How does clothing and body language affect a job interview?

Answers

Answer 1

Answer:

Answer:

please give me brainlist and follow

Explanation:

Whilst first impressions can be important, your body language during the interview can make or break your overall performance. With experts saying that between 75-90% of communication is non-verbal, it is important to think about what your body is saying about you during an interview.

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How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole
A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge 2e) which is also traveling perpendicular to the same field. The ratio of their speeds, vproton/valpha is:

What is the speed of an ocean wave if it’s wavelength is 5.0 m and it’s frequency is 3/s?

Answers

Answer:

15 m/s

Explanation:

We know that $v = f * d$ where f = frequency & d = wavelength .

So here.

Wavelength = 5 m

Frequency = 3 s⁻¹

Hence Speed = 5 * 3 = 15 m/s

A screw can be considered a type of

Answers

Fastener because a fastener is something that connects to objects and usually can come apart but can also be permanent

An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Determine the speed at this time and the maximum height at which it reaches.

Answers

Answer:43.34 m

Explanation:

Given

acceleration(a) $=2 m/s^2$

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

$v=0+2* 6=12 m/s$

After this object will start moving under gravity

height reached in first 6 s

$s=ut+(at^2)/(2)$

$s=0+(2* 6^2)/(2)$

s=36 m

After fuel run out distance traveled in upward direction is

$v^2-u^2=2as_0$

here v=0

u=12 m/s

$a=9.8 m/s^2$

$0-12^2=2(-9.8)(s)$

$s_0=(144)/(2* 9.8)=7.34 m$

$s+s_0=36+7.34=43.34 m$

An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and a quality of 0.25. An electric heater inside the tank is turned on to heat this H2O until the pressure increases to 200 kPa. Please determine the change in total entropy of water during this process. Hint: See if you can find the electrical work consumed during this process.

Answers

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

$v_1 = vf + x(vg - vf)$

$= 0.001053 + 0.25(1.1594 - 0.001053)$

$v_1 = 0.20964 m^3/kg$

$s_1 = Sf + x(Sfg)$

$= 1.4337 + 0.25 * 5.7894 = 2.88 kJ/kg K$

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume so $v_1 - v_2 = 0.29064 m^3/kg$

$v_2 = v_1 = vf + x(vg - vf)$

$=0.29064 = x_2(0.88578 - 0.001061)$

$x_2 = 0.327$

$s_2 = 1.5302 + 0.32 * 5.5968 = 3.36035 kJ/kg K$

Change in entropy $\Delta s = m(s_2 - s_1)$

=3( 3.36035 - 2.88) = 1.44 kJ/kg

g Adjacent rows in the first part of the experiment are found to have potentials of 3.66 V and 4.22 V. If the distance between rows is found to be 0.4 cm, what is the magnitude of the electric field at the location between the rows

Answers

Answer:

$E=140V/m$

Explanation:

If the electric field is uniform, the electric field between two points at potentials $V_1$ and $V_2$ which are separated by a distance d will be given by the formula:

$E=(\Delta V)/(d)$

So in our case, we have $E=(4.22V-3.66V)/(0.004m)=140V/m$

The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of the strings is then increased by 1.0%. If they are now struck, what is the beat frequency between the fundamentals of the two strings? SOLUTION Conceptualize As the tension in one of the strings is changed, its fundamental frequency changes. Therefore, when both strings are played, they will have different frequencies and beats be heard. Categorize We must combine our understanding of the waves model for strings with our new knowledge of beats.