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dimulka [17.4K]

3 years ago

11

What is the relationship between the internal energy of a substance and its state of matter? A) As a gas loses internal energy i

t becomes a plasma. B) Closer packed atoms generally have more internal energy. C) As the internal energy decreases a substance will spread further apart. D) As the internal energy increases a substance would go from solid to a liquid.

2 answers:

vodomira [7]3 years ago

8 0

Answer: Option (D) is the correct answer.

Explanation:

Internal energy is defined as the rapid and disordered notion of particles of an object or substance.

This means that particles of the object or substance has high kinetic energy.

Therefore, this energy will lead to change in state of matter.

Hence, we can conclude that the relationship between the internal energy of a substance and its state of matter is that as the internal energy increases a substance would go from solid to a liquid.

babymother [125]3 years ago

4 0

D)<span>As the internal energy increases a substance would go from solid to a liquid.</span>

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A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

ivolga24 [154]

Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

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Angle, $\theta = 50^{\circ}$

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Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

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Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

$\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}$

$\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}$

$\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km$

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$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

$\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}$

$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

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$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

$\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}$

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