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Ierofanga [76]
3 years ago
7

Describe how does the temperature relate to density and the movement of atmospheric gases? A) As the air warms expanding its vol

ume, decreasing its density, and the air molecules rise. B) UV light causes air to warm its volume declines, density increases, and the air molecules sink. C) The volume of air distends as it cools, while its density increases and the air molecules descend. D) Radiation cools the air and its volume extends, density diminishes, and the air molecules accelerate.
Physics
1 answer:
cluponka [151]3 years ago
8 0
The correct answer is A) As the air warms expanding its volume, decreasing its density, and the air molecules rise

The air molecules speed up and spread. If the container is closed then it may pop because the air expands. Hot air goes up and cold air goes down.

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The only force acting on a 3.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a
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Answer:

The work done on the canister by the 5.0 N force during this time is

54.06 Joules.

Explanation:

Let the initial kinetic energy of the canister be

KE₁ = \frac{1}{2} mv_1^{2} = \frac{1}{2} *3*3.6^{2} = 19.44 J in the x direction

Let the the final kinetic energy of the canister be

KE₂ = \frac{1}{2} mv_2^{2} = \frac{1}{2} *3*7.0^{2} = 73.5 J in the y direction

Therefore from the Newton's first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final

= 73.5 J - 19.44 J = 54.06 J

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3 years ago
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Why echos are undesirable in a big hall​
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Answer:

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Sound created in a big hall will persist by repeated reflection from the walls until it is reduced to a value where it is no longer audible. The repeated reflection that results in this persistence of sound is called reverberation. In an auditorium or big hall excessive reverberation is highly undesirable

Explanation:

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Creating plasma can be dangerous because of the high amount of ___ needed to create it.
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Anaerobic metabolism:
goblinko [34]

Explanation:

B. leads to muscle strain.

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3 years ago
You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts,
lbvjy [14]

Answer:

The value is T_c  =  12 .1 ^oC

Explanation:

From the question we are told that

The mass of the ice cube is m_i  =  7.50 *10^{-3} \  kg

The temperature of the ice cube is T_i = 0^o C

The mass of the copper cube is m_c  =  0.540 \  kg

The final temperature of both substance is T_f  =  0^oC

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

Q =  m_c  *  c_c *  [T_c  -  T_f ]

The specific heat of copper is c_c  = 385J/kg \cdot  ^oC

Generally the heat gained by the ice cube is mathematically represented as

Q_1 =  m_i * L

Here L is the latent heat of fusion of the ice with value L  =  3.34 * 10^{5} J/kg

So

Q_1 =  7.50 *10^{-3} * 3.34 * 10^{5}

=> Q_1 =  2505 \ J

So

2505  =  0.540  *  385 *  [T_c  - 0 ]

=>    T_c  =  12 .1 ^oC

4 0
3 years ago
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