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Sergio039 [100]
3 years ago
12

Simple Harmonic Motion

Physics
1 answer:
Delvig [45]3 years ago
7 0

- The net force is greatest at the position of maximum displacement

- The net force is zero when at the equilibrium position

Explanation:

The motion of a spring is a Simple Harmonic Motion, in which the displacement of the end of the spring is given by a periodic function of the form

x=Asin (\omega t)

where A is the amplitude (the maximum displacement), and \omega the angular frequency of the motion.

We can analyze the net force acting on the spring by looking at Hooke's law:

F=kx

where

F is the net force

k is the spring constant

x is the displacement

From the equation, we notice immediately that:

  • The net force is the greatest when the displacement x is the greates, so at the position in which the spring has maximum compression or stretching
  • The net force is zero when the displacement x is zero, so when the spring crosses the equilibrium position

Learn more about forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

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Answer:

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= 10000/300=33.33m/s

Explanation:

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A cannon of mass 6.43 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 7
Nata [24]

Answer:

The velocity of the shell when the cannon is unbolted is 500.14 m/s

Explanation:

Given;

mass of cannon, m₁ = 6430 kg

mass of shell, m₂ = 73.8-kg

initial velocity of the shell, u₂ = 503 m/s

Initial kinetic energy of the shell; when the cannon is rigidly bolted to the earth.

K.E = ¹/₂mv²

K.E = ¹/₂ (73.8)(503)²

K.E = 9336032.1 J

When the cannon is unbolted from the earth, we apply the principle of conservation of linear momentum and kinetic energy

change in initial momentum = change in momentum after

0 = m₁u₁ - m₂u₂

m₁v₁ = m₂v₂

where;

v₁ is the final velocity of cannon

v₂ is the final velocity of shell

v_1 = \frac{m_2v_2}{m_1}

Apply the principle of conservation kinetic energy

K = \frac{1}{2}m_1v_1^2 +  \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_1(\frac{m_2v_2}{m_1})^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_2v_2^2(\frac{m_2}{m_1}) + \frac{1}{2}m_2v_2^2 \\\\K = \frac{1}{2}m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\2K = m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\v_2^2 = \frac{2K}{M_2(\frac{m_2}{m_1} + 1)} \\\\v_2^2 = \frac{2*9336032.1}{73.8(\frac{73.8}{6430} + 1)}\\\\

v_2^2 = 250138.173\\\\v_2 = \sqrt{250138.173} \\\\v_2 = 500.14  \ m/s

Therefore, the velocity of the shell when the cannon is unbolted is 500.14 m/s

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Please see the attached file.

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