A 1300-kg car initially has a velocity of 22.2 m/s due south. It brakes to a stop over a 180 m distance.
1 answer:
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Answer:
<h2>workdone = force × distance </h2><h2>236J = 18.9cos(o) × 24.4</h2><h2>236/24.4 = 18.9cos(o)</h2><h2>(0.5117)cos^-1 = (o)</h2><h2><u>59.21°</u></h2>
Answer:
a) the one with a lower orbit b) the one with a higher orbit
Explanation:
Let's consider orbital mechanics. To get an object in orbit, we need it to fall to earth parallel to the earth's surface. To understand it easily imagine a projectile thrown horizontally further and further away, at one point, the projectile hits the cannon from behind. Considering there is no wind resistance, that would be a projecile in orbit.
In other words, the circular orbits of some objects around a massive body are due to the equality between centrifugal acceleration and gravity acceleration.
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so the velocity is
where "G" is the gravitational constant, "M" the mass of the massive body and "r" the distance between the object and the center of gravity of mass M. As you can note, if "r" increase, "v" decrease.
The orbital period of any object in orbit is
where "a" is length of semi-major axis (a = r in circular orbits). So if "r" increase, "T" increase.