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dangina [55]
3 years ago
15

In a weekly project update meeting, Liza asks the following questions of one of her employees: "Why were you late meeting your l

ast deadline? Were there external factors that delayed your work? Did other coworkers get their part of the assignment to you on time? Do you need more help from me?" What type of questions are these?
a. closed
b. rapport-building
c. solution-oriented
d. probing
e. funnel
Physics
1 answer:
Aleonysh [2.5K]3 years ago
7 0

Answer: This type of questions are called probing questions. The correct option is D.

Explanation:

Probing questions are the type of questions that are asked to investigate an ongoing event. It helps the investigator to know more about what is happening and how to obtain conclusive decisions through the personal opinions of the respondent . For example from the question, Liza wanted to know more about the project updates which was held in the weekly meetings. She asked her employees questions like:

- Why were you late meeting your last deadline?

-Were there external factors that delayed your work?

-Did other coworkers get their part of the assignment to you on time?

- Do you need more help from me?".

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An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 ​7 ​​ m/s over
stich3 [128]

Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :  

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron  

• a is the acceleration of the electron  

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.  

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m  

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

          a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

  = 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m  

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .  

xf =xi+vi*t+1/2*a*t

xf =xi+vi*t+1/2*a*t

  t=√2(xf-xi)/a

 t=3.765×10^-10 s

now we can use the power P Eq.  

 P=W/Δt => ΔK/Δt  

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

6 0
3 years ago
What elements are good conductors of electricity?
kramer
In solids: All metals are good conductors of electricity as they contain free moving electrons. Non-metals doesn't conduct , but we consider Graphite the only non-metal that can conduct electricity for the presence of free moving electrons.

In Liquids ; Ionic compunds contains free moving ions , so they conduct electricity as well .

6 0
3 years ago
Two horizontal forces,225N and 165N,are exerted on s canoe in the same direction. Find the net horizontal foce
Readme [11.4K]

Since they are in the same direction, you would add them together. Let’s also assume said direction is positive. 225 N + 165 N = 390 N

5 0
3 years ago
An object is located 70 cm from a concave mirror with a focal length of 15 cm. What is the image
Stels [109]

(a) The distance of the image formed by the concave mirror is 19.1 cm.

(b) The image formed is diminished and real.

<h3>Image distance </h3>

The distance of the image formed by the concave mirror is calculated as follows;

1/f = 1/v + 1/u

1/v = 1/f - 1/u

1/v = 1/15 - 1/70

1/v = 0.05238

v = 1/0.05238

v = 19.1 cm

The image distance is smaller than object distance, thus the image formed is diminished and real.

Learn more about concave mirror here: brainly.com/question/13164847

#SPJ1

5 0
1 year ago
Anyone know which wire matches the other one? for all 4. Need help! Thanks :-)
Fittoniya [83]

red goes to red, black goes to white, yellow goes to green, blue goes to blue.

3 0
3 years ago
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