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Marrrta [24]
3 years ago
14

What matter does not take up the space in the container?

Physics
1 answer:
son4ous [18]3 years ago
5 0

Answer:

I think it is solids

Explanation:

gaseous materials take any space in any container they are stored in the same is for liquids.

But solids can not take up the space of a contractor because they have a definite shape .

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El conductor de un coche de 650 kg que va a 90 km/h frena y reduce su
Ivanshal [37]
A. 90km/h = 25 m/s
W=1/2mv^2
=1/2* 650* 25^2
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b. v’= 90-50= 40km/h
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3 years ago
Define relative density​
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3 years ago
Why the distance between two loudspeakers have to be large??​
tangare [24]

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3 years ago
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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