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Stells [14]
2 years ago
15

Air at 27 ºC and 5 atm is throttled by a valve to 1 atm. If the valve is adiabatic and the change in kinetic energy is negligibl

e, what is the exit temperature of the air?
Physics
1 answer:
Citrus2011 [14]2 years ago
4 0

Answer:

27 °C

Explanation:

BY the statement of the question it is clear that it is about an ideal gas - and hence if change in KE is about zero - then there will be no change of temperature.

So, answer is 27 °C

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Which of the following best describes an action-reaction pair? A. The Moon Pulls on Earth, and Earth pulls back on the moon. B.
Papessa [141]
An action-reaction pair would be a pair in which one of the elements exerts a force on the other element (action), and then the other element would respond to this force by exerting another force in the opposite direction (reaction).

From the given choices, we will see that:
For choice A, the moon exerts a force on the earth by pulling it (action) and the earth responds to this force by pulling the moon (reaction in opposite direction of the action).

Therefore, the correct choice would be: 
A. <span>The Moon Pulls on Earth, and Earth pulls back on the moon.</span>
4 0
2 years ago
Read 2 more answers
A pizza delivery driver must make three stops on her route. She will first leave the restaurant and travel 4 km due north to the
topjm [15]
<h2>5.3 km</h2>

Explanation:

       This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.

       Let us denote each of the individual displacements by a vector. Consider the unit vectors \vec{i}\textrm{ and }\vec{j} as the unit vectors in the direction of East and North respectively.

       By simple calculations, we can derive the unit vectors \vec{j},\frac{-\vec{i}-\vec{j}}{2}\textrm{ and }\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2} in the directions North, 45^{o} South of West and 60^{o} North of West respectively.

       So Total displacement vector = Sum of individual displacement vectors.

       Displacement vector = 4(\vec{j})+6(\frac{-\vec{i}-\vec{j}}{2})+5(\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2})=-4.25\vec{i}+3.165\vec{j}

       Magnitude of Displacement = |-4.25\vec{i}+3.165\vec{j}|=5.3km

∴ Total displacement = 5.3km

4 0
2 years ago
The furnace keeps houseAat 25◦C, while thefurnace in houseBkeeps it at 20◦C. Which house requires heat to be supplied by its fur
EleoNora [17]

Answer:

House A requires heat at a slightest faster rate than B

Explanation:

House A requires heat at a slightest faster rate than B due to the slight high temperature the furnace A is.

5 0
2 years ago
Jeff puts on a leather jacket over his sweater. The sweater becomes negatively charged. Which statements about Jeff’s situation
nikdorinn [45]

b. The sweater has a tendency to attract protons.

8 0
2 years ago
Read 2 more answers
Kayla, a fitness trainer, develops an exercise that involves pulling a heavy crate across a rough surface. The exercise involves
Lisa [10]

Answer:

a = 3.784\,\frac{m}{s^{2}}

Explanation:

The Free Body Diagram of the system formed by the crate and the rope and the reference axis are presented below as an attached image. The equations of equilibrium are introduced hereafter:

\Sigma F_{x} = T\cdot \cos \theta - \mu\cdot N = m\cdot a

\Sigma F_{y} = T\cdot \sin \theta + N - m\cdot g = 0

After some algebraic handling, the system of equations is reduce to a sole expression:

N = m\cdot g - T\cdot \sin \theta

T\cdot \cos \theta - \mu \cdot (m\cdot g - T\cdot \sin \theta) =m\cdot a

T\cdot (\cos\theta + \mu \cdot \sin \theta) - \mu \cdot m \cdot g = m\cdot a

The minimum force to accelerate the crate from rest is:

T_{min} = \frac{m\cdot \mu_{s}\cdot g}{\cos \theta + \mu_{s} \cdot \sin \theta}

T_{min} = \frac{(45\,kg)\cdot (0.770)\cdot (9.807\,\frac{m}{s^{2}} )}{\cos 23^{\textdegree}+(0.770)\cdot \sin 23^{\textdegree}}

T_{min} = 278.223\,N

Since T > T_{min}, the crate will experiment an acceleration due to the tension exerted. The acceleration of the crate is:

a = \frac{T}{m}\cdot (\cos \theta + \mu_{k}\cdot \sin \theta)-\mu_{k}\cdot g

a = \frac{325\,N}{45\,kg}\cdot [\cos 23^{\textdegree}+(0.410)\cdot \sin 23^{\textdegree}]-(0.410)\cdot (9.807\,\frac{m}{s^{2}} )

a = 3.784\,\frac{m}{s^{2}}

5 0
2 years ago
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