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 equivalent distance in kilometers is 4012 × km.
Explanation:
It's known that 1 millimeter converts to meters. Then, 1 meter converts to
kilometers. Therefore, the conversion for 1 millimeter to kilometers can be stated as
1 mm = m
1 m = km
Thus, 1 mm = ×
km =
km.
Given the distance of 4012 mm, the corresponding distance in kilometers will be
4012 mm = 4012 × km.
The distance therefore is 4012 × km.
Answer:
Height (h) = 17 m
Velocity (v) = 18.6 m/s
Explanation: This problem can be solved using kinematic motion equations.
Given Data
Initial velocity (u) = 0
Acceleration (a) = g
Time (t) = 1.9 seconds
First, we calculate the height.
Then, we find the final velocity
The acceleration graph is a linear representation described by y=9.8, as it remains constant:
The velocity graph can be represented by y=9.8x (where y signifies velocity and x indicates time):
The displacement graph can be described as y=4.9x^2 (with x as time and y as displacement):
These graphs apply exclusively from x=0 to x=1.9, so disregard other sections of the graphs.
