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
First, you make a diagram of all the forces acting on the system. This is shown in the figure. We have to determine F1 and F4. Let's do a momentum balance. Momentum is conserved so the summation of all momentum is equal to zero. Momentum is force*distance.
To determine F1: (reference is F4, so F4=0)
∑Momentum = 0 = -F2 - F3 + F1
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.3m)+F1(0.5-0.1m)
F1 = 53.96 N (left knife-edge)To determine F4: (reference is F1, so F1=0)
∑Momentum = 0 = -F2 - F3 + F4
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.2m)+F4(0.5-0.1m)
F4 = 68.67 N (right knife-edge)
Answer:9.17 m/s^2
Explanation:
mass=1200kg
Force=11 x 10^3 N
Acceleration=force ➗ mass
Acceleration=11 x 10^3 ➗ 1200
Acceleration=9.17
Acceleration=9.17 m/s^2
Answer:
Explanation:
From the question we are told that:
Frictional force 
Coefficient of kinetic friction 
Generally the equation for Normal for is mathematically given by
Therefore
