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
The acceleration of the wagon along the ground is 3.6 m/s².
To solve the problem above, we need to use the formula of acceleration as related to force and mass.
Acceleration: This can be defined as the rate of change of velocity.
⇒ Formula:
- Fcos∅ = ma................. Equation 1
⇒ Where:
- F = Force
- ∅ = angle above the horizontal
- m = mass of the wagon
- a = acceleration of the wagon
⇒ make a the subject of equation 1
- a = Fcos∅/m..................... Equation 2
From the question,
⇒ Given:
⇒ Substitute these values into equation 2
- a = 44(cos35°)/10
- a = 44(0.8191)/10
- a = 3.6 m/s²
Hence, The acceleration of the wagon along the ground is 3.6 m/s²
Learn more about acceleration here: brainly.com/question/9408577
A is the answer I really don’t know the answer to that but if u can u can help me on my work
Answer:
false
Explanation:
discovered colours of the rainbow
