Answer:
The variation in apparent gravitational acceleration (g) at different locations on Earth is caused by two things (as you implied). ... The distance between the centers of mass of two objects affects the gravitational force between them, so the force of gravity on an object is smaller at the equator compared to the poles.
The Professor's centripetal acceleration is 0.044 m/s²
Centripetal acceleration is the acceleration of an object moving in circular motion. It is usually directed towards the center of the rotation.
It is given by:
a = v²/r
where v is the velocity and r is the radius.
Given that the radius (r) = 4 m, velocity (v) = 0.419 m/s, hence:
a = v²/r = 0.419²/4 = 0.044 m/s²
The Professor's centripetal acceleration is 0.044 m/s²
Find out more at: brainly.com/question/6082363
The work done by 
along the given path <em>C</em> from <em>A</em> to <em>B</em> is given by the line integral,
I assume the path itself is a line segment, which can be parameterized by
with 0 ≤ <em>t</em> ≤ 1. Then the work performed by <em>F</em> along <em>C</em> is
![\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E1%20%5Cleft%286x%28t%29%5E3%5C%2C%5Cvec%5Cimath-4y%28t%29%5C%2C%5Cvec%5Cjmath%5Cright%29%5Ccdot%5Cfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5Bx%28t%29%5C%2C%5Cvec%5Cimath%20%2B%20y%28t%29%5C%2C%5Cvec%5Cjmath%5Cright%5D%5C%2C%5Cmathrm%20dt%20%5C%5C%5C%5C%20%3D%20%5Cint_0%5E1%20%28288%283t-1%29%5E3-8%282t%2B5%29%29%20%5C%2C%5Cmathrm%20dt%20%3D%20%5Cboxed%7B312%7D)
Answer:
The equation used to calculate the work done is: work done = force × distance. W = F × d. This is when: work done (W) is measured in joules (J)
