In Figure 10-1, if the force exerted on a 3.0-kg backpack that is initally at rest is 20.0 N and the distance it acts over is 0.
1 answer:
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Answer:
the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.
Explanation:
Mass of satellite, m
orbit radius of first, r1 = r
orbit radius of second, r2 = 2r
Centripetal force is given by
Where v be the orbital velocity, which is given by
So, the centripetal force is given by
where, g bet the acceleration due to gravity
So, the centripetal force
Gravitational force on the satellite having larger orbit
.... (1)
Gravitational force on the satellite having smaller orbit
.... (2)
Comparing (1) and (2),
F' = 4 F
So, the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.
Answer:
The final velocity of the second car is 57 m/s south.
Explanation:
This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.
The formula to apply is :
where ;
Given in the question that;
Apply the formula as;
{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}
263700+42075=87900 + 3825v₂f
305775 =87900 + 3825v₂f
305775-87900 = 3825v₂f
217875=3825v₂f
217875/3825 =v₂f
56.96 = v₂f
<u>57 m/s = v₂f { nearest whole number}</u>