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marysya [2.9K]
2 years ago
5

Two cylinders contain the same ideal gas and have the same volume. If the gas molecules in cylinder B have twice the average kin

etic energy as those in cylinder A how does the internal energy, U, of cylinder B compare to that of cylinder A
Physics
1 answer:
ipn [44]2 years ago
7 0

Hence, internal energy in cylinder B is twice than that in A.

It is given that two cylinders contain the same ideal gas and have the same volume. If the gas molecules in cylinder B have twice the average kinetic energy as those in cylinder, we need to compare internal energy.

As kinetic energy is directly proportional to temperature, internal energy is also proportional to temperature.

We can say that more the average kinetic energy, more is the internal energy.

Hence, internal energy in cylinder B is twice than that in A.

The ratio in A:B is 1:2.

Learn more about kinetic energy click here brainly.com/question/11067389

#SPJ4

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Your neighbor’s 14-month-old toddler says things like "cup!" when he means that he would like a cup of milk. Your neighbor think
Serhud [2]

Answer:

The options are not properly punctuated.  

(a) cognitively impaired (b) using holophrases which is developmentally appropriate  (c) language-delayed and needs professional assistance (d)  trying to manipulate his father and needs to be disciplined

The correct answer is  (b) using holophrases which is developmentally appropriate

Explanation:

Holophrases are one of the single-word utterances characteristic of children in the early stages of language acquisition,   it is the use of a single word to express a complex idea. Holophrases are normal among toddlers within the  age of two years  below.

Hence the the use of "cup" by the toddler to express his intent for a cup of milk is referred to as holophrases which is developmentally appropriate

6 0
3 years ago
A sledgehammer hits a wall How do the hammer and the wall act on each other?
tigry1 [53]

We want to study the impact of a sledgehammer and a wall.

Before the sledgehammer hits the wall, it has a given velocity and a given mass, so it has momentum and it has kinetic energy.

When it hits the wall, the velocity of the hammer disappears, this means that the energy is transferred to the wall, this "transfer of energy" can be thought of a force applied for a really short time on the wall, which for the third law of Newton, the force is also applied on the hammer.

This is why you feel the impact on the handle when you hit something with a hammer, this also means that some of the energy is dissipated on your arms.

Now, because the wall is made of a material usually not as strong as the head of the sledgehammer, we will see that in this interaction the wall seems more affected than the hammer, but the forces that each one experiences are exactly equal in magnitude.

If you want to learn more, you can read:

brainly.com/question/13952508

7 0
3 years ago
The weight of a box having a mass of 100 kg is blank N
erastova [34]

Answer:980 N

Explanation:

4 0
3 years ago
What two sources of friction do you have to overcome when you are walking?
tatuchka [14]
<span>
Gravity is pushing you down on the earth while the friction from the ground is pushing you up. Those are the two frictions you must overcome when walking.






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3 0
3 years ago
Read 2 more answers
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
spayn [35]

Answer:

v_0 = 3.53~{\rm m/s}

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

6 0
3 years ago
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