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ASHA 777 [7]
3 years ago
13

PLEASE HELP!!!!

Physics
2 answers:
sveticcg [70]3 years ago
7 0
<h3><u>Answer and explanation</u>;</h3>
  • Energy of a shear wave relates to the direction of a particle motion is <u>upwards and downwards</u>.
  • <em><u>Shear wave or Transverse wave is a type of wave in which particles of the medium move in the direction perpendicular to the direction of the wave motion. </u></em>
  • The energy is transmitted from the source to other points in the medium. As the energy is being transported, the particles in the medium will be displaced upwards and downwards. The particles will move perpendicular to the direction of wave motion.
Gala2k [10]3 years ago
3 0
<span>Energy of a shear wave relates to the direction of a particle motion is downwards.</span>
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A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its veloci
Vlad [161]

Answer:

Approximately \rm 2.5\; m \cdot s^{-1}.

Explanation:

Let the increase in the rocket's velocity be \Delta v. Let v_0 represent the initial velocity of the rocket. Note that for this question, the exact value of  v_0 doesn't really matter.

The momentum of an object is equal to its mass times its velocity.

  • Mass of the rocket with the 5 kg of fuel: 1000.
  • Initial velocity of the rocket and the fuel: v_0.
  • Hence the initial momentum of the rocket: 1000\,v_0.
  • Mass of the rocket without that 5 kg of fuel: 1000 - 5 = 995.
  • Final velocity of the rocket: v_0 + \Delta v.
  • Hence the final momentum of the rocket: 995\,(v_0 + \Delta v).
  • Mass of the 5 kg of fuel: 5.
  • Final velocity of the fuel: v_0 - 500 (assuming that the the 500 m/s in the question takes the rocket as its reference.)
  • Hence the final momentum of the fuel: 5\,(v_0 - 500).

Momentum is conserved in an isolated system like the rocket and its fuel. That is:

Sum of initial momentum = Sum of final momentum.

1000\,v_0 = 995\,(v_0 + \Delta v) + 5\,(v_0 - 500).

Note that 1000\, v_0 appears on both sides of the equation. These two terms could hence be eliminated.

0 = 995\, \Delta v - 5\times 500.

\displaystyle \Delta v = \frac{5}{995}\times 500 \approx \rm 2.5\; m \cdot s^{-1}.

Hence, the velocity of the rocket increased by around 2.5 m/s.

5 0
3 years ago
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