1. <span>FX = 0 since the bottom charges are symmetrical about the y axis and the top charge is aligned with the y-axis
3. </span><span>FY > 0 since the bottom charges do not equate the top charge. The bottom charges have horizontal components which is less than q and adding them all, will not result to 3q.</span>
To get x on its own, you times the 3 over to the other side so the 3 cancels out on the LHS.
~ x greater than or equal to -18
(C)
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
Answer:
Explanation:
From the question we are told that:
Far point is 
Near point is 
Therefore
Focal Length
Generally the equation for the Lens is mathematically given by
Answer: The illusion of motion that occurs when a stationary object is first seen briefly in one location and, following a short interval, is seen in another location.
Explanation:
