The number of protons in the nucleus of the atom is equal to the atomic number (Z). The number of electrons in a neutral atom is equal to the number of protons. The mass number of the atom (M) is equal to the sum of the number of protons and neutrons in the nucleus..
MARK BRAINIEST PLEASE!!!
Answer:
3494444444.44444 J
-87077491.39453 J
Explanation:
M = Mass of Earth = 
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
R = Radius of Earth = 
h = Altitude = 
m = Mass of satellite = 629 kg
v = Velocity of spacecraft = 
The kinetic energy is given by
The spacecraft's kinetic energy relative to the earth is 3494444444.44444 J
Potential energy is given by
The potential energy of the earth-spacecraft system is -87077491.39453 J
Answer:
1. 0 J
2. 7500 J
3. 7500 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) of car = 600 Kg
Initial velocity (v₁) of car = 0 m/s
Final velocity (v₂) of car = 5 m/s
Original kinetic energy (KE₁) =?
Final kinetic energy (KE₂) =?
Work used =?
1. Determination of the original kinetic energy.
Mass (m) of car = 600 Kg
Initial velocity (v₁) of car = 0 m/s
Original kinetic energy (KE₁) =?
KE₁ = ½mv₁²
KE₁ = ½ × 600 × 0²
KE₁ = 0 J
Thus, the original kinetic energy of the car is 0 J.
2. Determination of the final kinetic energy.
Mass (m) of car = 600 Kg
Final velocity (v₂) of car = 5 m/s
Final kinetic energy (KE₂) =?
KE₂ = ½mv₂²
KE₂ = ½ × 600 × 5²
KE₂ = 300 × 25
KE₂ = 7500 J
Thus, the final kinetic energy of the car is 7500 J
3. Determination of the work used.
Original kinetic energy (KE₁) = 0
Final kinetic energy (KE₂) = 7500 J
Work used =?
Work used = KE₂ – KE₁
Work used = 7500 – 0
Work used = 7500 J
