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Lapatulllka [165]

3 years ago

9

When it comes to making a good impression in a work setting, it does not apply to an initial contact, since both people are meet

ing for an express purpose. you have a 5 to 10 minute grace period before people form their impression of you. your handshake and facial expression dominate all other elements of your appearance. you have only seconds in which a person will accept or reject an employee or firm.

1 answer:

Art [367]3 years ago

4 0

Answer:

you have only seconds in which a person will accept or reject an employee or firm

Explanation:

First impression matters that's why when looking for employment with an organisation, lack of a tie for men may lead to automatic rejection. You have to be smart both intellectually and physically. Therefore, it means that you have only seconds in which a person will accept or reject an employee or firm.

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Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large

salantis [7]

Answer:

$Q_{cv} = -1007.86kJ$

Explanation:

Our values are,

State 1

$V=3m^3\\P_1=1bar\\T_1 = 295K$

We know moreover for the tables A-15 that

$u_1 = 210.49kJ/kg\\h_i = 295.17kJkg$

State 2

$P_2 =6bar\\T_2 = 296K\\T_f = 320K$

For tables we know at T=320K

$u_2 = 228.42kJ/kg$

We need to use the ideal gas equation to estimate the mass, so

$m_1 = \frac{p_1V}{RT_1}$

$m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}$

$m_1 = 3.54kg$

Using now for the final mass:

$m_2 = \frac{p_2V}{RT_2}$

$m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}$

$m_2 = 19.59kg$

We only need to apply a energy balance equation:

$Q_{cv}+m_ih_i = m_2u_2-m_1u_1$

$Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i$

$Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)$

$Q_{cv} = -1007.86kJ$

The negative value indidicates heat ransfer from the system

7 0

3 years ago

The acceleration of a particle is given by a = 2t − 10, where a is in meters per second squared and t is in seconds. Determine t

tensa zangetsu [6.8K]

Answer

given,

a = 2 t - 10

velocity function

we know,

$\dfrac{dv}{dt}=a$

$\dfrac{dv}{dt}=(2t-10)$

integrating both side

$\int dv =\int (2t -10) dt$

v = t² - 10 t + C

at t = 0 v = 3

so, 3 = 0 - 0 + C

C = 3

Velocity function is equal to v = t² - 10 t + 3

Again we know,

$\dfrac{dx}{dt}=v$

$\dfrac{dx}{dt}=(t^2-10t + 3)$

integrating both side

$\int dx =\int (t^2-10t + 3)dt$

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t + C$

now, at t= 0 s = -4

$-4 = \dfrac{0^3}{3}- 10\dfrac{0^2}{2} + 0 + C$

C = -4

So,

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

Position function is equal to $x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

8 0

3 years ago

How many people made machines

olganol [36]

Answer:

The total number of people whom have made machines is not a recorded figure? Need to be more specific :/

Explanation:

Sorry not very helpful, your question is REALLY broad

3 0

2 years ago

Read 2 more answers

How many broadcast or vlan is in this switchs and router ? and why?

Mnenie [13.5K]

Answer:The move from hubs (shared networks) to switched networks was a big improvement. Control over collisions, increased throughput, and the additional features offered by switches all provide ample incentive to upgrade infrastructure. But Layer 2 switched topologies are not without their difficulties. Extensive flat topologies can create congested broadcast domains and can involve compromises with security, redundancy, and load balancing. These issues can be mitigated through the use of virtual local area networks, or VLANs. This chapter provides the structure and operation of VLANs as standardized in IEEE 802.1Q. This discussion will include trunking methods used for interconnecting devices on VLANs.

Problem: Big Broadcast Domains

With any single shared media LAN segment, transmissions propagate through the entire segment. As traffic activity increases, more collisions occur and transmitting nodes must back off and wait before attempting the transmission again. While the collision is cleared, other nodes must also wait, further increasing congestion on the LAN segment.

The left side of Figure 4-1 depicts a small network in which PC 2 and PC 4 attempt transmissions at the same time. The frames propagate away from the computers, eventually colliding with each other somewhere in between the two nodes as shown on the right. The increased voltage and power then propagate away from the scene of the collision. Note that the collision does not continue past the switches on either end. These are the boundaries of the collision domain. This is one of the primary reasons for switches replacing hubs. Hubs (and access points) simply do not scale well as network traffic increases.

7 0

2 years ago

Whats the thing in the picture? Its a part of a computer.

Digiron [165]

Answer:

Is it part of a computer?

Explanation:

5 0

2 years ago

Read 2 more answers