What is the vertical component of its velocity at launch? v0y = 20*sin(53) = 16 m/s
Its horizontal component of velocity? v0x = 20*cos(53) = 12 m/s Neglect icing air friction, which of these components remains constant throughout the flight path? The horizontal component. (the vertical component is affected by gravity)
Which of these components determines the projectile's time in the air? The vertical component. (because the projectile falls down after t = 2*v0*sin(53)/g, you can see it doesn't depend on the horizontal component)
C.
Because it’s falling it has acceleration in the y direction. If you have acceleration, you usually also have velocity, and since kinetic energy is KE= Mv^2 you know you have it. It also has potential energy because it has some height to it, and PE= Mgh.
Answer:
The electric potential is approximately 5.8 V
The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero
Explanation:
The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:
(1)
where is the charge of the particle, the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:
Substituting the values , and we obtain:
The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.