A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio
1 answer:
Answer:
Explanation:
According to the parameters provided,
mass of the clay lump, m₁ = 0.05 kg
initial velocity of the lump, u₁ = 12 m/s
mass of the cart, m₂ = 0.15 kg
initial speed of the cart, u₂ = 0
As the clay adheres to the cart, we have an inelastic collision scenario. Let v represent the combined speed of both the cart and lump post-collision. Given that momentum is conserved, we have:
The resultant speed is v = 3 m/s.
Thus, the final speed of both cart and lump following the collision is 3 m/s. This concludes the solution.
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Answer:
the lump descends
Explanation:
The full articulation of the query is
Upon fully submerging a 3 kg lump of material in a certain fluid, the fluid that would have occupied the space now filled by the lump weighs 2 kg. (a) When released, does the lump float up, sink, or remain steady
(a)
= Buoyant force acting upward on the lump
= mass of irregular lump = 3 kg
= mass of fluid displaced = 2 kg
The upward buoyant force on the lump is given by the weight of the displaced fluid, thus
the weight of the irregular lump of material is represented as
Given that the weight of the lump downward exceeds the upward buoyant force, the lump will indeed descend