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tatyana61 [14]
3 years ago
9

Practicing good human relations skills and having a positive attitude will

Physics
2 answers:
skad [1K]3 years ago
7 0
Have a positive effect on you as a human being. Your attitude will improve, your charisma will spread, and you will a better, more well rounded person.
goldfiish [28.3K]3 years ago
4 0

If you practice good human relation skills then you will be more happy, be open, talk more, you will think more positive and a lot more. (these are just <u>some </u>examples)


Hope this will help someone! ;)

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HELP PLEASE BOYLES LAW
kogti [31]

Answer:

3 L

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 2 L

Initial pressure (P₁) = 0.75 atm

Final pressure (P₂) = 0.5 atm

Final volume (V₂) =?

Using the Boyle's law equation, the new volume (i.e final volume) of the Ne gas can be obtained as:

Initial volume (V₁) = 2 L

Initial pressure (P₁) = 0.75 atm

Final pressure (P₂) = 0.5 atm

Final volume (V₂) =?

P₁V₁ = P₂V₂

0.75 × 2 = 0.5 × V₂

1.5 = 0.5 × V₂

Divide both side by 0.5

V₂ = 1.5 / 0.5

V₂ = 3 L

Thus, the new volume of the Ne gas is 3 L

7 0
3 years ago
What is loudness of sound?
dedylja [7]
The loudness of sound refer to how loud or soft a sound seems to a listener
6 0
3 years ago
Read 2 more answers
a mass of 0.75 kg is attached to a spring and placed on a horizontal surface. the spring has a spring constant of 180 N/m, and t
Artemon [7]

Answer:

6.57 m/s

Explanation:

First use Hook's Law to determine the F the compressed spring acts on the mass. Hook's Law F=kx; F=force, k=stiffnes of spring (or spring constant), x=displacement

F=kx; F=180(.3) = 54 N

Next from Newton's second law find the acceleration of the mass.

Newton's .2nd law F=ma; a=F/m ; a=54/.75 = 72m/s²

Now use the kinematic equation for velocity (or speed)

v₂²= v₀² + 2a(x₂-x₀); v₂=final velocity; v₀=initial velocity; a=acceleration; x₂=final displacement; x₀=initial displacment.

v₀=0, since the mass is at rest before we release it

a=72 m/s² (from above)

x₀=0 as the start position already compressed

x₂=0.3m (this puts the spring back to it's natural length)

v₂²= 0 + 2(72)(0.3) = 43.2 m²/s²

v₂=\sqrt{43.2)\\ = 6.57 m/s

5 0
3 years ago
A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back
elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity=9.8 m/s^2

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates=\mu mg=0.202\times 9.8\times 35

Force by which box accelerates=62.286 N

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

ma=\mu mg

a=\mu g=9.8\times 0.202=1.9796 m/s^2

Hence, the truck can have maximum acceleration before the box slide=1.9796 m/s^2

3 0
3 years ago
How is it that substances can have the same chemical formulas but make different types of matter?
damaskus [11]
There are several different formulars in the world
8 0
3 years ago
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