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3 years ago

Which of the following is a common human relations mistake?

2 answers:

8 0

The correct answer should be A. Misjudging others' abilities

Taking responsibility when you make a mistake and listening to your supervisor are a must in the adult world while saying thank you is not a mistake but rather a trait. Misjudging others' abilities can sometimes even be offensive and is common as a mistake.

Deffense [45]3 years ago

3 0

Misjudging others' abilities is <span>a common human relations mistake.

Choices B, C, and D are not mistakes.
They're desirable habits that should be developed intentionally.
</span>

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I NEED HELP PLEASE, THANKS! :)

makkiz [27]

Answer:

They move outwards.

They don't intersect each other at any point.

They show the electric field.

Explanation:

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3 years ago

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A batter swings and hits a pitched baseball far over the left field wall. at the moment the baseball contacts the bat, which obj

Lyrx [107]

The baseball bat not the baseball

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3 years ago

The volume of a cylindrical tin can with a top and a bottom is to be 16π cubic inches. If a minimum amount of tin is to be used

sergiy2304 [10]

Answer:

h = 4 in

Explanation:

GIVEN DATA:

volume of tin $= 16 \pi$

we know that

volume of cylinder is $v = \pi r^2 h$

so,

$16 \pi = \pi r^2 h$

$16 = r^2 h$

$r = \sqrt{\frac{16}{h}}$

construct formula for surface area

$S = 2\pi r^2 + 2\pi rh$

$S = \frac{2v}{h} + 2 \sqrt{v \pi h}$

minimize the function wrt h

$S' = \frac{2v}{h^2} + \sqrt{\frac{v \pi}{h} = 0$

solving for h we have

$h = [\frac{4 v}{\pi}]^{1/3}$

we kow $v = 16 \pi$ so

h = 4 in

8 0

3 years ago

during the course of songraphic exam, you notice lateral splaying of echoes in the far field. what can you do to improve the ima

horrorfan [7]

Answer:

lateral splaying of echoes in the far field can be improved by Increasing the maximum number of transmit focal zones and optimize their location.

8 0

3 years ago

A top-fuel dragster starts from rest and has a constant acceleration of 42.0 m/s2. What are (a) the final velocity of the dragst

disa [49]

Answer:

a) Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) The displacement of the dragster at the end of 1.8 s = 68.04 m

d) The displacement of the dragster at the end of 3.6 s = 272.16 m

Explanation:

a) We have equation of motion v = u + at

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 1.8 s

Substituting

v = u + at

v = 0 + 42 x 1.8 = 75.6 m/s

Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) We have equation of motion v = u + at

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 3.6 s

Substituting

v = u + at

v = 0 + 42 x 3.6 = 75.6 m/s

Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) We have equation of motion s= ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 1.8 s

Substituting

s= ut + 0.5 at²

s = 0 x 1.8 + 0.5 x 42 x 1.8²

s = 68.04 m

The displacement of the dragster at the end of 1.8 s = 68.04 m

d) We have equation of motion s= ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration , a = 42 m/s²

Time = 3.6 s

Substituting

s= ut + 0.5 at²

s = 0 x 3.6 + 0.5 x 42 x 3.6²

s = 272.16 m

The displacement of the dragster at the end of 3.6 s = 272.16 m

3 0

3 years ago