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.
A rocket moves upward, starting from rest with an acceleration of +29.4 for 3.98 s. it runs out of fuel at the end of the 3.98 s but does not stop. m/s2 how high does it rise above the ground?
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;
acceleration of the rocket, a = 29.4 m/s²time of motion of the rock, t = 3.98 sThe 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]v_f[/tex] is the final velocity of the rocket at maximum height = 0[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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