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solong [7]
3 years ago
14

A 2 kg mass is attached to a vertical spring and the spring stretches 15 cm from its original relaxed position.

Physics
1 answer:
Naya [18.7K]3 years ago
5 0

Answer:

A. 130.7 N/m

Explanation:

By Hook's law

Force(F) = Spring constant(k) × Extention (d)

F = k× d ⇒ k = F/d

To find Force(F) you know it is the weight of the object. Considering g = 9.8 ms⁻²

                   k  = 2×9.8/0.15 = 130.7 Nm⁻¹

Hooke’s law,

for small deformations of an object, the displacement is directly proportional to the deforming force or load.

Under these conditions the object returns to its original shape and size upon removal of the load.

The deforming force may be applied to a solid by stretching, compressing,etc. so a spring shows an elastic behavior according to Hooke’s law

Mathematically,

applied force( F) = constant (k)* displacement in length (x)

F = kx.

The value of k depends

  • on the kind of elastic material
  • on its dimensions and shape.
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Answer:

lead with Z = 82 is transformed into Bismuth with Z = 73

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Explanation:

Beta decay occurs when a neutron emits an electron and an anti neutrino from the atomic nucleus, therefore the atomic number of the material increases by one unit.

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lead with Z = 82 is transformed into Bismuth with Z = 73

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A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at
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Answer:

Explanation:

Given

mass of drop m=2 kg

height of fall h=1 m

ball leaves the foot with a speed of 18 m/s at an angle of 55^{\circ}

Velocity of ball just before the collision with the floor

u^=2gh

u=\sqrt{2gh}

u=\sqrt{2\times 9.8\times 1}=4.42 m/s

Impulse delivered in Y direction

J_y=m(v\sin (55)-(-u))

J_y=2(18\sin (55)+4.42)

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Impulse in x direction

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