Answer:
the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment.
Explanation:
Research has shown that this type of interview is the most effective in predicting later job
performance.
Answer:
Situational Interview
Explanation:
A situational interview is about as close to the real job as it gets. During this type of interview, candidates may be presented with a visual or audio simulation of a scenario and asked to respond to it. They are asked to analyze a problem and profer suggestions on how they would handle it.
If the candidate has solved similar problems in the past, it will come to the fore.
If they haven't then the best outcome is that it will tell the interviewers how well the candidate is able to solve similar problems.
An example of a Situational Interview question is this:
How would you handle an angry customer who for no justifiable reason has decided to create a problematic scene to disrupt the business?
Because Situational Interviews are about behavioral responses (present, past, and future), they are powerful tools in determining or predicting future job performance. An interviewing technique that is developed using this methodology is called the S.T.A.R.
This is an acronym for Situation, Task, Action, Result.
Situation: the candidate is asked to present a challenging situation that occurred recently. This tests what the candidate sees as a challenging situation.
Task: The candidate based on the situation is asked to identify what they need to do to remedy the problem. This tells the interviewer(s) whether or not the candidate can think up a solution for the problem.
Action: Here they define the actual steps taken to resolve the problem
Result: The candidate against the above is required to give the result gotten
Action and Result tell the interviewer the quality of the candidate's ability to follow through and achieve the intended results. This also judges the quality of execution in terms of cost and time. The candidate with the lowest cost and time and the highest quality of outcome is considered the best.
Cheers
The Earth's magnetic field is modeled as that of a bar magnet with the geographic poles being Magnetic poles of the bar magnet, Based on our definitions of Magnetic Poles, if you were to go to the Earth's Geographic North Pole, you would be at a Magnetic _______________ of the bar magnet.
Answer:
South pole
Explanation:
In a bar magnet, field lines go from the North Pole to the South Pole (outside the magnet).
As the earth magnetic field lines go from South Pole (geographic) to the North one, this means that the North pole (geographic) really behaves as a South Pole (magnetic).
One airplane is approaching an airport from the north at 181 kn/hr. A second airplane approaches from the east at 278 km/hr. Find the rate at which the distance between the planes changes when the southbound plane is 30 km away from the airport and the westbound plane is 15 km from airport.
Answer:
The value is [tex] \frac{dR}{dt} = -286.2 \ km/hr [/tex]
Explanation:
From the question we are told that
The speed of the airplane from the north is [tex]\frac{dN}{dt} = -181 \ km /hr[/tex]
The negative sign is because the direction is towards the south
The speed of the airplane from the east is [tex]\frac{dE}{dt} = -278 \ km/hr[/tex]
The negative sign is because the direction is towards the west
The distance of the southbound plane from the airport is [tex]N = 30 \ km[/tex]
The distance of the westbound plane is [tex]E = 15 \ km[/tex]
Generally the distance between the plane is mathematically represented using Pythagoras theorem as
[tex]R^2 = N^2 + E^2[/tex]
Next differentiate implicitly this equation to obtain the rate at which the distance between the planes changes
So
[tex]2R\frac{dR}{dt} = 2N \frac{dN}{dt} + 2E\frac{dE}{dt}[/tex]
Here
[tex]R = \sqrt{N^2 + E^2}[/tex]
=> [tex]R = \sqrt{30^2 + 15^2}[/tex]
=> [tex]R = \sqrt{30^2 + 15^2}[/tex]
=> [tex]R =33.54 \ m [/tex]
[tex]2(33.54) * \frac{dR}{dt} = 2( 30)*(-181) + 2*15*(-278)[/tex]
=> [tex] 67.08 * \frac{dR}{dt} = -19200[/tex]
=> [tex] \frac{dR}{dt} = -286.2 \ km/hr [/tex]
The rate of change of the distance between the planes is 286.23 km/hr.
