The safety risks are the same for technicians who work on hybrid electric vehicles (HEVs) or EVs as those who work on conventional gasoline vehicles.

Answers

Answer 1

Answer:

The safety risks are the same for technicians who work on hybrid electric vehicles (HEVs) or EVs as those who work on conventional gasoline vehicles: False.

Safety risks can be defined as an assessment of the risks and occupational hazards associated with the use, operation or maintenance of an equipment or automobile vehicle that is capable of leading to the;

Harm of a worker (technician).

Injury of a worker (technician).

Illness of a worker (technician).

Death of a worker (technician).

Hybrid electric vehicles (HEVs) or EVs are typically designed and developed with parts or components that operates through the use of high voltageelectrical systems ranging from 100 Volts to 600 Volts. Also, these type of vehicles have an in-built HEV batteries which are typically encased in sealed shells so as to mitigate potential hazards to a technician.

On the other hand, conventional gasoline vehicles are typically designed and developed with parts or components that operates on hydrocarbon such as fuel and motor engine oil. Also, conventional gasoline vehicles do not require the use of high voltage electrical systems and as such poses less threat to technicians, which is in contrast with hybrid electric vehicles (HEVs) or EVs.

This ultimately implies that, the safety risks for technicians who work on hybrid electric vehicles (HEVs) or EVs are different from those who work on conventional gasoline vehicles due to high voltage electrical systems that are being used in the former.

In conclusion, technicians who work on hybrid electric vehicles (HEVs) or EVs are susceptible (vulnerable) to being electrocuted to death when safety risks are not properly adhered to unlike technicians working on conventional gasoline vehicles.

Find more information: brainly.com/question/2878752

Answer 2

Answer:

Answer:

Batteries are safe when handled properly.

Explanation:

Just like the battery in your phone, the battery in some variant of an electric car is just as safe. If you puncture/smash just about any common kind of charged battery, it will combust. As long as you don't plan on doing anything extreme with the battery (or messing with high voltage) you should be fine.

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8.2.1: Function pass by reference: Transforming coordinates. Define a function CoordTransform() that transforms the function's first two input parameters xVal and yVal into two output parameters xValNew and yValNew. The function returns void. The transformation is new

Answers

Answer:

The output will be (3, 4) becomes (8, 10)

Explanation:

#include <stdio.h>

//If you send a pointer to a int, you are allowing the contents of that int to change.

void CoordTransform(int xVal,int yVal,int* xNew,int* yNew){

*xNew = (xVal+1)*2;

*yNew = (yVal+1)*2;

}

int main(void) {

int xValNew = 0;

int yValNew = 0;

CoordTransform(3, 4, &xValNew, &yValNew);

printf("(3, 4) becomes (%d, %d)\n", xValNew, yValNew);

return 0;

}

Write a C++ program to display yearly calendar. You need to use the array defined below in your program. // the first number is the month and second number is the last day of the month. into yearly[12][2] =

Answers

Answer:

//Annual calendar

#include <iostream>

#include <string>

#include <iomanip>

void month(int numDays, int day)

{

int i;

string weekDays[] = {"Su", "Mo", "Tu", "We", "Th", "Fr", "Sa"};

// Header print

cout << "\n----------------------\n";

for(i=0; i<7; i++)

{

cout << left << setw(1) << weekDays[i];

cout << left << setw(1) << "|";

}

cout << left << setw(1) << "|";

cout << "\n----------------------\n";

int firstDay = day-1;

//Space print

for(int i=1; i< firstDay; i++)

cout << left << setw(1) << "|" << setw(2) << " ";

int cellCnt = 0;

// Iteration of days

for(int i=1; i<=numDays; i++)

{

//Output days

cout << left << setw(1) << "|" << setw(2) << i;

cellCnt += 1;

// New line

if ((i + firstDay-1) % 7 == 0)

{

cout << left << setw(1) << "|";

cout << "\n----------------------\n";

cellCnt = 0;

}

// Empty cell print

if (cellCnt != 0)

{

// For printing spaces

for(int i=1; i<7-cellCnt+2; i++)

cout << left << setw(1) << "|" << setw(2) << " ";

cout << "\n----------------------\n";

}

int main()

