Answer:
option B
Explanation:
given,
diameter of the rotating space = 2 Km
Force exerted at the edge of the space = 1 g
force experienced at the half way = ?
As the object is rotating in the circular part
Force is equal to centripetal acceleration.
at the edge
g = ω² r
ω is the angular velocity of the particle
r is the radius.
now, acceleration at the half way
g' = ω² r'
People at the halfway experience g/2
hence, the correct answer is option B
The rate of acceleration of the crate would be 1 m/s^2 because the equation for force is F=ma and when you plug in your numbers you get 10=10a so a=1
Maybe you can split up the questions. I will try to answer your first question.
1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)
2. Momentum: p = m₁v₁ + m₂v₂
m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B
Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0
Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0
3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.
4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7
5.You figure out.
Answer:
N = 337.96 N
Explanation:
∅ = 32º
F = 249 N
m = 21 Kg
N = ?
We can apply:
∑ F = 0 (↑)
- Fy - W + N = 0 ⇒ N = Fy + W
⇒ F*Sin ∅ + m*g = N
⇒ N = (249 N*Sin32º) + (21 Kg*9.81 m/s²)
⇒ N = 337.96 N (↑)
