Answer:
The speed and direction of the two players immediately after the tackle are 3.3 m/s and 53.4° South of West
Explanation:
given information:
mass of fullback, 
= 92 kg
speed of full back, 
= 5.8 to south
mass of lineman, 
=110 kg
speed of lineman, 
= 3.6
according to conservation energy,
assume that the collision is perfectly inelastic, thus
initial momentum = final momentum
= 
'
m₁v₁ = (m₁+m₂)'
' = m₁v₁/(m₁+m₂)
= (92) (5.8)/(92+110)
= 2.64 m/s
= 
'
m₂v₂ = (m₁+m₂)'
' = m₁v₁/(m₁+m₂)
= (110) (3.6)/(92+110)
= 1.96 m/s
thus,
' = √'²+'²
= 3.3 m/s
then, the direction of the two players is
θ = 90 - tan⁻¹('/')
= 90 - tan⁻¹(1.96/2.64)
= 53.4° South of West
Divide the change in speed by the time for the change.
By Snell's law:
η = sini / sinr. i = 25, η = 1.33
1.33 = sin25° / sinr
sinr = sin25° / 1.33 = 0.4226/1.33 = 0.3177 Use a calculator.
r = sin⁻¹(0.3177)
r ≈ 18.52°
Option A.
God's grace.
Since the y axis stayed consistent, we can assume it did not move at all.
(So your answer would be A)
Answer:
In physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion
In classical mechanics, the gravitational potential at a location is equal to the work (energy transferred) per unit mass that would be needed to move an object to that location from a fixed reference location. It is analogous to the electric potential with mass playing the role of charge. The reference location, where the potential is zero, is by convention infinitely far away from any mass, resulting in a negative potential at any finite distance.
In mathematics, the gravitational potential is also known as the Newtonian potential and is fundamental in the study of potential theory. It may also be used for solving the electrostatic and magnetostatic fields generated by uniformly charged or polarized ellipsoidal bodies
