Answer:
Explanation:
Mercury, the first planet from the Sun, endures drastic temperature changes from day to night. During the day, the planet is incredibly near to the Sun, with temperatures reaching 430°C.
In fact, the force Rahul exerts on Earth corresponds to the force of gravity. But Rahul's weight is, in fact, the force of gravity exerted by the Earth on Rahul, and these two forces correspond to the action-reaction pair of Newton's third law, which states that the two forces are equal.
The amount of energy lost to air friction, given the data is 37.71 J
<h3>How to obtain the initial energy</h3>
- Initial velocity (u) = 15.1 m/s
- Mass (m) = 450 g = 450 / 1000 = 0.45 Kg
- Initial Energy (E₁) = ?
E₁ = ½mu²
E₁ = ½ × 0.45 × 15.1²
E₁ = 51.3 J
<h3>How to obtain the final energy</h3>
- Final velocity (u) = 19.89 m/s
- Mass (m) = 450 g = 450 / 1000 = 0.45 Kg
- Final Energy (E₂) = ?
E₂ = ½mv²
E₂ = ½ × 0.45 × 19.89²
E₂ = 89.01 J
<h3>How to determine the energy lost</h3>
- Initial Energy (E₁) = 51.3 J
- Final Energy (E₂) = 89.01 J
- Energy lost =?
Energy lost = E₂ - E₁
Energy lost = 89.01 - 51.3
Energy lost = 37.71 J
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Answer:
Explanation:
<u>Dynamics</u>
When a particle of mass m is subject to a net force F, it moves at an acceleration given by
The particle has a mass of m=2.37 Kg and the force is horizontal with a variable magnitude given by
The variable acceleration is calculated by:
The instant velocity is the integral of the acceleration:
Integrating
