The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.
At time
, the acceleration vector has direction
such that

which indicates the particle is situated at a point on the lower left half of the circle, while at time
the acceleration has direction
such that

which indicates the particle lies on the upper left half of the circle.
Notice that
. That is, the measure of the major arc between the particle's positions at
and
is 270 degrees, which means that
is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.
Recall that

where
is the radius of the circle and
is the period. We have

and the magnitude of the particle's acceleration toward the center of the circle is

So we find that the path has a radius
of

Answer:
They are separated by a distance of 18 m. Find the gravitational attraction between them. r= 1 m r= 1 m 18 m. Mass = 1.5 kg. Mass = 8.5 kg.
<u>Answer
</u>
D. Salt water is denser than freshwater.
<u>Explanation</u>
A boat is able to float on water because it experiences an upthrust upwards.
The magnitude of the upthrust depends on the density of the liquid.
When the liquid is denser the boat will experience a great upthrust as compared to when in a less dense liquid.
If the boat sinks lower in the freshwater than in salty water, then Salt water is more dense than freshwater.
<span>The amount of dissolved oxygen in water may decrease
because of the increase in organic matter in the water. <span>Aquatic organisms breathe and use oxygen. Large amounts of
oxygen are consumed by the decomposition of bacteria (when there are large
amounts of dead matter to decompose, there will be a significant number of
bacteria). Examples: dead organic matter (algae), wastewater, garden waste,
oils and fats, all this results in a decrease in dissolved oxygen in the water.</span></span>