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3 years ago

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A 65-kg person stands on a scale in a moving elevator while holding a 5.0 kg mass suspended from a massless spring with spring c

onstant 1.08 kN/m . None of the objects in the elevator is moving relative to the elevator, but the spring is stretched by 6.0 cm .?

1 answer:

disa [49]3 years ago

7 0

Answer:

The question is incomplete. However, I believe, it is asking for the acceleration of the elevator. This is 3.16 m/s².

Explanation:

By Hooke's law, $F = ke$

F is the force on a spring, k is the spring constant and e is the extension or compression.

From the question,

$F = (1.08\text{ kN/m}) \times (6.0 \times 10^{-2}\text{ m}) = 64.8 \text{ N}$

This is the force on the mass suspended on the spring. Its acceleration, a, is given by

$F = ma$

$a = \dfrac{F}{m}$

$a = \dfrac{64.8 \text{ N}}{5\text{ kg}} = 12.96\text{ m/s}^2$

This acceleration is more than the acceleration due to gravity, g = 9.8 m/s². Hence the elevator must be moving up with an acceleration of

12.96 - 9.8 m/s² = 3.16 m/s²

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Answer:

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Pepsi [2]

Answer:

$\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}$

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Time in which train will come to rest (t).

Explanation:

$\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a: \\ \sf \implies 0 = 30 + ( - 1.5)(t) \\ \sf \implies 0 = 30 - 1.5(t) \\ \sf \implies 30 - 1.5(t) = 0 \\ \\ \sf Subtract \: 30 \: from \: both \: sides: \\ \sf \implies (30 - \boxed{ \sf 30}) - 1.5(t) = \boxed{ \sf - 30} \\ \\ \sf 30 - 30 = 0 : \\ \sf \implies - 1.5(t) = - 30 \\ \\ \sf Divide \: both \: sides \: of \: - 1.5(t) = - 30 \: by \: - 1.5 : \\ \sf \implies \frac{ - 1.5(t)}{ \boxed{ \sf - 1.5}} = \frac{ - 30}{ \boxed{ \sf -1.5 }} \\ \\ \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}} = 1 : \\ \sf \implies t = \frac{ - 30}{ - 1.5} \\ \\ \sf \frac{ - 30}{ - 1.5} = \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}} = 20 : \\ \sf \implies t = 20 \: sec$

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