{

int i, day=1;

int yearly[12][2] = {{1,31},{2,28},{3,31},{4,30},{5,31},{6,30},{7,31},{8,31},{9,30},{10,31},{11,30},{12,31}};

string months[] = {"January",

"February",

"March",

"April",

"May",

"June",

"July",

"August",

"September",

"October",

"November",

"December"};

for(i=0; i<12; i++)

{

//Monthly printing

cout << "\n Month: " << months[i] << "\n";

month(yearly[i][1], day);

if(day==7)

{

day = 1;

}

else

{

day = day + 1;

}

cout << "\n";

}

return 0;

}

//end

___________ are used in an automotive shop and can be harmful to the environment if not disposed of properly.Oil filters
Fluorescent lamps
Mercury-containing lamps
All of the above

Answers

Answer: D all above

Explanation:

Jus done it

A rectangular channel with a width of 2 m is carrying 15 m3/s. What are the critical depth and the flow velocity

Answers

Answer:

The critical depth of the rectangular channel is approximately 1.790 meters.

The flow velocity in the rectangular channel is 4.190 meters per second.

Explanation:

From Open Channel Theory we know that critical depth of the rectangular channel ( $y_(c)$ ), measured in meters, is calculated by using this equation:

$y_(c) = \sqrt[3]{(\dot V^(2))/(g\cdot b^(2)) }$ (Eq. 1)

Where:

$\dot V$ - Volume flow rate, measured in cubic meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

$b$ - Channel width, measured in meters.

If we know that $\dot V = 15\,(m^(3))/(s)$ , $g = 9.807\,(m)/(s^(2))$ and $b = 2\,m$ , then the critical depth is:

$y_(c) = \sqrt[3]{(\left(15\,(m^(3))/(s) \right)^(2))/(\left(9.807\,(m)/(s^(2)) \right)\cdot (2\,m)^(2)) }$

$y_(c) \approx 1.790\,m$

The critical depth of the rectangular channel is approximately 1.790 meters.

Lastly, the flow velocity ( $v$ ), measured in meters, is obtained from this formula:

$v = (\dot V)/(b\cdot y_(c))$ (Eq. 2)

If we know that $\dot V = 15\,(m^(3))/(s)$ , $b = 2\,m$ and $y_(c) = 1.790\,m$ , then the flow velocity in the rectangular channel is:

$v = (15\,(m^(2))/(s) )/((2\,m)\cdot (1.790\,m))$

$v = 4.190\,(m)/(s)$

The flow velocity in the rectangular channel is 4.190 meters per second.

If new research showed that the standard heights for crest vertical curve design were H1=3.0 ft and H2=1.5 ft, respectively, by what percent would the minimum length of curvature increase or decrease for a crest curve with a stopping sight distance of 500 ft and grades of +3% and -2%, respectively.

Answers

Answer:

A = 5

S<L, L = 714.89ft

S>L, L = 650.29ft

L = 115.85ft

Percentage min. Length of curvature = 6.2 %

Explanation: see explanation at the attached file

"Given a nodal delay of 84.1ms when there is no traffic on the network (i.e. usage = 0%), what is the effective delay when network usage = 39.3% ? (Give answer is miliseconds, rounded to one decimal place, without units. So for an answer of 0.10423 seconds you would enter "104.2" without the qu"

Answers

Answer:

Explanation:

effective delay = delay when no traffic x $(100)/(100- network\r usage)$

effective delay = $84.1 * (100)/(100-39.3)=138.55024711697ms$

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The substance called olivine may have any composition between Mg2SiO4 and Fe2SiO4, i.e. the Mg atoms can be replaced by Fe atoms in any proportion without altering the crystal structure except by expanding it slightly: this is an example of a binary solid solution series. For different compositions, the lines in the powder diffraction patterns are in slightly different positions, because of the cell expansion, but the overall pattern remains basically the same. The spacing of the lattice planes varies linearly with composition, and this can be used in a rapid and non- destructive method of analysis. a. The (062) reflection from olivine is strong and well resolved from other lines. Calculate d062 for an olivine that displays its (062) reflection at a Bragg angle of 37.21° (i.e., a diffraction angle of 74.42°) when x-rays with a wavelength of 0.1790 nm are used. b. The d062 spacing as measured accurately for synthetic materials is 0.14774 nm for Mg2SiO4 and 0.15153 nm for Fe2SiO4. What would be the approximate composition, expressed in mol.% Mg2SiO4, of an olivine material for which do62 has the value obtained in part 2.1 above?