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The correct answer of this question is : Molecular motion or vibration slows down

EXPLANATION :

The kinetic energy of the molecule depends on the molecular motion. Molecules might have different types of kinetic energy depending on the type of substance i.e solid,liquid or gases.

During cooling, we are reducing the temperature of the substance. Due to the decrease of temperature, the molecular speed is also decreased. It is so because the molecular speed is dependent on the temperature of the substance.

Hence, the vibration, rotation or translation motion of the molecule slows down during cooling, which in turn, decreases the kinetic energy of the molecule.

(b). The gravitational force between the objects when they are 150 km apart is [tex]\boxed{889\,{\text{N}}}[/tex].

(c). The gravitational force between the objects when they are 50 km apart is [tex]\boxed{8000\,{\text{N}}}[/tex] .

Further Explanation:

The gravitational force of attraction between the two bodies is given by the Newton’s Law of Gravitation. According to Newton’s law of Gravitation, the gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

The gravitational force is expressed mathematically as:

[tex]F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}[/tex]

Here, [tex]G[/tex] is the gravitational constant, [tex]{m_1}[/tex] is the mass of first body, [tex]{m_2}[/tex] is the mass of second body and [tex]r[/tex] is the distance between two bodies.

For two bodies kept at a distance of 100 Km , the gravitational force of attraction is 2000 N.

Substitute 200 N for [tex]F[/tex] and 100 Km for r in above equation.

[tex]\begin{aligned}2000=\frac{{G{m_1}{m_2}}}{{{{\left( {100\times {{10}^3}}\right)}^2}}}\hfill\\G{m_1}{m_2} = 2\times {10^{13}}\hfill\\\end{aligned}[/tex]

Part (b):

Now, the force experienced by the bodies when they are 150 Km apart is:

[tex]F = \dfrac{{G{m_1}{m_2}}}{{\left( {150 \times {{10}^3}}\right)}}[/tex]

Substitute [tex]2 \times{10^8}[/tex] for [tex]G{m_1}{m_2}[/tex] in above equation.

[tex]\begin{aligned}F&=\frac{{2 \times {{10}^{13}}}}{{{{\left( {150 \times {{10}^3}}\right)}^2}}}\\&= 888.9\,{\text{N}}\\&\approx {\text{889}}\,{\text{N}}\\\end{aligned}[/tex]

Thus, the gravitational force between the objects when they are 150 Km  apart is [tex]\boxed{889\,{\text{N}}}[/tex].

Part (c):

Now, the force experienced by the bodies when they are 50 km apart is:

[tex]F =\Dfrac{{G{m_1}{m_2}}}{{\left( {50 \times {{10}^3}}\right)}}[/tex]

Substitute [tex]2 \times {10^8}[/tex] for [tex]G{m_1}{m_2}[/tex] in above equation.

[tex]\begin{aligned}F &=\frac{{2 \times {{10}^{13}}}}{{{{\left({50 \times {{10}^3}}\right)}^2}}}\\&= 8000\,{\text{N}}\\\end{aligned}[/tex]

Thus, the gravitational force between the objects when they are 50 Km  apart is  [tex]\boxed{8000\,{\text{N}}}[/tex].

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Answer Details:

Grade: College

Subject: Physics

Chapter: Newton’s law of Gravitation

Keywords:  Gravitation, newton’s law, force of attraction, 2000N, 100km, 150 km, 50 km, gravitational force, two spherical objects, 889 N, 8000 N.

consider the motion of rocket until it runs out of fuel

v₀ = initial velocity = 0 m/s

v = final velocity when it runs out of fuel = ?

t = time after which fuel is finished = 3.98 sec

a = acceleration = 29.4 m/s²

Y₀ = height gained when the fuel is finished = ?

using the kinematics equation

v = v₀ + a t

v = 0 + (29.4) (3.98)

v = 117.01 m/s

using the equation

v² = v²₀ + 2 a Y₀

(117.01)² = 0² + 2 (29.4) Y₀

Y₀ = 232.85 m

consider the motion of rocket after fuel is finished till it reach the maximum height.

Y₀ = initial position = 232.85 m

Y = final position at maximum height

v₀ = initial velocity just after the fuel is finished = 117.01 m/s

v = final velocity after it reach the maximum height = 0 m/s

a = acceleration due to gravity = - 9.8 m/s²

using the kinematics equation

v² = v²₀ + 2 a (Y - Y₀)

inserting the values

0² = (117.01)² + 2 (- 9.8) (Y - 232.85)

Y = 931.4 m

The total distance traveled by the rocket above the ground is 931.4 m.

The given parameters;

The distance traveled by the rocket during the 3.98 s is calculated as follows;

[tex]h_1 = v_0t + \frac{1}{2} at^2\\\\h_1 = 0 + \frac{1}{2} (29.4)(3.98)^2\\\\h_1 = 232.85 \ m[/tex]

The final velocity of the rocket after 3.98 s is calculated as follows;

[tex]v_i= v_0 + at\\\\v_i= 0 + (29.4 \times 3.98)\\\\v_i = 117.01 \ m/s[/tex]

"when the rocket runs out of fuel, it moves at a constant speed and the acceleration is zero. The rocket will be moving against gravity."

The distance traveled by the rocket when it runs out of fuel is calculated as follows;

[tex]v_f^2 = v_i^2 - 2gh_2[/tex]

where;

[tex]0 = (117.01)^2 -2(9.8)h_2 \\\\2(9.8)h_2 = (117.01)^2\\\\h_2 = \frac{ (117.01)^2}{2(9.8)} \\\\h_2 = 698.54 \ m[/tex]

Total distance traveled by the rocket above the ground;

H = h₁ + h₂

H = 232.85 m + 698.54 m

H = 931.4 m

Thus, the total distance traveled by the rocket above the ground is 931.4 m.

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