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3 years ago

Which best explains why a wood burning fireplace represents an open system

Physics

2 answers:

REY [17]3 years ago

5 0

Answer:

Explanation:

An open system is defined as the a system in which energy or mass can be gained or loss from the environment.

Here in this case a wood burning fireplace represents an open system as it is directly taking oxygen from the atmosphere.

During this direct exchange of gases heat is produced hence mass of wood is lost and oxygen is gained from the atmosphere.

So as fire is required from atmosphere to burn the wood, it best explains why a wood burning represent an open system.

-Dominant- [34]3 years ago

4 0

Because a fire requires oxygen from the atmosphere to burn

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An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

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Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

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3 years ago

If a farsighted person has a near point that is 0.600 mm from the eye, what is the focal length f2f2f_2 of the contact lenses th

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Answer:

0.22mm

Explanation:

A far sighted person is a person suffering from long sightedness i.e such individual can only see far distant object clearly but not near distant object. The defect is corrected using convex lens.

Since convex lens is used, the focal (f) length of the lens is positive and the image distance (v) is also positive.

Using the lens formula,

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f = 1/4.53

f = 0.22mm

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