The force is gravitational because when something is falling is call gravitational
Over time, the types of technology can vary and be improved upon so that more advanced techniques become more valued. This could be the situation with mining whereby back in the 1500's in underground mines the rock was broken by fire setting ie lighting a fire below the rock face to heat up the rock and then throwing cold water on it to crack it, so that it could be dug by hand. With the advent of explosives, this all changed so that the rock could be blasted. The increase in advance rates for an underground heading have thus gone from 5-20 feet per month to up to 300meters (984 ft) per month for a 24/7 mining operation, which is a huge improvement.
The object is fixed relative to the motion you are trying to describe.
Four of them pass a point every second.
Answer:
![|\vec r|=339.82\ m](https://tex.z-dn.net/?f=%7C%5Cvec%20r%7C%3D339.82%5C%20m)
![\theta=-6.67^o](https://tex.z-dn.net/?f=%5Ctheta%3D-6.67%5Eo)
Explanation:
<u>Displacement
</u>
It's a vector magnitude that measures the space traveled by a particle between an initial and a final position. The total displacement can be obtained by adding the vectors of each individual displacement. In the case of two displacements:
![\vec r=\vec r_1+\vec r_2](https://tex.z-dn.net/?f=%5Cvec%20r%3D%5Cvec%20r_1%2B%5Cvec%20r_2)
Given a vector as its polar coordinates (r,\theta), the corresponding rectangular coordinates are computed with
![x=rcos\theta](https://tex.z-dn.net/?f=x%3Drcos%5Ctheta)
![y=rsin\theta](https://tex.z-dn.net/?f=y%3Drsin%5Ctheta)
And the vector is expressed as
![\vec z==](https://tex.z-dn.net/?f=%5Cvec%20z%3D%3Cx%2Cy%3E%3D%3Crcos%5Ctheta%2Crsin%5Ctheta%3E)
The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is
![\vec r_1==\ km=\ m](https://tex.z-dn.net/?f=%5Cvec%20r_1%3D%3C0.198cos0%5Eo%2C0.198sin0%5Eo%3E%3D%3C0.198%2C0%3E%5C%20km%3D%3C198%2C0%3E%5C%20m)
The second move is (145 m , -15.8°). The angle is negative because it points South of East. The second displacement is
![\vec r_2==\ m](https://tex.z-dn.net/?f=%5Cvec%20r_2%3D%3C145cos%28-15.8%5Eo%29%2C145sin%28-15.8%5Eo%29%3E%3D%3C139.52%2C-39.48%3E%5C%20m)
The total displacement is
![\vec r=\ m+\ m](https://tex.z-dn.net/?f=%5Cvec%20r%3D%3C198%2C0%3E%5C%20m%2B%3C139.52%2C-39.48%3E%5C%20m)
![\vec r=\ m](https://tex.z-dn.net/?f=%5Cvec%20r%3D%3C337.52%2C-39.48%3E%5C%20m)
In (magnitude,angle) form:
![|\vec r|=\sqrt{337.52^2+(-39.48)^2}=339.82\ m](https://tex.z-dn.net/?f=%7C%5Cvec%20r%7C%3D%5Csqrt%7B337.52%5E2%2B%28-39.48%29%5E2%7D%3D339.82%5C%20m)
![\boxed{|\vec r|=339.82\ m}](https://tex.z-dn.net/?f=%5Cboxed%7B%7C%5Cvec%20r%7C%3D339.82%5C%20m%7D)
![\displaystyle tan\theta=\frac{-39.48}{337.52}=-0.1169](https://tex.z-dn.net/?f=%5Cdisplaystyle%20tan%5Ctheta%3D%5Cfrac%7B-39.48%7D%7B337.52%7D%3D-0.1169)
![\boxed{\theta=-6.67^o}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctheta%3D-6.67%5Eo%7D)