Answer:
The speed of the piñata immediately after being cracked by the stick is
.
Explanation:
Using the conservation of linear momentum:
(1)
Here:
m(s) is the mass of the stick
m(p) is the mass of the piñata
v(is) is the initial velocity of the stick
v(fp) is the final velocity of the piñata
So, we just need to solve the equation (1) to v(fp).


I hope it helps you!
No sorry, wish I could help
<h2>
Answer:Protons interact in ways that electrons do not. ... Electrons are not affected by the strong force, and so they only get trapped by the electrical attraction to the nucleus which is much weaker in ionized atoms.</h2><h2 /><h2>
Explanation:Therefore it is easier for electrons to move away from one atom to another, transferring charge.</h2>
Answer:
The statement "if the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line" is false.
Explanation:
The equation for the magnetic force on a charge q moving at velocity v on a magnetic field B is given by the (vectorial) Lorentz Force Law 
From it we can clearly see that the <em>magnitude of the magnetic force </em>exerted on the particle is <em>proportional to the magnitude of the charge q and to the speed v of the particle</em>, and that it is also <em>perpendicular to the particle's velocity</em>. This means that at each instant it moves perpendicularly to the force, so <em>the work done by the magnetic force on the particle is zero</em>.
The statement "if the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line" is false not only for this but for any force, a force always perpendicular to a velocity will curve the trajectory.