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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.

You might be interested in

Is the COP of a heat pump always larger than 1?

Liono4ka [1.6K]

Answer:

Yes

Explanation:

Yes it is true that COP of heat pump always greater than 1.But the COP of refrigeration can be greater or less than 1.

We know that

COP of heat pump= 1 + COP of refrigeration

It is clear that COP can not be negative .So from the above expression we can say that COP of heat pump is always greater than one.

3 0

3 years ago

The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature

Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

$P_{1}$ = 5 Mpa

$T_{1}$ = 1623°C

= 1896 K

$V_{1}$ = 0.05 $m^{3}$

Also given $\frac{V_{2}}{V_{1}} = 20$

Therefore, $V_{2}$ = 1 $m^{3}$

R = 0.27 kJ / kg-K

$C_{V}$ = 0.8 kJ / kg-K

Also given : $P_{1}V_{1}^{1.25}=C$

Therefore, $P_{1}V_{1}^{1.25}$ = $P_{2}V_{2}^{1.25}$

$5\times 0.05^{1.25}=P_{2}\times 1^{1.25}$

$P_{2}$ = 0.1182 MPa

a). Work transfer, δW = $\frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

$\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1} \right ]\times 10^{6}$

= 527200 J

= 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

= $\frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

= $\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W$

= $\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200$

= 197.7 kJ

6 0

3 years ago

What type of engineer makes sure equipment is safe and operational

zvonat [6]

Answer:

mechanical engineer is the best answer

8 0

3 years ago

A circular specimen of MgO is loaded in three-point bending. Calculate the minimum possible radius of the specimen without fract

Hitman42 [59]

Answer:

radius = 9.1 × $10^{-3}$ m

Explanation:

given data

applied load = 5560 N

flexural strength = 105 MPa

separation between the support = 45 mm

solution

we apply here minimum radius formula that is

radius = $\sqrt[3]{\frac{FL}{\sigma \pi}}$ .................1

here F is applied load and is length

put here value and we get

radius = $\sqrt[3]{\frac{5560\times 45\times 10^{-3}}{105 \times 10^6 \pi}}$

solve it we get

radius = 9.1 × $10^{-3}$ m

8 0

3 years ago

(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5

bezimeni [28]

Answer:

a.) 1.453MW/m2, b.) 2,477,933.33 BTU/hr c.) 22,733.33 BTU/hr d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, $Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\$

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, $Q = \frac{109*0.5*100}{0.0075} = 726,666.667W$

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, $Q=\frac{1*0.5*100}{0.0075} = 6,666.67W$

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, $Q=\frac{109*0.5*100}{0.015} =363,333.33W$

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0

3 years ago