All categories

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>

You might be interested in

If there's air in a ball and it bounces how come a room doesn't bounce

finlep [7]

Answer:

Funny answer but works!

Explanation:

Not enough air. Go outside, crack open the window and blow air into the room like you would a balloon then close the window quickly. If it doesn't start bouncing, try giving the house a push. Good luck.

4 0

3 years ago

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3 km/h.

Amanda [17]

Answer:

$F_{1}=\frac{1}{5}F_{2}$ or $F_{2}=5F_{1}$

In other words, $F_{1}$ is one fifth of $F_{2}$ or $F_{2}$ is five times as big as $F_{1}$

Explanation:

In order to solve this problem we must start by sketching the situation (refer to the attached picture).

When the ship is pulled only by force 1, it will change its speed by 3km/hr in 10 seconds. So in order to use these values we need to either turn the km/hr in km/s or turn the seconds to hours. Let's turn the seconds to hours:

$10s*\frac{1hr}{3600s}=\frac{1}{360} hr$

so we can now use the acceleration formula to find the acceleration of the boat so we get:

$a=\frac{\Delta v}{\Delta t}$

which will give us an accceleration of:

$a=\frac{3km/hr}{\frac{1}{360}hr}=1080km/hr^{2}$

once we got the acceleration we can for sure say taht:

$F_{1}=ma=m*1080\frac{km}{hr^{2}}$

Now, if we take a look at the second drawing we can see that the resultant force applied to the boat is found by adding the two forces, force one and force two, so we get:

$F_{1}+F_{2}=ma$

in this case the acceleration changes because the change in velocity is of 18km/hr in the same 10 seconds, so we get that:

$a=\frac{\Delta v}{\Delta t}$

$a=\frac{18km/hr}{\frac{1}{360}hr}=6480km/hr^{2}$

so we can say that:

$F_{1}+F_{2}=m*6480km/hr^{2}$

we can substitute the first force into this equation so we get:

$m*1080km/hr^{2}+F_{2}=m*6480km/hr^{2}$

and solve for the second force, so we get:

$F_{2}=m*6480km/hr^{2}-m*1080km/hr^{2}$

which yields:

$F_{2}=m*5400km/hr^{2}$

Now we can compare theh two forces, force 1 and force 2 by dividing them:

$\frac{F_{1}}{F_{2}}=\frac{m*1080km/hr^{2}}{m*5400km/hr^{2}}$

which yields:

$\frac{F_{1}}{F_{2}}=\frac{1}{5}$

when solving for the first force we get:

$F_{1}=\frac{1}{5}F_{2}$

which tells us that the second force is one fifth of the first force.

and when solving for the second force we get that:

$F_{2}=5F_{1}$

which means that the second force is 5 times as big as the first force.

8 0

3 years ago

How much potential energy foes a 5kg mass have when its 2 meters above the ground?(Hint :PE=m*g*h)

frez [133]

Answer:

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

PE = 5 × 9.8 × 2

We have the final answer as

Hope this helps you

5 0

2 years ago

Scientists think that a few hundred million years or so after the Big Bang, space cooled off enough for matter to form. This mat

Eva8 [605]

This matter spread out around theuniverse with no structure. But each little bit of matter exerted the force of gravity on each other little bit. Gravity is the only type of force connected with <span>the Big Bang as it is a basic force that was to create our world in th way we can see and discover it today.</span>

6 0

3 years ago

Read 2 more answers

LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Svetradugi [14.3K]

Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$

8 0

3 years ago