F=ma
Therefore the net force = 1000kg × 2 metres per second per second
So F=2000 N
Answer:
v₁f = 0.5714 m/s (→)
v₂f = 2.5714 m/s (→)
e = 1
It was a perfectly elastic collision.
Explanation:
m₁ = m
m₂ = 6m₁ = 6m
v₁i = 4 m/s
v₂i = 2 m/s
v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i
v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)
v₁f = 0.5714 m/s (→)
v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i
v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)
v₂f = 2.5714 m/s (→)
e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1
It was a perfectly elastic collision.
Answer:
ω = 0.05 rad/s
Explanation:
We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,
where,
ω = angular velocity of cylinder = ?
g = required acceleration = 9.8 m/s²
r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m
Therefore,
<u>ω = 0.05 rad/s</u>
In Physics, 'work' has a very clear definition:
It's (strength of a force) times (distance through which the force acts).
'Work' has the units of Energy.
If you push against a shopping cart with 30 newtons of force, and
you keep pushing while the cart moves 4 meters, then you have
done (30 x 4) = 120 newton-meters of work = 120 "Joules".
