Answer: A car slows from 22 m/s to 3.0 m/s at a constant rate of 2.1 m/s2. How many seconds are required before the car is a r2.1 m/s2 traveling at 3.0 m/s?
5 pages
Explanation:
The positive charge is strongest in the middle, because that is were the charges are going off from.
In both graphs, speed will be measured in meters/second.
In graph A, the object travelled a distance of 40 meters in 4 seconds
In graph B, the object is at rest for 5 seconds.
These are the correct answers
Answer:
Magnitude of the hiker's total displacement = 3.5 km
Angle and direction of the hiker's total displacement = 22 degree south of west
Explanation:
The first Displacement in vector form is
= -4.5 cos 45i +4.5 sin 45j
= -3.182i + 3.182j
The second Displacement in vector form is
= -4.5j
The total displacement is ,
![s = s_1+ s_2](https://tex.z-dn.net/?f=s%20%3D%20s_1%2B%20s_2)
= -3.182i + 3.182j - 4.5j
=-3.182i-1.318j
The magnitude of the displacement is,
![s= \sqrt{(-3.182)^2 + (-1.318)^2\\](https://tex.z-dn.net/?f=s%3D%20%5Csqrt%7B%28-3.182%29%5E2%20%2B%20%28-1.318%29%5E2%5C%5C)
![s=\sqrt{10.125+1.7371}](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B10.125%2B1.7371%7D)
![s=\sqrt{11.862}](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B11.862%7D)
s= 3.444
The direction is,
![\theta = tan^{-1}(\frac{1.318}{3.182})](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%28%5Cfrac%7B1.318%7D%7B3.182%7D%29)
![\theta = tan^{-1}(0.4142)](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%280.4142%29)
![\theta = 22.49^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2022.49%5E%7B%5Ccirc%7D)
_______________________//__/1 amu