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2 years ago

Which two factors decrease as the kinetic energy of the particles in an object decrease

1 answer:

3 0

The particle temperature and the valency of the atom in the particles are the two parameters that decrease with the kinetic energy of the particles in an item.

<h3>What is temperature?</h3>

Temperature directs to the hotness or coldness of a body. In clear terms, it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles more the temperature.

Because an item possesses kinetic energy when it is moving, anything that prevents that motion will reduce the kinetic energy of the object.

A valence electron is an electron in the outermost shell associated with an atom that can experience the innovation of a chemical bond if the outer shell is not closed. Both atoms contribute one valence electron to create a shared pair.

More the valence electron more the motion of atoms results in more kinetic energy,

Hence the two factors decrease as the kinetic energy of the particles in an object will be the particle temperature and the valency of the atom in the particles.

To learn more about the temperature refer to the link;

brainly.com/question/7510619

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a ferryboat has an acceleration of 0.50 m/s. If the ferry travels 125m in 20 seconds while accelerating, what was its initial ve

Papessa [141]

as we know that

acceleration = 0.50 m/s^2

distance traveled = 125 m

time taken = 20 s

now we will use the equation of kinematics

$d = v_i*t + \frac{1}{2}at^2$

$125 = v_i* 20 + \frac{1}{2}*0.50*20^2$

$125 = 20*v_i + 100$

$125 - 100 = 20*v_i$

$v_i = 1.25 m/s$

so its initial speed must be 1.25 m/s

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3 years ago

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Fantom [35]

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4 years ago

No file link please

musickatia [10]

Rock 3, because it has the most mass while still moving at the same speed, it is kinetic because there is movement and energy being used

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3 years ago

The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio

LUCKY_DIMON [66]

Answer:

29274.93096 m/s

$2.73966\times 10^{33}\ J$

$-5.39323\times 10^{33}\ J$

$2.56249\times 10^{33}\ J$

$-5.21594\times 10^{33}$

Explanation:

$r_p$ = Distance at perihelion = $1.471\times 10^{11}\ m$

$r_a$ = Distance at aphelion = $1.521\times 10^{11}\ m$

$v_p$ = Velocity at perihelion = $3.027\times 10^{4}\ m/s$

$v_a$ = Velocity at aphelion

m = Mass of the Earth = 5.98 × 10²⁴ kg

M = Mass of Sun = $1.9889\times 10^{30}\ kg$

Here, the angular momentum is conserved

$L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s$

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J$

Kinetic energy at perihelion is $2.73966\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}$

Potential energy at perihelion is $-5.39323\times 10^{33}\ J$

$K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J$

Kinetic energy at aphelion is $2.56249\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}$