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alexgriva [62]
3 years ago
14

What does the term equilibrium refer to

Physics
2 answers:
igor_vitrenko [27]3 years ago
6 0
Equilibrium<span>-the condition of a system when no observable change is taken place or the kinetic energy is equal. </span>Equilibrium means<span> to stay balanced or equal.</span>
Morgarella [4.7K]3 years ago
5 0
<span>A condition in which all influences acting cancel each other

</span>In physics equilibrium<span> results from the cancellation of forces acting on an object.</span>
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What is air masses? And 5 effect of it​
zhannawk [14.2K]

Answer:

an air mass is a volume of air defined by its temperature and water vapor content. Air masses cover many hundreds or thousands of miles, and adapt to the characteristics of the surface below them. They are classified according to latitude and their continental or maritime source regions. Colder air masses are termed polar or arctic, while warmer air masses are deemed tropical. Continental and superior air masses are dry while maritime and monsoon air masses are moist. Weather fronts separate air masses with different density (temperature and/or moisture) characteristics. Once an air mass moves away from its source region, underlying vegetation and water bodies can quickly modify its character.When winds move air masses, they carry their weather conditions (heat or cold, dry or moist) from the source region to a new region. When the air mass reaches a new region, it might clash with another air mass that has a different temperature and humidity. This can create a severe storm.

Air masses can affect the weather because of different air masses that are different in temperature, density, and moisture. When two different air masses meet a front forms. This is one way air masses effect our weather.

5 0
3 years ago
A 2.0-kilogram ball traveling north at 4.0 meters per second collides head on with a 1.0-kilogram ball traveling south at 8.0 me
enyata [817]

Answer:

We know the momentum after the collision MUST be equal to the momentum BEFORE the collision.  

Momentum is a VECTOR quantity having both magnitude and direction.  The first ball has momentum P =m*v = 2*4 = 8 at 90degrees.  The second ball has momentum P = 1*8 = 8 at -90 or 270 degrees.  They sum to zero when you perform vector addition.

Explanation:

6 0
2 years ago
In a candy factory, the nutty chocolate bars contain 20.0 % pecans by mass. If 4.0 kg of pecans were used for candy last Tuesday
ladessa [460]

Answer:

44.09 pounds

Explanation:

We got that 20 % of the mass of a nutty chocolate bar its pecans, if 4.0 kg of pecans were used, we need to find the X in the equation

0.2 * X = 4.0 \ kg

where X its the total mass of nutty chocolate bars produced. So, we can just divide by 0.2 on both sides, and we find:

X = 4.0 kg / 0.2

X = 20.0 \ kg

Of course, we need the total mass produced in pounds, and not in kilograms.  Looking at an conversion table, we can find that 1 kg its 2.20462 pounds, multiplying the value for total mass produced by the conversion factor we get:

X = 20.0 \ kg * 2.20462 \ \frac{pounds}{kg}

X = 44.0924 \ pounds

Now, we just need to round off to two significant figures. This is:

X = 44.09 \ pounds,

the total mass of nutty chocolate bars  made last Tuesday to two significant figures.

5 0
3 years ago
Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103
son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

PE = -\frac{GMm}{R}

As M=m, then

PE = -\frac{Gm^2}{R}

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

U = -\frac{2Gm^2}{R}

dU = -\frac{2Gm^2}{R}

dKE = -dU

Therefore replacing we have that,

\frac{1}{2}mv^2 =\frac{Gm^2}{2R}

Re-arrange to find v,

v= \sqrt{\frac{Gm}{R}}

v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}

v = 816.7m/s

Therefore the  velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

2r = 2*10^3m

Therefore

dU = Gm^2(\frac{1}{R}-\frac{1}{2r})

v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}

v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}

v = 1.83*10^7m/s

Therefore the velocity when they are about to collide is 1.83*10^7m/s

7 0
3 years ago
PLEASE HELP ASAP!!!
kirza4 [7]

Answer:

0.5mv^2=50, v=5, 25/2×m=50, m=50×2/25, So, the answer is 4

6 0
3 years ago
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