The given parameters;
speed of the airplane from North, dn/dt = 181 Km/hspeed of the airplane from the East, de/dt = 278 km/hnorth distance, n = 30 kmeast distance, e= 15 kmThe distance between the two planes is calculated by applying Pythagoras theorem as shown below;
[tex]d^2 = n^2 + e^2\\\\d = \sqrt{n^2 + e^2} \\\\d = \sqrt{30^2 + 15^2} \\\\d = 33.54 \ km[/tex]
The rate of change of the distance between the planes is calculated as follows;
[tex]d^2 = e^2 + n^2\\\\2\frac{dd}{dt} = 2e\frac{de}{dt} + 2n\frac{dn}{dt} \\\\d\frac{dd}{dt} = e\frac{de}{dt} + n\frac{dn}{dt}\\\\(33.54) \frac{dd}{dt} = (15)(278) \ + (30)(181)\\\\(33.54) \frac{dd}{dt} = 9600\\\\\frac{dd}{dt} = \frac{9600}{33.54} \\\\\frac{dd}{dt} = 286.23 \ km/hr[/tex]
Thus, the rate of change of the distance between the planes is 286.23 km/hr.
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when hydrogen shares electrons with oxygen the outermost shell of the hydrogen atoms are full with how many electrons? and oxygens valence shell is full with how many electrons? because the valence shells of these atoms are full,the atoms are stable.
Answer:
2 and 8
Explanation:
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Silly Goose falls 1.0 m to the floor. How long does the fall take
Answer:You need to give more explanation sorry
Explanation:
Answer:
4.20 seconds
Explanation:
Supposing that silly goose weighs 69 pounds, we need to start on the math.
Simple maths, truly and really. 69/1=69, of course.
Therefore it will take 4.20 seconds for silly goose to hit the ground. if he is going to be a silly goose though, he can just go in the pond, instead of wasting his time.
A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east. what is the distance and displacement
Explanation:
Let east = E, and, west = opposite to east = - E.
Here, displacement:
=> 2m east + 4m west + 1m east
=> 2E + 4(-E) + 1E
=> 2E - 4E + 1E
=> - 1E
=> 1(-E)
=> 1m west
And, distance,
=> 2m + 4m + 1m = 7m
The distance of a person is 7 m and the displacement of the person is 1m west.
To find the distance and displacement, the given values are,
A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east.
What is the distance and the displacement?Displacement:
The displacement is shortest distance between initial and final position or we can say it is the straight line distance between initial and final position.If object moves in a straight line path without any turn then the path length and the displacement is always same.Distance:
The distance is the total path length of the object while it will move from initial to final position.If the object move on curved path then displacement is smaller than the distance moved by the object.Let us consider East = E and west = opposite to east = - E.
Calculating the displacement:
= 2m east + 4m west + 1m east
= 2E + 4(-E) + 1E
= 2E - 4E + 1E
= - 1E
= 1(-E)
= 1m west.
The displacement is 1m west.
Now calculating the distance,
= 2m + 4m + 1m
= 7m
The distance is 7m.
Thus, the displacement and the distance is found as 1 m west and 7m.
Learn more about distance and displacement,
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what is the meaning of the word physics
Answer:
the scientific study of natural forces such as light, sound, heat, electricity, pressure, etc.
Explanation:
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A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it going after that acceleration?
Answer:68.15m/s
Explanation:
Given:
v₁=15m/s
a=6.5m/s²
v₁=?
x=340m
Formula:
v₁²=v₁²+2a (x)
Set up:
=[tex]\sqrt{15m/s} ^{2} +2(6.5m/s^2)(340m)[/tex]
Solution:68.15m/s
A soccer ball accelerates from rest and rolls 6.5m down a hill in 3.1 s. It then bumps into a tree. What is the speed of the ball just before it hits the tree.
