $kvar$ and $C=5.66$ $uF$

(b) $pf=0.90$ lagging

Explanation:

Given Data:

$P=600kW$

$V=12.47kV$

$f=60Hz$

$pf_{old} =0.75$

$pf_{new} =0.95$

(a) Find the required kVAR rating of a capacitor

$\alpha _{old}=cos^{-1}(0.75) =41.41$ °

$\alpha _{new}=cos^{-1}(0.95) =18.19$ °

The required compensation reactive power can be found by

$Q=P(tan(\alpha_{old}) - tan(\alpha_{new}))$

$Q=600(tan(41.41) - tan(18.19))$

$Q=332$ $kvar$

The corresponding capacitor value can be found by

$C=Q/2\pi fV^{2}$

$C=332/2*\pi *60*12.47^{2}$

$C=5.66$ $uF$

(b) calculate the resultant supply power factor

First convert the hp into kW

$P_{mech} =250*746=186.5$ $kW$

Find the electrical power (real power) of the motor

$P_{elec} =P_{mech}/n$

where $n$ is the efficiency of the motor

$P_{elec} =186.5/0.80=233.125$ $kW$

The current in the motor is

$I_{m} =(P/\*V*pf)$

The pf of motor is 0.85 Leading

Note that represents the angle in complex notation (polar form)

$I_{m} =(233.125/12.47*0.85)$

$I_{m}=18.694+11.586$ $j$ $A$

Now find the Load current

pf of load is 0.75 lagging (notice the minus sign)

$I_{load} =(600/12.47*0.75)$

$I_{load} =48.115-42.433$ $j$ $A$

Now the supply current is the current flowing in the load plus the current flowing in the motor

$I_{supply} =I_{m} + I_{load}$

$I_{supply}= (18.694+11.586)+(48.115-42.433)$

$I_{supply} =66.809-30.847j$ $A$

or in polar form

$I_{supply} =73.58$ °

Which means that the supply current lags the supply voltage by 24.78

therefore, the supply power factor is

$pf=cos(24.78)=0.90$ lagging

Which makes sense because original power factor was 0.75 then we installed synchronous motor which resulted in improved power factor of 0.90

8 0

3 years ago