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3 years ago

Cause and Effect: How do changes in temperature affect air? Pls hurry I’m being timed

Physics

2 answers:

expeople1 [14]3 years ago

7 0

When gas molecules are heated, the molecules move more quickly, and the increased velocity causes more collisions. As a result, more force is exerted on each molecule and air pressure increases. Temperature affects air pressure at different altitudes due to a disparity in air density.

Elan Coil [88]3 years ago

6 0

As the molecules heat and move faster, they are moving apart. So air, like most other substances, expands when heated and contracts when cooled. Because there is more space between the molecules, the air is less dense than the surrounding matter and the hot air floats upward.

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3 years ago

A moving particle encounters an external electric field that decreases its kinetic energy from 9970 eV to 6340 eV as the particl

marusya05 [52]

Answer:

q = -1.61x10⁻¹⁷ C

Explanation:

The charge of the particle can be found using the definition of the work done by electric force:

$W = -q\Delta V$ (1)

Where:

q: is the charge

ΔV: is the difference in electric potential

The work is also equal to:

$W = E_{p_{A}} - E_{p_{B}}$ (2)

Where:

$E_{p_{A}}$ and $E_{p_{B}}$ are the electric potential energy of the points A and B, respectively.

Now, by conservation of energy we have:

$K_{A} + E_{p_{A}} = K_{B} + E_{p_{B}}$ (3)

Where:

$K_{A}$ and $K_{B}$ are the kinetic energy of the points A and B, respectively.

Rearranging equation (3):

$K_{B} - K_{A} = E_{p_{A}} - E_{p_{B}}$

$K_{B} - K_{A} = W$

$K_{B} - K_{A} = -q\Delta V$

Solving the above equation for q:

$q = -\frac{K_{B} - K_{A}}{V_{B} - V_{A}} = -\frac{6340 eV - 9970 eV}{19.0 V - 55.0 V} = -100.83 e \cdot \frac{1.6 \cdot 10^{-19} C}{1 e} = -1.61 \cdot 10^{-17} C$

Therefore, the charge of the particle is -1.61x10⁻¹⁷ C.

I hope it helps you!

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3 years ago