I think we will use the law of conservation of linear momentum;
M1V1 = M2V2
M1 = 4 kg (mass of the water balloon launcher)
V1=?
M2= 0.5 kg ( mass of the balloon)
V2 = 3 m/s
Therefore; 4 V1 = 0.5 × 3
4V1= 1.5
V1= 1.5/4
= 0.375 m/s
Answer: I don't know this one but I'm just came here for points
Explanation:
The ball's vertical velocity at the time it just passes over the goal is 0 m/s. Its initial vertical velocity is unknown and we denote it by 
, where 
here is the ball's initial speed. Vertically, the only force acting on the ball is gravity, which attributes a downward acceleration of 9.8 m/s^2. We expect the maximum height achieved by the ball to be 2.4 m, so we can find the initial speed by solving
Answer:
the switching circuitry is opened the soft iron armature is replaced to tasty le macha the switching circuitry is closed explain what happens when the switch circuit is a wonder you know about the characteristics of open to close this which right take the example of an electric generator for an electric motor which rotates in half rotation and change the armature and just try it ok I am sorry
4. The Coyote has an initial position vector of 
.
4a. The Coyote has an initial velocity vector of 
. His position at time 
is given by the vector
where 
is the Coyote's acceleration vector at time . He experiences acceleration only in the downward direction because of gravity, and in particular 
where 
. Splitting up the position vector into components, we have 
with
The Coyote hits the ground when 
:
4b. Here we evaluate 
at the time found in (4a).
5. The shell has initial position vector 
, and we're told that after some time the bullet (now separated from the shell) has a position of 
.
5a. The vertical component of the shell's position vector is
We find the shell hits the ground at
5b. The horizontal component of the bullet's position vector is
where 
is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for :
