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2 years ago

What advanced safety features are now standard on 2023 Z?

Engineering

1 answer:

Anettt [7]2 years ago

3 0

The advanced safety features that are now standard on 2023 Z are:

Pedestrian Detection and Automatic Emergency Braking
Intelligent Forward Collision Warning (IFCW).
Blind Spot Alert.

The above is the short name or nickname for the All New 2023 Nissan Z coupe Sports car.

The care also features Advanced driver assist and safety technology which relieves the driver's everyday workload, allowing you to focus on what matters most.

Learn more about safety features in cars:
brainly.com/question/24078882
#SPJ1

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At an impaired driver checkpoint, the time required to conduct the impairment test varies (according to an exponential distribut

professor190 [17]

Answer:

Option (d) 2 min/veh

Explanation:

Data provided in the question:

Average time required = 60 seconds

Therefore,

The maximum capacity that can be accommodated on the system, μ = 60 veh/hr

Average Arrival rate, λ = 30 vehicles per hour

Now,

The average time spent by the vehicle is given as

⇒ $\frac{1}{\mu(1-\frac{\lambda}{\mu})}$

thus,

on substituting the respective values, we get

Average time spent by the vehicle = $\frac{1}{60(1-\frac{30}{60})}$

Average time spent by the vehicle = $\frac{1}{60(1-0.5)}$

Average time spent by the vehicle = $\frac{1}{60(0.5)}$

Average time spent by the vehicle = $\frac{1}{30}$ hr/veh

Average time spent by the vehicle = $\frac{1}{30}\times60$ min/veh

[ 1 hour = 60 minutes]

thus,

Average time spent by the vehicle = 2 min/veh

Hence,

Option (d) 2 min/veh

7 0

3 years ago

Let A→=(150iˆ+270jˆ) mm , B→=(300iˆ−450jˆ) mm , and C→=(−100iˆ−250jˆ) mm . Find scalars r and s, if possible, such that R→=rA→+s

ioda

Answer: r = 0.8081; s = -0.07071

Explanation:

A = (150i + 270j) mm

B = (300i - 450j) mm

C = (-100i - 250j) mm

R = rA + sB + C = 0i + 0j

R = r(150i + 270j) + s(300i - 450j) + (-100i - 250j) = 0i + 0j

R = (150r + 300s - 100)i + (270r - 450s - 250)j = 0i + 0j

Equating the i and j components;

150r + 300s - 100 = 0

270r - 450s - 250 = 0

150r + 300s = 100

270r - 450s = 250

solving simultaneously,

r = 0.8081 and s = -0.07071

QED!

5 0

3 years ago

Regeneration can only increase the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than the worki

storchak [24]

Answer:

True, Regeneration is the only process where increases the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than working fluid leaving the compressor.

Option: A

Explanation:

To increase the efficiency of brayton cycle there are three ways which includes inter-cooling, reheating and regeneration. Regeneration technique is used when a turbine exhaust fluids have higher temperature than the working fluid leaving the compressor of the turbine.

Thermal efficiency of a turbine is increased as the exhaust fluid having higher temperatures are used in heat exchanger where the fluids from the compressor enters and increases the temperature of the fluids leaving the compressor.

6 0

3 years ago

Read 2 more answers

Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank is fitted with a paddle whee

Juli2301 [7.4K]

Answer:

a) 1 m^3/Kg

b) 504 kJ

c) 514 kJ

Explanation:

Given

-The mass of C_o2 = 1 kg

-The volume of the tank V_tank = 1 m^3

-The added energy E = 14 W

-The time of adding energy t = 10 s

-The increase in specific internal energy Δu = +10 kJ/kg

-The change in kinetic energy ΔKE = 0 and The change in potential energy

ΔPE =0

Required

(a)Specific volume at the final state v_2

(b)The energy transferred by the work W in kJ.

(c)The energy transferred by the heat transfer W in kJ and the direction of

the heat transfer.

Assumption

-Quasi-equilibrium process.

Solution

(a) The volume and the mass doesn't change then, the specific volume is constant.

v= V_tank/m ---> 1/1= 1 m^3/Kg

(b) The added work is defined by.

W =E * t ---> 14 x 10 x 3600 x 10^-3 = 504 kJ

(c) From the first law of thermodynamics.

Q - W = m * Δu

Q = (m * Δu) + W--> (1 x 10) + 504 = 514 kJ

The heat have (+) sign the n it is added to the system.

7 0

3 years ago

A counter-flow double pipe heat exchanger is heat heat water from 20 degrees Celsius to 80 degrees Celsius at the rate of 1.2 kg

lakkis [162]

Answer:

$L=107.6m$

Explanation:

Cold water in: $m_{c}=1.2kg/s, C_{c}=4.18kJ/kg\°C, T_{c,in}=20\°C, T_{c,out}=80\°C$

Hot water in: $m_{h}=2kg/s, C_{h}=4.18kJ/kg\°C, T_{h,in}=160\°C, T_{h,out}=?\°C$

$D=1.5cm=0.015m, U=649W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2},L=?m$

Step 1: Determine the rate of heat transfer in the heat exchanger

$Q=m_{c}C_{c}(T_{c,out}-T_{c,in})$

$Q=1.2*4.18*(80-20)$

$Q=300.96kW$

Step 2: Determine outlet temperature of hot water

$Q=m_{h}C_{h}(T_{h,in}-T_{h,out})$

$300.96=2*4.18*(160-T_{h,out})$

$T_{h,out}=124\°C$

Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)

$dT_{1}=T_{h,in}-T_{c,out}$

$dT_{1}=160-80$

$dT_{1}=80\°C$

$dT_{2}=T_{h,out}-T_{c,in}$

$dT_{2}=124-20$

$dT_{2}=104\°C$

$LMTD = \frac{dT_{2}-dT_{1}}{ln(\frac{dT_{2}}{dT_{1}})}$

$LMTD = \frac{104-80}{ln(\frac{104}{80})}$

$LMTD = \frac{24}{ln(1.3)}$

$LMTD = 91.48\°C$

Step 4: Determine required surface area of heat exchanger

$Q=UA_{s}LMTD$

$300.96*10^{3}=649*A_{s}*91.48$

$A_{s}=5.07m^{2}$

Step 5: Determine length of heat exchanger

$A_{s}=piDL$

$5.07=pi*0.015*L$

$L=107.57m$

7 0

3 years ago