Engineering High School

Of these common qualifications for Agriculture, Food, and Natural Resources careers, select those you already have or would like to gain.


A) physical strength, coordination, and dexterity

B) skills in working safely with equipment and machinery

C) accuracy and attention to detail

D) problem-solving skills

E) organizational skills

F) skills in communicating with, teaching, and supervising others

G) reliability and responsibility

Answer: C) accuracy and attention to detail. D) problem-solving skills. G) reliability and responsibility. I didn't know the answer at first but then I figured it out and got it right.

Of These Common Qualifications For Agriculture, Food, And Natural Resources Careers, Select Those You

Answers

Answer 1

The answers to the given questions are:

C) accuracy and attention to detail. D) problem-solving skills. G) reliability and responsibility

What are the important traits?

It is my belief that possessing qualities such as precision, meticulousness, ability to troubleshoot, reliability and responsibility, could prove indispensable for one's career in Agriculture, Food and Natural Resources.

The ability to minutely observe details and ensure exactness assumes significance towards ensuring excellence in the realm of agricultural products, as well as safeguarding appropriate standards for food safety. Responding effectively to any potential issue or obstacle via thoughtful problem resolution is fundamental to surmounting challenges within related industries.

Undoubtedly, personnel exhibiting traits harmonious with reliability and responsibility are prone to cultivating trust-worthy relationships amongst clients, comrades, and superiors.

In essence, an amalgamation of these multifaceted skills and interlinked attributes can mark a requirement when seeking employment within the domains as mentioned above.

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Related Questions

How tall was President Abraham Lincoln?

Answers

6 feet 4 inches

6 feet 4 inchesThe tallest U.S. president was Abraham Lincoln at 6 feet 4 inches (193 centimeters)

make me brainalist

President Abraham Lincoln was 1.93 m tall

A small block sits at one end of a flat board that is 4.00 m long. The coefficients of friction between the block and the board are μs = 0.550 and μk = 0.400. The end of the board where the block sits is slowly raised until the angle the board makes with the horizontal is α0 , and then the block starts to slide down the board.
Part A
If the angle is kept equal to α0 as the block slides, what is the speed of the block when it reaches the bottom of the board?

Answers

The speed of the block when it reaches the bottom of the board would be 3.19 m/s.

How to calculate the speed?

Let's start by finding the height of the board when the block starts to slide. The maximum angle α0 at which the block will remain stationary on the board is given by:

μs = tan(α0)

Solving for α0, we get:

α0 = tan^-1(μs) = tan^-1(0.550) = 29.59 degrees

The height of the board when the block starts to slide is given by:

h = 4.00 m sin(α0)

h = 4.00 m sin(29.59 degrees)

h = 2.09 m

Now let's find the work done by friction. The friction force is given by:

f= μk * N

where N is the normal force acting on the block. The normal force is equal in magnitude and opposite in direction to the component of the gravitational force acting perpendicular to the board, which is given by:

N = mg cos(α)

where m is the mass of the block, g is the acceleration due to gravity, and α is the angle of the board. The kinetic friction force does negative work, which is equal to:

Wf = -f * d

where d is the distance the block travels down the board. The work done by gravity is equal to the change in potential energy, which is given by:

Wg = mgh

where h is the height of the board. At the bottom of the board, all of the potential energy has been converted to kinetic energy, so we have:

1/2 * m * v^2 = mgh - f * d

Solving for v, we get:

v = sqrt(2gh - 2μk(N/m) * d)

We can now substitute the values we have found:

h = 2.09 m

μk = 0.400

N = mg cos(α0) = mg cos(29.59 degrees)

d = 4.00 m

Substituting these values and solving for v, we get:

v = sqrt(2gh - 2μk(N/m) * d)

v = sqrt(2 * 9.81 m/s^2 * 2.09 m - 2 * 0.400 * (m * g * cos(29.59 degrees)/m) * 4.00 m)

v = 3.19 m/s

Therefore, the speed of the block when it reaches the bottom of the board is 3.19 m/s.

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