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Zanzabum
2 years ago
6

If there is 8 g of a substance before a physical change, how much will there be afterwards?

Physics
2 answers:
grigory [225]2 years ago
8 0
There will still be 8g due to the law of conservation of mass. An object undergoing a physical change (change in state, color change..) will have the same mass before and after.
Lerok [7]2 years ago
3 0

Answer:

There will be 8 g.

Explanation:

Physical changes keeps intact the molecular structure of substances.

A typical example of physical change is changing the state of matter of a substance: although its density, temperature or shape may vary, the mass will remain the same.

Back in the 18th century, Antoine Lavoisier, noticed that, in physical process, non mass nor energy went missing before and after.

<em>"Nothing is lost, nothing is created, everything is transformed"</em>

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<span>Like charges repel and opposite charges attract.
The further away two charged objects are the weaker the electrical force between them.
The closer two charged objects are the stronger the electrical force between them.
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The table shows data for the planet Uranus. A 2 column table with 4 rows. The first column is labeled Quantity with entries, Esc
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The answer is 218

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a vehicle travels at a constant speed of 65 mph for 4 hours how far has this vehicle travelled in this time
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3 years ago
Where did we use rotational and irrotational flow​
WARRIOR [948]

Answer:

The term rotational and irrotational flow is associated withe the flow of particles in fluid.

The common example of irrrotational flow can be seen on the carriages of the Ferris wheel (giant wheel).

Explanation:

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  • Similarly if fluid particles do not rotate along its axis while flowing in a stream line flow then it is considered as the irrotational flow.
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3 0
3 years ago
A 544 g ball strikes a wall at 14.3 m/s and rebounds at 14.4 m/s. The ball is in contact with the wall for 0.042 s. What is the
Nezavi [6.7K]

Answer:

F = 371.738\,N

Explanation:

Let assume that ball strikes a vertical wall in horizontal direction. The situation can be modelled by the appropriate use of the definition of Moment and Impulse Theorem, that is:

(0.544\,kg)\cdot (14.3\,\frac{m}{s})-F\cdot \Delta t = -(0.544\,kg)\cdot (14.4\,\frac{m}{s} )

F\cdot \Delta t = 15.613\,\frac{kg}{m\cdot s}

The average force acting on the ball during the collision is:

F = \frac{15.613\,\frac{kg}{m\cdot s} }{0.042\,s}

F = 371.738\,N

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3 years ago
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