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Anon25 [30]
4 years ago
8

What is the star cycle?

Physics
1 answer:
Serjik [45]4 years ago
5 0
♥ A star cycle is the "life cycle" of a star, like the phases it goes through and more.
Similar to humans, we have a life cycle.
Everyone and everything has it. <span />
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The record of an earthquake obtained from a seismic instrument is a(n) ________.
serious [3.7K]
<span>The record of an earthquake obtained from a seismic instrument is a </span>
SEISMOGRAM 
6 0
3 years ago
An incidence that is caused by sleep and also occurs during sleep is known as a __________.
Setler79 [48]

Answer:

parasomnia

Explanation:

6 0
3 years ago
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List the names, charges and locations of the three kinds of particles that make up an atom
Marina86 [1]

The atom contains three kinds of particles:

  1. <u>Electrons</u>: They are negative charged particles with charge of 1.6×10^{-19} C and they spin around the nucleus in specific orbits
  2. <u>Protons</u>: They are positive charged particles with the same charge of the electrons (but in positive sign) and they are located inside the nucleus
  3. <u>Neutrons</u>:  They are uncharged particles and they are located inside the nucleus with the protons

Then the atom contains positive particles and negative particles, it is electrically neutral.

5 0
4 years ago
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

3 0
3 years ago
A box is pulled with a horizontal force of 500N and moves 5m. What is the work done?
shepuryov [24]
The formula for work is N x m, which will equal in joules.
500 x 5 = 2,500 joules
8 0
4 years ago
Read 2 more answers
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