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Virty [35]
3 years ago
8

A. Which of the four thermal processes describes the pressure-volume relationship at a constant temperature?

Physics
2 answers:
ANTONII [103]3 years ago
8 0

Answer:

Boyle's law, which describes the relationship between the pressure and the volume of a gas when the temperature remains constant.

Explanation:

Boyle's law says that the pressure of a gas in a closed container is inversely proportional to the volume of the container that contains it, when the temperature is constant. This means that if the pressure increases, the volume decreases, and if the pressure decreases, the volume increases.

The formula that describes this law is as follows:

P1 * V1 = P2 * V2

Where P1 is the gas pressure in the initial state, V1 is the volume of the gas in the initial state and P2 and V2 is the pressure and volume in the final state.

zhannawk [14.2K]3 years ago
4 0
Boyles law describes pressure volume relationship at constant temperature  pressure is directly proportional  to volume at constant tamparature
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An asteroid is on a collision course with Earth. An astronaut lands on the rock to bury explosive charges that will blow the ast
forsale [732]

Answer:

The maximum radius the asteroid can have for her to be able to leave it entirely simply by jumping straight up is approximately 1782.45 meters

Explanation:

Whereby the height the astronaut can jump on Earth = 0.500 m, we have the following kinematic equation;

v² = u² - 2·g·h

Where;

v = The final velocity

u = The initial velocity

g = The acceleration due to gravity ≈ 9.8 m/s²

h = The height she jumps

At the maximum height, h_{max} = 0.500 m, she jumps, v = 0, therefore, we have;

0² = u² - 2·g·h_{max}

u² = 2 × 9.8 × 0.5 = 9.8

u = √9.8 ≈ 3.13

u = 3.13 m/s

Her initial jumping velocity ≈ 3.13 m/s

Escape velocity, v_e = \sqrt{\dfrac{2 \cdot G \cdot M}{r} }

Where;

M = The mass of the asteroid

G = The Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

r = The radius of the asteroid

The average density of the Earth = 5515 kg/m³

The mass of the asteroid, M = Density × Volume = 5515 kg/m³× 4/3 × π × r³

The escape velocity, she has, v_e ≈ 3.13 m/s is therefore;

3.13 = \sqrt{\dfrac{2 \times 6.67408 \times 10^{-11} \times 5515 \times \frac{4}{3} \times \pi \times r^3}{r} } = r \times \sqrt{3.084 \times 10^{-6}}

r = \dfrac{3.13}{ \sqrt{3.084 \times 10^{-6}}} \approx 1782.45

Therefore, the maximum radius of the asteroid can have for her jumping velocity to be equal to the escape velocity for her to be able to leave it entirely simply by jumping straight up = r ≈ 1782.45 meters.

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3 years ago
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notka56 [123]

The frequency of the wave will not change. Since the change in medium doesn't affect the source of the waves, the frequency of those waves do not change.

Hope this helps! :)

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Mass and distance are the two factors
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vf = - 46 m/s since up is positive

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