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MissTica
3 years ago
14

light spring of force constant k = 158 N/m rests vertically on the bottom of a large beaker of water (Figure a). A 4.36-kg block

of wood (density = 650 kg/m3) is connected to the spring, and the block-spring system is allowed to come to static equilibrium (Figure b). What is the elongation ΔL of the spring?
Physics
1 answer:
Fantom [35]3 years ago
4 0

Answer:

0.146 m

Explanation:

f = -KΔL according to Hooke's law

volume of water displaced = mass / density of block since a body will displace equal volume of its own

weight of water displaced = mass of water × acceleration due to gravity

and mass of water = volume of water / density of water

weight of water displaced = Vw × dw × g = mg (dw / dblock)

net force = mg - mg (dw / dblock) = 42.728  - 65.74 = -23.00

it will be balanced by a restoring force of 23 N

ΔL = F / k = 23 / 158 = 0.146 m

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La masa de la mascota es de 11,24 kg.

Explanation:

Dado que la masa total de = 80 kg

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La mascota salta horizontalmente con una velocidad, v, de 6.0 m / s para dar un movimiento de 1 m a la caja de masa y a la persona

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La energía del movimiento del perro = 1/2 × m × v² = Trabajo realizado al saltar de la mascota

1/2 × m × 6² = (80 - m) × 9.81 × 0.3

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m = 235440/20943 = 11,24 kg

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Answer:

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