Answer:
To calculate displacement, simply draw a vector from your starting point to your final position and solve for the length of this line. If your starting and ending position are the same, like your circular 5K route, then your displacement is 0. In physics, displacement is represented by Δs.
Explanation:
hope helps
<span>1 C = 6.24150965(16)×10^18 electrons
31.25 x 10^18 electrons / (6.24150965(16)×10^18 electrons / C) = 5.007 Coulombs
</span><span>I hope this helps. </span>
<h2>
Answer: 540 J</h2>
Explanation:
The Work 
done by a Force 
refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.
Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:
(1)
In this case both (the force and the distance in the path) are parallel (this means they are in the same direction), so the work performed is the product of the force exerted to push the box 
by the distance traveled 
.
Hence:
(2)
<span>C. switching to cheaper fuel. Physical capital pertains to non-human asset. Under this type were the asset that use to process goods and services like machinery, buildings etc.</span>
According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>
In other words, this law states a relation between the orbital period 
of a body (moon, planet, satellite) orbiting a greater body in space with the size 
of its orbit.
This Law is originally expressed as follows:
<h2>
(1)
</h2>
Where;
is the Gravitational Constant and its value is 
is the mass of Jupiter
is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)
If we want to find the period, we have to express equation (1) as written below and substitute all the values:
<h2>
(2)
</h2>
Then:
<h2>
(3)
</h2>
Which is the same as:
<h2>
</h2>
Therefore, the answer is:
The orbital period of Io is 42.482 h
