Answer:
- d = 2; m = 2; m = 2
- d= 1; m = 1; m = 2
- d = 1; m = 1; m = 1
- d = 2; m = 1; m = 2
- d = 2; m = 1; m = 1
Explanation:
The following diagrams show five pairs of asteroids, labeled with their relative masses (M) and distances (d) between them. For example, an asteroid with M=2 has twice the mass of one with M=1 and a distance of d=2 is twice as large as a distance of d=1. Rank each pair from left to right based on the strength of the gravitational force attracting the asteroids to each other, from strongest to weakest.
- d = 2; m = 2; m = 2
- d= 1; m = 1; m = 2
- d = 1; m = 1; m = 1
- d = 2; m = 1; m = 2
- d = 2; m = 1; m = 1
2.12pa of pressure is generated on the ground as a result of the weight.
<h3>How do you determine the force applied to the ground?</h3>
The formula for average pressure, P = F/A, can be used to determine average ground pressure. This is just the object's weight divided by the contact area in the ideal scenario, which is a static, uniform net force normal to level ground.
<h3>How much pressure is there at 1,000 feet?</h3>
In ideal circumstances, every 1000 feet of elevation gain results in a 1 inHg (inch of mercury) decrease in air pressure (barometer). The pressure will therefore drop to 28.92 inHg at 1000 ft MSL from 29.92 inHg at sea level (0 ft MSL) (1000 feet higher).
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No additional force is required because it's already going downhill