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3 years ago

Which statement best describes an isolated system?

Physics

2 answers:

NARA [144]3 years ago

4 0

Answer:

C. Neither energy nor matter are exchanged with the surroundings.

Explanation:

Isolated system is a system that is completely cut off from its surroundings.

In an isolated system, both energy and matter are said to be conserved.

It cannot release energy to its surroundings or it cannot take in energy from the surroundings. Apart from energy, even matter cannot be exchanged .

For example if there is a hot box which is insulated and some contents like hot water or milk are poured into it, the heat energy of the water/milk cannot be lost to the surroundings and also it will not take in any heat energy from the surroundings; and the actual amount of water/milk content doesn't change. Such a system is called an isolated system.

charle [14.2K]3 years ago

3 0

It is c
because Mrs.Brown told us in class that <span>Neither energy nor matter are exchanged with the surroundings.</span>

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3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

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Answer:

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Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

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$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

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$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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