Answer:
- <em>Abbie’s acceleration is (1/2) Zak’s acceleration.</em>
Explanation
1. <u>Data</u>:
a) ω = constant
b) Abbie: r₁ = 1 m
c) Zak: r₂ = 2 m
d) Ac₁ = ? Ac₂
2. <u>Formulae</u>
3. <u>Solution</u>:
a) Abbie:
b) Zack:
c) Divide Ac₁ / Ac₂
- Ac₁ / Ac₂ = ω² (1m) / [ω² (2m) ] = 1/2
⇒ Ac₁ = (1/2) Ac₂ = Ac₂ / 2 = 0.5 Ac₂
Answer:
I believe the correct answer would be A :)
Explanation:
<u>Momentum</u>
- a vector quantity; has both magnitude and direction
- has the same direction as object's velocity
- can be represented by components x & y.
Find linebacker momentum given m₁ = 120kg, v₁ = 8.6 m/s north
P₁ = m₁v₁
P₁ = (120)(8.6)
[ P₁ = 1032 kg·m/s ] = y-component, linebacker momentum
Find halfback momentum given m₂ = 75kg, v₂ = 7.4 m/s east
P₂ = m₂v₂
P₂ = (75)(7.4)
[ P₂ = 555 kg·m/s ] = x-component, halfback momentum
Find total momentum using x and y components.
P = √(P₁)² + (P₂)²
P = √(1032)² + (555)²
[[ P = 1171.77 kg·m/s ]] = magnitude
!! Finally, to find the magnitude of velocity, take the divide magnitude of momentum by the total mass of the players.
P = mv
P = (m₁ + m₂)v
1171.77 = (120 + 75)v <em>[solve for v]</em>
<em />v = 1171.77/195
v = 6.0091 ≈ 6.0 m/s
If asked to find direction, take inverse tan of x and y components.
tanθ = (y/x)
θ = tan⁻¹(1032/555)
[ θ = 61.73° north of east. ]
The magnitude of the velocity at which the two players move together immediately after the collision is approximately 6.0 m/s.
Answer:
Explanation:
According to the free body diagram, in this case, we have:
Recall that the force of friction is given by:
Replacing and solving for the coefficient of kinetic friction:
We have an uniformly accelerated motion. Thus, the acceleration is defined as:
Finally, we calculate 
:
