The answer is A. <span>Some work input is used to overcome friction. </span>
Answer:
5.23km/s
Explanation:
Given
Radius of Earth = 6.37 * 10^6 m
Altitude of Satellite = 8200km = 8200 * 10³m = 8.2 * 10^6 m
Gravity Acceleration on Satellite Altitude = 1.87965m/s²
For a satellite to remain in circular orbit, then it means the acceleration of gravity must be exact as the centripetal acceleration.
Centripetal Acceleration = V²/R
So, Acceleration of Gravity (A)= Centripetal Acceleration = V²/R
Make V the subject of formula
A = V²/R
V² = AR
V = √AR
Where R = (radius of earth) + (altitude of satellite)
R = 6.37 * 10^6 + 8.2 * 10^6
R = 14.57 * 10^6m
A = 1.87965m/s²
V = √(1.87965 * 14.57x10^6)
V = √27386500.5
V = 5233.211299001789
V = 5233.2113 m/s ------- Approximated
V = 5.23km/s
To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.
The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,
M,m = Counts per second
Our radios are given by
Therefore replacing we have that,
Therefore the number of counts expect at a distance of 20 cm is 19.66cps
Answer:
1.33×10⁻¹⁰ N
Explanation:
F = GMm / r²
where G is the gravitational constant,
M and m are the masses of the objects,
and r is the distance between them.
F = (6.67×10⁻¹¹ N/m²/kg²) (1000 kg) (2000 kg) / (1000 m)²
F = 1.33×10⁻¹⁰ N