Answer:
2.096m/s
Explanation:
The speed of this soccer ball can be calculated using the formula;
Speed = distance/time
According to this question, the distance of the ball before it hits the tree is 6.5m, the time it takes is 3.1s, hence;
Speed = 6.5/3.1
Speed of the ball = 2.096m/s
Therefore, the speed of the ball before hitting the tree is 2.096m/s
How much work is done lifting a 5 kg ball from a height of 2 m to a height of 6 m? (Use 10 m/s2 for the acceleration of gravity.)
A) 100 J B) 200 J C) 300 J D) 400 J
Answer:
B) 200 [J]
Explanation:
In order to solve this problem we must remember the definition of work which tells us that it is equal to the product of force by a distance, in this case, the force is the weight of the ball. The distance traveled is 4 [m] since 6-2 = 4[m]
F = m*g
where:
m = mass = 5 [kg]
g = gravity acceleration = 10 [m/s^2]
F = 5*10 = 50 [N]
w = F*d
where:
F = force = 50 [N]
d = 4 [m]
w = 50*4 = 200 [J]
A
6. All other changeable factors that must
be kept the same to ensure a fair test
(what you keep the same).
Answer:
a constant variable?
Explanation:
A constant variable is any aspect of an experiment that a researcher intentionally keeps unchanged throughout an experiment.
Experiments are always testing for measurable change, which is the dependent variable. You can also think of a dependent variable as the result obtained from an experiment. It is dependent on the change that occurs
Compare and contrast the CONFLICT (choose one) in the short story you read with the elements appearing in The Watsons Go to Birmingham—1963. Explain how they are similar or different in a few sentences.
Answer:
they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta
Explanation:
use in your own words teachers know when your not trust me.
Please provide explanation!!!
Thank you.
Answer:
(a) 102 cm/s
(b) 0.490 cm²
Explanation:
(a) Use Bernoulli equation.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
0 + ½ ρ v₁² + ρgh₁ = 0 + ½ ρ v₂² + 0
½ ρ v₁² + ρgh₁ = ½ ρ v₂²
½ v₁² + gh₁ = ½ v₂²
½ (25.0 cm/s)² + (980 cm/s²) (5.00 cm) = ½ v²
v = 102 cm/s
(b) The flow rate is constant.
v₁ A₁ = v₂ A₂
(25.0 cm/s) (2.00 cm²) = (102 cm/s) A
A = 0.490 cm²
block of mass m sits at rest on a rough inclined ramp that makes an angle with the horizontal. What must be true about normal force F on the block due to the ramp
Answer:
Explanation:
For a body on a ramp with mass m, the forces acting on the body along the vertical component are the weight and the normal reaction.
The weight of the body acts in the negative y direction while the normal reaction acts in the positive y direction
Taking the sum of forces along the y component
Sum Fy = -W+R = ma
Since acceleration is zero
-W+R = m(0)
-W+R = 0
-W = -R
W = R
Hence the Normal reaction force acting on the on the body is equal to normal force
A chef places an open sack of flour on a kitchen scale. The scale reading of
40 N indicates that the scale is exerting an upward force of 40 N on the sack. The magnitude of this force equals the magnitude of the force of Earth’s gravitational attraction on the sack. The chef then exerts an upward force of
10 N on the bag and the scale reading falls to 30 N.Draw a free-body diagram of the latter situation.
Answer:
Explanation:
Given
Initial reading on scale =40 N
So, we can conclude that weight of the sack is 40 N
After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)
This net downward force is the resultant of earth graviational pull and the applied upward force.
So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.
This situation can be diagramatically represented by figure given below
Answer:
40N
Explanation:
trust
What is the key for a successful relationship? and Why?
Answer:
communication, if you don't talk you'll never know what's going on.
Explanation:
The boys are finally old enough to compete in the box car derby race at the local fair. They have been working on their cars since the conclusion of the race last year. One boy's car raced down the track and placed 2nd in his race. However, the other boy's car started well but half-way through the race a wheel came off and his car came to a complete stop. The boy was very disappointed and the other boy felt horrible for his friend. Which of the following graphs best represents the motion of boy's car that stopped?