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An inverted garbage can of weight W is suspended in air by water from a geyser. The water shoots up from the ground with a speed v0, at a constant rate K kg/s. The problem is to find the maximum height at which the garbage can rides. Neglect the effect of the water falling away from the garbage can.

Respuesta :

Answer:

Your clue is impossible, water release at a velocity of 20m/s cannot lift a 10 kg garbage can under earth's gravity.

F=lambda 2v^2 (assuming elastic collision)

= 2mv^2/d

= 2(dm/dt)(v^2)/u

= 20N

which can lift a can of 20/9.81 kg

so your water can only lift a max of 2.04 kg garbage can.

Explanation:

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