Answer: Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN
Explanation: To find the answer we need to know more about the Newton's law of gravitation.
<h3>What is Newton's law of gravitation?</h3>
- Gravitation is the force of attraction between any two bodies.
- Every body in the universe attracts every other body with a force.
- This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between these two masses.
- Mathematically we can expressed it as,
<h3>How to solve the problem?</h3>
- Here, we have given with the data's,
- Thus, the force of attraction between these two bodies will be,
Thus, if two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg and, If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN.
Learn more about the Newton's law of gravitation here:
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The answer is 165.3 cm³.
P1 * V1 / T1 = P2 * V2 / T2
The initial sample:
P1 = 84.6 kPa
V1 = 215 cm³
T1 = 23.5°C = 23.5 + 273 K = 296.5 K
At STP:
P2 = 101.3 kPa
V2 = ?
T2 = 273 K
Therefore:
84.6 * 215 / 296.5 = 101.3 * V2 / 273
61.34 = 101.3 * V2 / 273
V2 = 61.34 * 273 / 101.3
V2 = 165.3 cm³
Answer:
a) m=20000Kg
b) v=0.214m/s
Explanation:
We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.
For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, 
, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as 
, and the mass of the first and second coals as 
and 
respectively
We start with the transition between parts A and B, so we have:
Which means
And since we want the mass of the first coal thrown () we do:
Substituting values we obtain
For the transition between parts B and C, we can write:
Which means
Since we want the new final speed of the car (
) we do:
Substituting values we obtain
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