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valkas [14]
2 years ago
9

Are the communication skills required for a team leader in a manufacturing plant different from those of a customer service exec

utive?
What would you do to improve your communication skills?

Discuss the importance of customer service and suggest ways to ensure good customer experience.
Engineering
1 answer:
Svetllana [295]2 years ago
5 0

Answer:

Explanation: Excellent customer service isn’t just down to your frontline staff, but a customer might be won or lost there. That’s why it’s so important every employee works together to resolve issues and create memorable moments – and empower your frontline agents to be as valuable to the customer as possible. It could be the difference between the customer coming back again, or going elsewhere.

The following tips are designed to help both customer service representatives, customer service management, and operations staff to work together to make experiences that matter. Developing customer service skills is important for the whole team to thrive – and to ensure customers keep coming back.

Customer service representative tips

Practice active listening

Learn to empathize with your customers

Use positive language

Improve your technical skills

Know your products and services

Look for common ground

Communicate clearly

Be solutions-focused

Admit mistakes

Be willing to learn

Customer service team management and operations tips

Provide first-class training

Set your standards high

Have a clear escalation pathway

Align your touchpoints

Create a culture of excellence

Be smart about automation

Use tools that boost speed and efficiency

Measure and analyze customer feedback

Use closed-loop feedback

Listen, understand and take action

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Strands of materials A and B are placed under a tensile force of 10 Newtons. Material A deforms more than Material B.
nadezda [96]

Answer:

True

Explanation:

6 0
2 years ago
Read 2 more answers
Firefighters are holding a nozzle at the end of a hose while trying to extinguish a fire. The nozzle exit diameter is 8 cm, and
ivanzaharov [21]

Question

Determine the average water exit velocity

Answer:

53.05 m/s

Explanation:

Given information

Volume flow rate, Q=16 m^{3}/min

Diameter d= 8cm= 0.08 m

Assumptions

  • The flow is jet flow hence momentum-flux correction factor is unity
  • Gravitational force is not considered
  • The flow is steady, frictionless and incompressible
  • Water is discharged to the atmosphere hence pressure is ignored

We know that Q=AV and making v the subject then

V=\frac {Q}{A} where V is the exit velocity and A is area

Area, A=\frac {\pi d^{2}{4} where d is the diameter

By substitution

V=\frac {16\times 4}{\pi 0.08^{2}}=3183.098862 m/min

To convert v to m/s from m/s, we simply divide it by 60 hence

V=\frac {3183.098862  m/min}{60 s}=53.0516477 m/s\approx 53.05 m/s

3 0
2 years ago
When removing a diesel engine from a truck, Technician A says it is OK to disconnect an air con­ditioning hose, but the refriger
agasfer [191]

Answer:

Technician B is right.

Explanation:

Air conditioning refrigerant contains Freon R22 and R410a, which have been linked to environmental damages, including ozone depletion, global warming, and energy-inefficiency.  For environmentally-savvy entities and individuals, there is the modern move to a more environment-friendly refrigerant, known as R-32.   Technician A's advice to vent the refrigerant outside the shop is in bad taste.  He does not seem to be aware of the environmental footprint of such an action.  Venting gas outside, in addition to the environmental damages, is also a waste of resources, and therefore, costly.  This is why Technician B's advice should be preferred.

5 0
3 years ago
Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging wat
zlopas [31]

Answer: 10.631\times 10^3\ N/m^2

Explanation:

Given

Discharge is Q=12.5\ L

Diameter of pipe d=150\ mm

Distance between two ends of pipe L=800\ m

friction factor f=0.008

Average velocity is given by

\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s

Pressure difference is given by

\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow  \Delta P=10.631\ kPa

8 0
2 years ago
A geothermal heat pump absorbs 15 KJ/s of heat from the Earth 15 m below a house. This heat pump uses a 7.45 kJ/s compressor.
Anna007 [38]

Answer:

COP of the heat pump is 3.013

OP of the cycle is  1.124

Explanation:

W = Q₂ - Q₁

Given

a)

Q₂ = Q₁ + W

     = 15 + 7.45

     = 22.45 kw

COP = Q₂ / W = 22.45 / 7.45 = 3.013

b)

Q₂ = 15 x 1.055 = 15.825 kw

therefore,

Q₁ = Q₂ - W

Q₁ = 15.825 - 7.45 = 8.375

∴ COP = Q₁ / W = 8.375 / 7.45 = 1.124

4 0
3 years ago
Read 2 more answers
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