Answer:
Speed of the satellite V = 6.991 × 10³ m/s
Explanation:
Given:
Force F = 3,000N
Mass of satellite m = 500 kg
Mass of earth M = 5.97 × 10²⁴
Gravitational force G = 6.67 × 10⁻¹¹
Find:
Speed of the satellite.
Computation:
Radius r = √[GMm / F]
Radius r = √[(6.67 × 10⁻¹¹ )(5.97 × 10²⁴)(500) / (3,000)
Radius r = 8.146 × 10⁶ m
Speed of the satellite V = √rF / m
Speed of the satellite V = √(8.146 × 10⁶)(3,000) / 500
Speed of the satellite V = 6.991 × 10³ m/s
Answer:
120,000J
Corrected question;
In one hour, coal supplies 500 000 J of energy. The wasted energy amounts to 380 000 J. How much useful energy is produced in one hour?
Explanation:
Given;
Total energy Et = 500,000 J
Wasted Energy Ew = 380,000J
The amount useful energy is the amount of energy that is available for supply.
This can be derived by subtracting the wasted energy from the total energy.
Useful energy = Total Energy - wasted energy
Eu = Et - Ew
Substituting the given values;
Eu = 500,000J - 380,000
Eu = 120,000 J
The amount of useful energy produced in one hour is 120,000 J
Protons and neutrons are located in the nucleus, a dense central core in the middle of the atom, while the electrons are located outside the nucleus.
let Coefficients of Friction of Rubber on asphalt (dry) =0.7
F= Coefficients of Friction * normal force = 0.7 * 60 =42 N
so the net force of the rubber is zero, meaning it will travel at a constant speed.
When the rubber is travel at 2m/s, 42N is required to keep moving at constant speed
