Answer:
d= 64.7 km

displacement vector
Explanation:
total distance = 40 + 30 = 70 km
during 1st flight


during 2nd flight



the two component of r are:


Geographical Direction ![\theta = tan^{-1}\frac{r_y}{r_x} [tex]\theta = 40.9^{o}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%5Cfrac%7Br_y%7D%7Br_x%7D%20%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Ctheta%20%3D%2040.9%5E%7Bo%7D)
Displacement d

d= 64.7 km
displacement vector
Answer:
Beth went back toward her origin
Explanation:
Please, see the attached image that represent's Beth's path.
She started her walk 2 meters from the origin (0) and after 6 seconds she headed back to position zero (0) the origin. Notice the second half of the graph as a segment going down, thus reducing her distance from the origin, and getting closer to it.
Answer:
The altitude of the plane is 379.5 m.
Explanation:
Initial horizontal velocity, u = 59.1 m/s
Horizontal distance, d = 521 m
let the time taken by the packet to cover the distance is t.
Horizontal distance = horizontal velocity x time
521 = 59.1 x t
t = 8.8 s
let the vertical height is h .
Use second equation of motion in vertical direction.

Answer:
α= 15.4[rad/s^2], 500 [rev/min]
Explanation:
La aceleración angular constante esta definida por la siguiente ecuación:
![w=w_{0} + \alpha *t\\Donde:\\w=2000 [\frac{rev}{min}]*[\frac{1min}{60s} ] *[\frac{2\pi rad}{1 rev} ]=209.4[\frac{rad}{s} ]\\w_{0} =1500 [\frac{rev}{min}]*[\frac{1min}{60s} ] *[\frac{2\pi rad}{1 rev} ]=157.1[\frac{rad}{s} ]\\t=3.4[s]\\reemplazando:\\209.4-157.1=\alpha *(3.4)\\\\alpha =15.4[\frac{rad}{s^{2} } ]](https://tex.z-dn.net/?f=w%3Dw_%7B0%7D%20%2B%20%5Calpha%20%2At%5C%5CDonde%3A%5C%5Cw%3D2000%20%5B%5Cfrac%7Brev%7D%7Bmin%7D%5D%2A%5B%5Cfrac%7B1min%7D%7B60s%7D%20%5D%20%2A%5B%5Cfrac%7B2%5Cpi%20rad%7D%7B1%20rev%7D%20%5D%3D209.4%5B%5Cfrac%7Brad%7D%7Bs%7D%20%5D%5C%5Cw_%7B0%7D%20%3D1500%20%5B%5Cfrac%7Brev%7D%7Bmin%7D%5D%2A%5B%5Cfrac%7B1min%7D%7B60s%7D%20%5D%20%2A%5B%5Cfrac%7B2%5Cpi%20rad%7D%7B1%20rev%7D%20%5D%3D157.1%5B%5Cfrac%7Brad%7D%7Bs%7D%20%5D%5C%5Ct%3D3.4%5Bs%5D%5C%5Creemplazando%3A%5C%5C209.4-157.1%3D%5Calpha%20%2A%283.4%29%5C%5C%5C%5Calpha%20%3D15.4%5B%5Cfrac%7Brad%7D%7Bs%5E%7B2%7D%20%7D%20%5D)
Para poder llegar a las 2000 [revoluciones /min] estando en 1500 [rev/min], la cuchilla tuvo que incrementar su numero de revoluciones a 500 [rev/min]. Mientras aceleraba.