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earnstyle [38]
3 years ago
12

On Earth, a spring stretches by 5.0 cm when a mass of 3.0 kg is suspended from one end.

Physics
2 answers:
Neko [114]3 years ago
8 0

Answer:

Mass = 18.0 kg

Explanation:

From Hooke's law,

F = ke

where: F is the force, k is the spring constant and e is the extension.

But, F = mg

So that,

mg = ke

On the Earth, let the gravitational force be 10 m/s^{2}.

3.0 x 10 = k x 5.0

30 = 5k

⇒ k = \frac{30}{5} ................ 1

On the Moon, the gravitational force is \frac{1}{6} of that on the Earth.

m x \frac{10}{6} = k x 5.0

\frac{10m}{6} = 5k

⇒ k = \frac{10m}{30} ............. 2

Equating 1 and 2, we have;

\frac{30}{5}  = \frac{10m}{30}

m = \frac{900}{50}

    = 18.0

m = 18.0 kg

The mass required to produce the same extension on the Moon is 18 kg.

riadik2000 [5.3K]3 years ago
4 0

Answer:

18 kg

Explanation:

  • weight (N) = mass (kg) × gravitational acceleration (m/s²)
  • force (N) = k (spring constant) × extension (m)

On Earth, acceleration of gravity is 10 m/s²

  • weight = 3.0 (kg) × 10 (m/s²)
  • weight = 30 (N)

Since weight is a force, the force is 30 N. The value of spring constant is unknown

  • 30 (N) = k × 5 (m)
  • k = 6 (m/N)

Spring constant is 6. Now let's find the mass on the Moon

  • mass (kg) × gravitational acceleration (m/s²) = k (spring constant) × extension (m)

Gravitational acceleration of the moon is 1/6 of that on Earth. Earth's g = 10, so Moon's g = 10/6

  • m × 10/6 = 6 × 5
  • m = 30/(10/6)
  • m = 18

The mass is 18 kg

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Answer:

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Explanation:

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\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}.

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Substitute this value into the Snell's Law equation:

\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}.

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The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For 0 < \theta < 90^{\circ}, \sin(\theta) is monotonically increasing with respect to \theta. In other words, for \!\theta in that range, the value of \sin(\theta)\! increases as the value of \theta\! increases.

Therefore, compared to the medium in this question with \theta_{c} = 27^{\circ}, the medium with the larger critical angle \theta_{c} = 32^{\circ} would have a larger \sin(\theta_{c}). such that light would travel faster in that medium.

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