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katrin [286]
4 years ago
5

Two Technicians are discussing shock absorbers. Technician A says most shock absorbers help support vehicle weight. Technician B

says a light coating of oil around the shock absorber shaft would require a shock replacement. Which Technician would be correct?
Question 4 options:

Technician A only


Technician B only


Both Technician A and B


Neither Technician A nor B
Engineering
1 answer:
yKpoI14uk [10]4 years ago
6 0

Answer:

Both Technician A and B

Explanation:

In a vehicle suspension system, a shock absorber has the functions such as; limiting the movement of the vehicle, stabilizing the ride, stabilizing the tires of the vehicle, minimizing wear and tear in the vehicle and decreasing the overall suspension tear.  Some of the ways of maintaining shock absorbers is to inspect the strut for leaks.Shock oil sometimes leaks to cover the seals with a transparent liquid.

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Write a Pig script and run it in local mode on this data to find out the top 10 states according to the land area. Since you wan
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Explanation:

Apache Pig script execution modes

Local mode: In 'local mode', you can run the pig script on the local file system. In this case, you don't need to store the data in the Hadoop HDFS file system, instead you can work with the data stored in the local file system.

MapReduce mode: In 'MapReduce mode', the data must be stored in the HDFS file system and you can process the data with the help of pig script.

Apache Pig Script in MapReduce mode

Let's say our task is to read data from a data file and display the required contents in the terminal as output.

The sample data file contains the following data:

Txt information file - Apache Pig Script - Edureka

Save the text file with the name 'information.txt'

The sample data file contains five First Name, Last Name, Mobile Number, City, and Profession columns separated by the tab key. Our task is to read the contents of this HDFS file and display all the columns of these records.

To process this data using Pig, this file must be present in Apache Hadoop HDFS.

Command: hadoop fs –copyFromLocal /home/edureka/information.txt / edureka

3 0
3 years ago
The electron beam in a TV picture tube carries 1015 electrons per second. As a design engineer, determine the voltage needed to
leonid [27]

Answer:

The voltage needed to accelerate the electron beam is 2.46 x 10^16 Volts

Explanation:

The rate of electron flow is given as:

q = 1015 electrons per second

The total current is given by:

Total Current = (Rate of electron flow)(Charge on one electron)

Total Current = I = (1015 electrons/s)(1.6 x 10^-19 C/electron)

I = 1.624 x 10^-16 A

Now, we know that electric power is given as:

Electric Power = Current x Voltage

P = IV

V = P/I

V = 4 W/1.624 X 10^-16 A

<u>V = 2.46 x 10^16 Volts</u>

6 0
3 years ago
An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures befo
SVEN [57.7K]

This question is incomplete, the complete question is;

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.

Answer:

a) the maximum pressure in the cycle is 30.01 Mpa

b) the heat transfer to air is 0.7058 KJ

c) mass of Air is 0.002957 kg

Explanation:

Given the data in the question;

We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)

so from the "ideal gas properties of air table"

Pr1 = 238

Pr4 = 2.379

we know that Pressure P1 is only maximum at the beginning of the expansion process,

so

now we express the relative pressure and pressure relation for the process 4-1

P1 = (Pr2/Pr4)P4

so we substitute

P1 = (238/2.379)300 kPa

P1 = 30012.6 kPa = 30.01 Mpa

Therefore the maximum pressure in the cycle is 30.01 Mpa

b)

the Thermal heat efficiency of the Carnot cycle is expressed as;

ηth = 1 - (TL/TH)

we substitute

ηth = 1 - (350K/1200K)

ηth = 1 - 0.2916

ηth = 0.7084

now we find the heat transferred

Qin = W_net.out / ηth

given that the net work output per cycle is 0.5 kJ

we substitute

Qin = 0.5 / 0.7084

Qin = 0.7058 KJ

Therefore, the heat transfer to air is 0.7058 KJ

c)

first lets express the change in entropy for process 3 - 4

S4 - S3 = (S°4 - S°3) - R.In(P4/P3)

S4 - S3 = - (0.287 kJ/Kg.K) In(300/150)kPa

= -0.1989 Kj/Kg.K = S1 - S2

so that; S2 - S1 = 0.1989 Kj/Kg.K

Next we find the net work output per unit mass for the Carnot cycle

W"_netout = (S2 - S1)(TH - TL)

we substitute

W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K

= 169.065 kJ/kg

Finally we find the mass

mass m = W_ net.out /  W"_netout

we substitute

m = 0.5 / 169.065

m = 0.002957 kg

Therefore, mass of Air is 0.002957 kg

5 0
3 years ago
A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the centr
seropon [69]

Answer:

1) The normal reactions at the front wheel is 9909.375 N

The normal reactions at the rear wheel is 8090.625 N

2) The least coefficient of friction required between the tyres and the road is 0.625

Explanation:

1) The parameters given are as follows;

Speed, u = 90 km/h = 25 m/s

Distance, s it takes to come to rest = 50 m

Mass, m = 1.8 tonnes = 1,800 kg

From the equation of motion, we have;

v² - u² = 2·a·s

Where:

v = Final velocity = 0 m/s

a = acceleration

∴ 0² - 25² = 2 × a × 50

a = -6.25 m/s²

Force, F =  mass, m × a = 1,800 × (-6.25) = -11,250 N

The coefficient of friction, μ, is given as follows;

\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625

Weight transfer is given as follows;

W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N

Therefore, we have for the car at rest;

Taking moment about the Center of Gravity CG;

F_R × 1.3 = 1.7 × F_F

F_R + F_F = 18000

F_R + \dfrac{1.3 }{1.7} \times  F_R = 18000

F_R = 18000*17/30 = 10200 N

F_F = 18000 N - 10200 N = 7800 N

Hence with the weight transfer, we have;

The normal reactions at the rear wheel F_R  = 10200 N - 2109.375 N = 8090.625 N

The normal reactions at the front wheel F_F =  7800 N + 2109.375 N = 9909.375 N

2) The least coefficient of friction, μ, is given as follows;

\mu = \dfrac{F}{R} =  \dfrac{11250}{18000} = 0.625

The least coefficient of friction, μ = 0.625.

3 0
4 years ago
I have to find the critical points of this function of two variables <img src="https://tex.z-dn.net/?f=%5C%5Cf%28x%2Cy%29%3Dx%5E
liraira [26]

Answer:

no i dont think there is

Explanation:

because theres not

4 0
4 years ago
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