Answer: c
Explanation:
The total distance the ultrasound pulse travelled is, twice the distance of the sea floor
Because it went back and forth
Answer:
El alcance de la bala es 3464,1 m.
Explanation:
El alcance de la bala se puede calcular como sigue:
![y = y_{0} + tan(\theta)*x - \frac{g}{2}*\frac{x^{2}}{(v_{0}cos(\theta))^{2}}](https://tex.z-dn.net/?f=y%20%3D%20y_%7B0%7D%20%2B%20tan%28%5Ctheta%29%2Ax%20-%20%5Cfrac%7Bg%7D%7B2%7D%2A%5Cfrac%7Bx%5E%7B2%7D%7D%7B%28v_%7B0%7Dcos%28%5Ctheta%29%29%5E%7B2%7D%7D)
En donde:
y: es la altura final = 0
: es la altura inicial = 0
x: es el alcance
θ: es el angulo respecto a la horizontal = 30°
: es la velocidad inicial = 200 m/s
g: es la gravedad = 10 m/s²
Entonces, tenemos:
![y = y_{0} + tan(\theta)*x - \frac{g}{2}*\frac{x^{2}}{(v_{0}cos(\theta))^{2}}](https://tex.z-dn.net/?f=%20y%20%3D%20y_%7B0%7D%20%2B%20tan%28%5Ctheta%29%2Ax%20-%20%5Cfrac%7Bg%7D%7B2%7D%2A%5Cfrac%7Bx%5E%7B2%7D%7D%7B%28v_%7B0%7Dcos%28%5Ctheta%29%29%5E%7B2%7D%7D%20)
![x = \frac{2tan(\theta)*(v_{0}cos(\theta))^{2}}{g} = \frac{2tan(30)*(200 m/s*cos(30))^{2}}{10 m/s^{2}} = 3464,1 m](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cfrac%7B2tan%28%5Ctheta%29%2A%28v_%7B0%7Dcos%28%5Ctheta%29%29%5E%7B2%7D%7D%7Bg%7D%20%3D%20%5Cfrac%7B2tan%2830%29%2A%28200%20m%2Fs%2Acos%2830%29%29%5E%7B2%7D%7D%7B10%20m%2Fs%5E%7B2%7D%7D%20%3D%203464%2C1%20m%20)
Por lo tanto, el alcance de la bala es 3464,1 m.
Espero que se te sea de utilidad!
Answer:
Part a)
![t = 2.83 s](https://tex.z-dn.net/?f=t%20%3D%202.83%20s)
Part b)
Ball thrown downwards =![v_f = 23.8 m/s](https://tex.z-dn.net/?f=v_f%20%3D%2023.8%20m%2Fs)
Ball thrown upwards =![v_f = 23.8 m/s](https://tex.z-dn.net/?f=v_f%20%3D%2023.8%20m%2Fs)
Part c)
![d = 22.24 m](https://tex.z-dn.net/?f=d%20%3D%2022.24%20m)
Explanation:
Part a)
Since both the balls are projected with same speed in opposite directions
So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection
Afterwards the motion will be same as the first ball which is projected downwards
so here the time difference is given as
![\Delta y = 0 = v_y t + \frac{1}{2}at^2](https://tex.z-dn.net/?f=%5CDelta%20y%20%3D%200%20%3D%20v_y%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E2)
![0 = 13.9 t - \frac{1}{2}(9.81) t^2](https://tex.z-dn.net/?f=0%20%3D%2013.9%20t%20-%20%5Cfrac%7B1%7D%7B2%7D%289.81%29%20t%5E2)
![t = 2.83 s](https://tex.z-dn.net/?f=t%20%3D%202.83%20s)
Part b)
Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls
so it is given as
![v_f^2 - v_i^2 = 2 a \Delta y](https://tex.z-dn.net/?f=v_f%5E2%20-%20v_i%5E2%20%3D%202%20a%20%5CDelta%20y)
![v_f^2 - (13.9)^2 = (2)(-9.81)(-19.1)](https://tex.z-dn.net/?f=v_f%5E2%20-%20%2813.9%29%5E2%20%3D%20%282%29%28-9.81%29%28-19.1%29)
![v_f^2 = 567.9](https://tex.z-dn.net/?f=v_f%5E2%20%3D%20567.9)
![v_f = 23.8 m/s](https://tex.z-dn.net/?f=v_f%20%3D%2023.8%20m%2Fs)
Part c)
Relative speed of two balls is given as
![v_{12} = v_1 - v_2](https://tex.z-dn.net/?f=v_%7B12%7D%20%3D%20v_1%20-%20v_2)
![v_{12} = (13.9) - (-13.9) = 27.8 m/s](https://tex.z-dn.net/?f=v_%7B12%7D%20%3D%20%2813.9%29%20-%20%28-13.9%29%20%3D%2027.8%20m%2Fs)
now the distance between two balls in 0.8 s is given as
![d = v_{12} t](https://tex.z-dn.net/?f=d%20%3D%20v_%7B12%7D%20t)
![d = 27.8 \times 0.8](https://tex.z-dn.net/?f=d%20%3D%2027.8%20%5Ctimes%200.8)
![d = 22.24 m](https://tex.z-dn.net/?f=d%20%3D%2022.24%20m)