The elastic potential energy stored in the car's spring during the process is 3.75 J
<h3>Determination of the spring constant</h3>
From the question given above, the following data were obtained:
K = F/e
K = 15 / 0.5
K = 30 N/m
<h3>Determination of the potential energy</h3>
- Spring constant (K) = 30 N/m
PE = ½Ke²
PE = ½ × 30 × 0.5²
PE = 15 × 0.25
PE = 3.75 J
Therefore, the elastic potential energy stored in the car's spring during the process is 3.75 J
Learn more about energy stored in spring:
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Answer:
Explanation:
The work-energy theorem states that the work done on the system is equal to its change of kinetic energy: . This refers to the center of mass in the case of a system of particles. To calculate this work, we need to use the net Force on the system and the displacement of its center of mass:
<>"Topographic maps conventionally show topography, or land contours, by means of contour lines. Contour lines are curves that connect contiguous points of the same altitude (isohypse). In other words, every point on the marked line of 100 m elevation is 100 m above mean sea level."<> I hope this helps.
Answer:
M = 1.38 10⁵⁹ kg
Explanation:
For this problem we will use the law of universal gravitation
F = G m₁ m₂ / r²
Where G is the gravitation constant you are value 6.67 10⁻¹¹ N m2 / kg2, m are the masses and r the distance
In this case the mass of the planet is m = 3.0 10²³ kg and the mass of the start is M
Let's write Newton's second law
F = m a
The acceleration is centripetal
a = v² / r
The speed module is constant, so we can use the kinematic relationship
v = d / t
The distance remembered is the length of the circular orbit and the time in this case is called the period
d = 2π r
a = 2π r / T
Let's replace Newton's second law
G m M / r² = m (4π² r² / T²) / r
G M = 4 π² r³ / T²
M = 4 π² r³ / T² G
Let's calculate
M = 4 π² (3.0 10²³)³ / (3.4 10¹¹)² 6.67 10⁻¹¹
M = 13.82 10⁵⁸ kg
M = 1.38 10⁵⁹ kg