Answer: I am not sure if you wanted me to answer this or not.
Explanation:
Answer:
let's say you're on a bus going 50 km/hr, you are moving at a velocity of zero relative to the bus. however, relative to the ground you are moving at the same velocity as the bus.
Explanation:
physics
Explanation:
It is given that, two teams are playing tug of war.
Force applied by Team A, 
Force applied by Team B, 
We need to find the net force acting on the rope. It is equal to :



So, the net force acting on the rope is 35 N and it is acting toward right. Hence, this is the required solution.
Answer:
El mango llega al suelo a una velocidad de 329.982 metros por segundo.
Explanation:
El mango experimenta un movimiento de caída libre, es decir, un movimiento uniformemente acelerado debido a la gravedad terrestre, despreciando los efectos de la viscosidad del aire y la rotación planetaria. Entonces, la velocidad final del mango, es decir, la velocidad con la que llega al suelo, se puede determinar mediante la siguiente fórmula cinemática:
(1)
Donde:
- Velocidad inicial, en metros por segundo.
- Velocidad final, en metros por segundo.
- Aceleración gravitacional, en metros por segundo al cuadrado.
- Tiempo, en segundos.
Si sabemos que
,
y
, entonces la velocidad final del mango es:



El mango llega al suelo a una velocidad de 329.982 metros por segundo.
Hi there!
A.
Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.
Thus, the time to its highest point:

Now, we can determine the velocity at which the can was launched at using the following equation:

In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.
Therefore:

***vsinθ is the vertical component of the velocity.
Solve for 'v':

Now, recall that:

Plug in the expression for velocity:

B.
We can use the same process as above, where T' = 2T and Th = T.

C.
The work done in part B is 4 times greater than the work done in part A.
