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earnstyle [38]
2 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]2 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]2 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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5 0
3 years ago
A cannonball is fired perfectly horizontally from the top of a 210 m tall cliff. It is fired with an initial velocity of 50 m/s.
pochemuha

Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

Explanation:

Given;

height of the cliff, h = 210 m

initial horizontal velocity of the cannonball, Ux = 50 m/s

initial vertical velocity of the cannonball, Uy = 0

The time for the cannonball to reach the ground is calculated as;

h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t  = 6.55 \ s

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8 0
3 years ago
How does a parallel circuit differ from a series circuit?
8_murik_8 [283]
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3 years ago
A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.
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Answer:

t = 6.09 seconds

Explanation:

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Distance, d = 269 cm

We need to find the time interval of the marble. Speed is distance per unit time.

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Hence, the time interval of the marble is 6.09 seconds.

6 0
3 years ago
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VikaD [51]

Answer:

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Explanation:

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