Near Earth's surface, gravitational acceleration is approximately 9.81 m/s2, which means that, ignoring the effects of air resistance, the speed of an object falling freely will increase by about 9.81 metres per second every second.
Answer: The gravitational force Fg exerted on the orbit by the planet is Fg = G 4/3πr3rhom/ (R1 + d+ R2)^2
Explanation:
Gravitational Force Fg = GMm/r2----1
Where G is gravitational constant
M Mass of the planet, m mass of the orbit and r is the distance between the masses.
Since the circular orbit move around the planet, it means they do not touch each other.
The distance between two points on the circumference of the two massesb is given by d, while the distance from the radius of each mass to the circumferences are R1 and R2 from the question.
Total distance r= (R1 + d + R2)^2---2
Recall, density rho =
Mass M/Volume V
Hence, mass of planet = rho × V
But volume of a sphere is 4/3πr3
Therefore,
Mass M of planet = rho × 4/3πr3
=4/3πr3rho in kg
From equation 1 and 2
Fg = G 4/3πr3rhom/ (R1 + d+ R2)^2
Answer:
the red at the bottom should not count, but the red at the top is the least dense because it floats upon the other liquids
Explanation:
hope this helps
Answer:
False
Explanation:
The formula of force that exists between two charges is expressed as;
F = kq1q2/r²
If two charges separated by one meter exert a 9 N force on each other, the;
9 = kq1q2/1²
9 = kq1q2 ..... 1
If the charges are pushed to a 3 meter separation, then;
F = kq1q2/3²
F = kq1q2/9 .... 2
Divide both equations;
9/F = (kq1q2)/ kq1q2/9
9/F = kq1q2 * 9/ kq1q2
9/F = 9
F = 9/9
F = 1N
Hence if the charges are pushed to a 3 meter separation, then the force on EACH charge will be 1N. Hence the answer is False
