hmax = 5740.48 m. The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.
This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.
V₀ = 420m/s and θ₀ = 53.0°
So, when the cannonball is fired it has horizontal and vertical components:
V₀ₓ = V₀ cos θ₀ = (420m/s)(cos 53°) = 252.76 m/s
V₀y = V₀ cos θ₀ = (420m/s)(cos 53°) = 335.43m/s
When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:
Vy = V₀y - g tₐ = 0
tₐ = V₀y/g
tₐ = (335.43m/s)/(9.8m/s²) = 34.23s
Then, the maximum height is reached in the instant tₐ = 34.23s:
h = V₀y tₐ - 1/2g tₐ²
hmax = (335.43m/s)(34.23s)-1/2(9.8m/s²)(34.23s)²
hmax = 11481.77m - 5741.29m
hmax = 5740.48m
The answer is the last choice.
Its electrical potential energy stays the same because it has the same electric potential. The reason why is that moving the charge towards X does not change the distance of the negative charge between the plates. The Electrical potential energy of a particle is the result energy by virtue of its position from the electrical fields produce by the plates both positive and negative. Since the charge is still equidistant to each other (assuming based from the diagram) no change in terms of electrical energy consumption or work was done.
That could be an electromagnet,
and it could also be an solenoid.
describes the volumetric flow rate (
) through tubular sections.
represent the injection pressure difference, the length of the section, the radius of the section and the viscosity index of the fluid that flows through the section respectively.