How does clothing and body language affect a job interview?


Answer 1


please give me brainlist and follow


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.

Related Questions

A hot-air balloon is descending at a rate of 2.3 m>s when a pas- senger drops a camera. If the camera is 41 m above the ground when it is dropped, (a) how much time does it take for the cam- era to reach the ground, and (b) what is its velocity just before it lands? Let upward be the positive direction for this problem.
A 0.26-kg stone is held 1.2 m above the top edge of a water well and then dropped into it. The well has a depth of 4.5 m. (a) Relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stone−Earth system before the stone is released?
Neptunium. In the fall of 2002, scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237 is about 60 kg. The critical mass of a fissionable material is the minimum amount that must be brought together to start a nuclear chain reaction. Neptunium-237 has a density of 19.5 g/cm3. What would be the radius of a sphere of this material that has a critical mass?
Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits
Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus the bullet rises to a height of 0.13 m along a circular arc with a 0.23 m radius. Assume: The entire track is frictionless. A bullet with a m1 = 30 g mass is fired horizontally into a block of wood with m2 = 4.2 kg mass. The acceleration of gravity is 9.8 m/s2 . Calculate the total energy of the composite system at any time after the collision. Answer in units of J. Taking the same parameter values as those in Part 1, determine the initial velocity of the bullet. Answer in units of m/s.

The direct sunlight at Earth's surface is about 1050 W/m2 . Compute mass lost by Sun in a thousand years as a fraction of Earth Mass? The lost Mass of sun/in a million yrs = ……?…..% of Earth Mass.




Energy falling on 1 m² surface of earth per second = 1050

Energy in one million years on 1 m²

= 1050 x 60 x 60 x 24 x 365 x 10⁶ = 3.311 x 10¹⁶ J

In order to calculate total energy coming out of the surface of the sun , we shall have to sum up this energy for the while spherical surface of imaginary sphere having radius equal to distance between sun and earth.

Area of this surface = 4π R² = 4 X 3.14 X (149.6 X 10⁹ )²

= 2.8 X 10²³ m²

So total energy coming out of the sun = 2.8 x 10²³ x 3.311 x 10¹⁶

= 9.271 x 10³⁹ J

From the formula

E = mc² { energy mass equivalence formula }

m = E / c² = (9.271 *10^(39))/(9 * 10^(16))

1.03 x 10²³ kg

mass of earth = 5.972 x 10²⁴

Answer in percentage of mass of earth

= (1.03*10^23)/(5.972*10^(24))*100

= 1.72  %

When the sun and moon pull at right angles to the earth wat kinda tide can yu expect



A neap tide. Hope this helps


Two narrow slits separated by 0.30 mm are illuminated with light of wavelength 496 nm. (a) How far are the first three bright fringes from the center of the pattern if observed on the screen 130 cm distant? (b) How far are the first three dark fringes from the center of the pattern?





d = separation of the slits = 0.30 mm = 0.30 x 10⁻³ m

λ = wavelength of the light = 496 nm = 496 x 10⁻⁹ m

n = order of the bright fringe

D = screen distance = 130 cm = 1.30 m

x_(n) = Position of nth bright fringe

Position of nth bright fringe is given as

x_(n) =( n D \lambda )/(d)

For n = 1

x_(1) =( (1) (1.30)(496* 10^(-9)))/(0.30* 10^(-3))

x_(1) = 2.15* 10^(-3)m

For n = 2

x_(2) =( (2) (1.30)(496* 10^(-9)))/(0.30* 10^(-3))

x_(2) = 4.30* 10^(-3)m

For n = 3

x_(2) =( (2) (1.30)(496* 10^(-9)))/(0.30* 10^(-3))

x_(2) = 6.45* 10^(-3)m


Position of nth dark fringe is given as

y_(n) =( (2n+1) D \lambda )/(2d)

For n = 1

y_(1) =( (2(1)+1) (1.30)(496* 10^(-9)))/(2(0.30* 10^(-3)))

y_(1) = 3.22* 10^(-3)m

For n = 2

y_(2) =( (2(2)+1) (1.30)(496* 10^(-9)))/(2(0.30* 10^(-3)))

y_(2) = 5.4* 10^(-3)m

For n = 3

y_(3) =( (2(3)+1) (1.30)(496* 10^(-9)))/(2(0.30* 10^(-3)))

x_(3) = 7.5* 10^(-3)m

The wind-chill index is modeled by the function W = 13.12 + 0.6215T − 11.37v0.16 + 0.3965Tv0.16 where T is the temperature (°C) and v is the wind speed (km/h). When T = 12°C and v = 18 km/h, by how much would you expect the apparent temperature W to drop if the actual temperature decreases by 1°C? (Round your answers to two decimal places.)


This question involves the concepts of derivative, apparent temperature, actual temperature,and wind speed.

The drop in apparent temperature will be "1.25°C".

The apparent temperature (W) is given in terms of actual temperature (T) and wind speed (v) is given by the following function:

W = 13.12 + 0.6215\ T-11.37\ v^(0.16)+0.3965\ Tv^(0.16)

Taking the derivative with respect to actual temperature, we get:

(dW)/(dT)=0.6215+0.3965\ v^(0.16)\n\n


dW = drop in apparent temperatures = ?

dT = drop in actual temperature = - 1°C

v = wind speed = 18 km/h



dW = - 1.25°C

Learn more about derivatives here:




Given that

w=13.12 +0.6215 T-11.37 v^(0.16)+0.3965 T v^(0.16)

Given that T= 12°C and v=19 km/h

Now to find the drop in the apparent temperature w

(dw)/(dT)=0.6215 +0.3965v^(0.16)


(\Delta w)/(\Delta T)= 0.6215 +0.3965 v^(0.16)

Now by putting the values v=19 km/hr and ΔT=1

(\Delta w)/(1)=0.6215 +0.3965* 18^(0.16)


So we can say that when temperature is decrease by 1°C then apparent temperature will decrease by 1.25°C at given velocity.

While the block hovers in place, is the density of the block (top left) or the density of the liquid (bottom center) greater?



for the body to float, the density of the body must be less than or equal to the density of the liquid.


For a block to float in a liquid, the thrust of the liquid must be greater than or equal to the weight of the block.

Weight is

        W = mg

let's use the concept of density

        ρ_body = m / V

        m = ρ_body V

        W = ρ_body V g

The thrust of the body is given by Archimedes' law

        B = ρ_liquid g V_liquid


as the body floats the submerged volume of the liquid is less than or equal to the volume of the block

       ρ_body V g = ρ_liquid g V_liquid


       ρ_body = ρ liquid Vliquido / V_body

As we can see, for the body to float, the density of the body must be less than or equal to the density of the liquid.

The main force(s) acting on the puck after receiving the kick is (are):_________.A) a downward force of gravity and an upward force exerted by the surfaceB) a downward force of gravity, and a horizontal force in the direction of motionC) a downward force of gravity, an upward force exerted by the surface, and a horizontal force in the direction of motionD) a downward force of gravityA) a downward force of gravity and an upward force exerted by the surface



the statements, the correct one is A

a downward force of gravity and an upward force exerted by the surface


When the disc is hit, a thrust force is exerted in the direction of movement, at the moment the disc moves this force loses contact and becomes zero.

When the movement is already established there are two main forces: gravity that acts downwards and the reaction force to the support of the disk called normal that acts upwards.

As it is not mentioned that there is friction, this force that opposes the movement is zero.

Analyzing the statements, the correct one is A