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torisob [31]
2 years ago
13

List the three components of effective communication and give an example of each

Physics
1 answer:
Marrrta [24]2 years ago
5 0

Answer:

The three components of effective communication are:

  1. Sending Communication
  2. Receiving Communication
  3. Feedback

Explanation:

<h3>1) Sending Communication</h3>

One should think about what he is going to say, structure his message and focus clearly on a the purpose of sending a message. For example if an employer wants to communicate an employee to improve his performance, he should focus hid communication on the results rather than on his failures.

<h3 /><h3>2) Receiving Communication</h3>

The person on the other end of the communication should be a good listener. For example in the example given, even if the employer effectively communicates his side of the communication, but the employee is not paying his full attention to what his boss is saying, he wouldn't be able to make much of a difference.

<h3>3) Feedback</h3>

After sending and receiving communication, feedback should be given to complete the communication effectively. For example the employer may tell his boss about what he understood from his boss's speech, or what will he do to improve his performance.

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What is the velocity of a beam of electrons that goes undeflected when passing through perpendicular electric and magnetic field
goldenfox [79]
F = qE + qV × B
where force F, electric field E, velocity V, and magnetic field B are vectors and the × operator is the vector cross product. If the electron remains undeflected, then F = 0 and E = -V × B
which means that |V| = |E| / |B| and the vectors must have the proper geometrical relationship. I therefore get
|V| = 8.8e3 / 3.7e-3
= 2.4e6 m/sec
Acceleration a = V²/r, where r is the radius of curvature.
a = F/m, where m is the mass of an electron,
so qVB/m = V²/r.
Solving for r yields
r = mV/qB
= 9.11e-31 kg * 2.37e6 m/sec / (1.60e-19 coul * 3.7e-3 T)
= 3.65e-3 m
6 0
2 years ago
A radio technician measures the frequency of an AM radio transmitter. The frequency is . What is the frequency in megahertz? Wri
sukhopar [10]

Complete Question

A radio technician measures the frequency of an AM radio transmitter. The frequency is 14603 kHz . What is the frequency in megahertz? Write your answer as a decimal.

Answer:

The value is  x =  14.6 \  MHz

Explanation:

From the question we are told that

  The  frequency is  f =  14603  \  kHz = 14603 *1000 = 14603000 \ Hz

Generally  

       1 Hz \to  1.0 *10^{-6} \  MHz

       14603000 \ Hz  \to x MHz

=>   x =  \frac{14603000 *  1.0*10^{-6}}{1 }

=>    x =  14.6 \  MHz

8 0
3 years ago
If there were no external forces acting on the two pucks, their complex motion could be described as the combination of the unif
mote1985 [20]

The absence of external forces will make the pucks move in the form of a uniform circular motion.

<h3>What is a circular motion?</h3>

It should be noted that a circular motion simply means the movement of an object along the circumference of the circle.

In this case, the absence of external forces will make the pucks move in the form of a uniform circular motion.

If the friction is absent, the pucks will continue to move on the same path due to the first law of Newton and the law of conversation of energy. In this case,the results will match the predictions until there's loss in energy.

Learn more about circular motion on:

brainly.com/question/106339

3 0
2 years ago
A rectangular coil of dimensions 5.40cm x 8.50cm consists of25 turns of wire. The coil carries a current of 15.0 mA.
Kazeer [188]

Answer:

(a) Magnetic moment will be 17.212\times 10^{-4}A-m^2

(b) Torque will be 6.024\times 10^{-4}N-m

Explanation:

We have given dimension of the rectangular 5.4 cm × 8.5 cm

So area of the rectangular coil A=5.4\times 8.5=45.9cm^2=45.9\times 10^{-4}m^2

Current is given as i=15mA=15\times 10^{-3}A

Number of turns N = 25

(A) We know that magnetic moment is given by magnetic\ moment=NiA=25\times 45.9\times 10^{-4}\times 15\times 10^{-3}=17.212\times 10^{-4}A-m^2

(b) Magnetic field is given as B = 0.350 T

We know that torque is given by \tau =BINA=0.350\times 15\times 10^{-3}\times 25\times 45.9\times 10^{-4}=6.024\times 10^{-4}N-m

4 0
2 years ago
PLEASE HELP!! ITS DUE AT 11:59!!
DENIUS [597]

Answer:

14.1 po

sana makatulong

3 0
3 years ago
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