Answer:
Magnitude of resultant = 131.15
Direction of resultant = 3.97°
Explanation:
||u|| = 70
θ = 40°
![\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62](https://tex.z-dn.net/?f=%5Cvec%7Bu%7D_x%3D%7C%7Cu%7C%7Ccos%5Ctheta%20%5C%5C%5CRightarrow%20%5Cvec%7Bu%7D_x%3D70cos40%3D53.62)
![\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99](https://tex.z-dn.net/?f=%5Cvec%7Bu%7D_y%3D%7C%7Cu%7C%7Csin%5Ctheta%20%5C%5C%5CRightarrow%20%5Cvec%7Bu%7D_y%3D70sin40%3D44.99)
||v|| = 85
θ = 335°
![\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03](https://tex.z-dn.net/?f=%5Cvec%7Bv%7D_x%3D%7C%7Cv%7C%7Ccos%5Ctheta%20%5C%5C%5CRightarrow%20%5Cvec%7Bv%7D_x%3D85cos335%3D77.03)
![\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92](https://tex.z-dn.net/?f=%5Cvec%7Bv%7D_y%3D%7C%7Cv%7C%7Csin%5Ctheta%20%5C%5C%5CRightarrow%20%5Cvec%7Bv%7D_y%3D85sin335%3D-35.92)
Resultant
![R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15](https://tex.z-dn.net/?f=R%3D%5Csqrt%7BR_x%5E2%2BR_y%5E2%7D%5C%5C%5CRightarrow%20R%3D%5Csqrt%7B%28%5Cvec%7Bu%7D_x%2B%5Cvec%7Bv%7D_x%29%5E2%2B%28%5Cvec%7Bu%7D_y%2B%5Cvec%7Bv%7D_y%29%5E2%7D%5C%5C%5CRightarrow%20R%20%3D%5Csqrt%7B%2870cos40%2B85cos335%29%5E2%2B%2870sin40%2B85sin335%29%5E2%7D%5C%5C%5CRightarrow%20R%20%3D131.15)
![\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3Dtan%5E%7B-1%7D%5Cfrac%7BR_y%7D%7BR_x%7D%5C%5C%5CRightarrow%20%5Ctheta%3Dtan%5E%7B-1%7D%5Cfrac%7B70sin40%2B85sin335%7D%7B70cos40%2B85cos335%7D%5C%5C%5CRightarrow%20%5Ctheta%3Dtan%5E%7B-1%7D0.069%3D3.97%5E%7B%5Ccirc%7D)
Magnitude of resultant = 131.15
Direction of resultant = 3.97°
The time when the particle is at rest is at 1.63 s or 3.36 s.
The velocity is positive at when the time of motion is at
.
The total distance traveled in the first 10 seconds is 847 m.
<h3>When is a particle at rest?</h3>
- A particle is at rest when the initial velocity of the particle is zero.
The time when the particle is at rest is calculated as follows;
s(t) = 2t³ - 15t² + 33t + 17
![v = \frac{ds}{dt} = 6t^2 -30t + 33\\\\at \ rest, \ v = 0\\\\6t^2 - 30t + 33 = 0\\\\6(t- \frac{5}{2} )^2- \frac{9}{2} = 0\\\\t = 1.63\ s \ \ or \ 3.36 \ s](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bds%7D%7Bdt%7D%20%3D%206t%5E2%20-30t%20%2B%2033%5C%5C%5C%5Cat%20%5C%20rest%2C%20%5C%20v%20%3D%200%5C%5C%5C%5C6t%5E2%20-%2030t%20%2B%2033%20%3D%200%5C%5C%5C%5C6%28t-%20%5Cfrac%7B5%7D%7B2%7D%20%29%5E2-%20%5Cfrac%7B9%7D%7B2%7D%20%3D%200%5C%5C%5C%5Ct%20%3D%201.63%5C%20s%20%5C%20%5C%20or%20%5C%203.36%20%5C%20s)
The velocity is positive at when the time of motion is as follows;
.
The total distance traveled in the first 10 seconds is calculated as follows;
![2(10)^3 - 15(10)^2 + 33(10) + 17 = 847 \ m](https://tex.z-dn.net/?f=2%2810%29%5E3%20-%2015%2810%29%5E2%20%2B%2033%2810%29%20%2B%2017%20%3D%20847%20%5C%20m)
Learn more about motion of particles here: brainly.com/question/11066673
Answer:
Since this is old, im just gonna get these points, don't wan't them to go to waste lm.ao
Explanation:
No, the building's size in comparison to the earth would have no change or change so increadibly miniscule, like if you were told to spin slowly and an and was placed on top of your head
Answer:
hello your question is poorly written and I have tried to understand it hence I will give a general diagram as related to your question
answer : attached below
Explanation:
Attached below is the required diagram of the situation you are trying to describe
let the wide receiver be ; W
Corner back = C
westward velocity = Vw
eastward velocity = Vc
at initial position X = Xw
x = 0 ( initial position )