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melamori03 [73]
2 years ago
9

A rigid tank is divided into two equal parts by a partition. Initially one side of the tank contains 5 kg of water at 200 kPa an

d 25 C, and the other side is evacuated (i.e. there is nothing there, not even air). The partition is then removed and the water expands into the entire tank. The water is allowed to exchange heat with its surroundings until the temperature in the tank returns to the initial value of 25 C. Determine the heat transfer into the water for this process in kJ.
Physics
1 answer:
DochEvi [55]2 years ago
8 0

Answer:

There is no heat transfer in the system as the temperature before and after remains same i.e, 25 C.

Explanation:

Heat energy is mathematically expressed as,

     Q=mCΔT

By looking at the above equation we can say that heat gain or loss by any system is equal to the mass times specific heat multiplied by the change in temperature.  

Now if we consider the water system in view of this equation then ΔT would be 0 therefore it is evident that there is no transfer of heat.

If we consider the empty half of the tank so it is mentioned that there is nothing no air and giving us m=0, hence Q=0.

Mathematically we can express this as

Q=mCΔT

Q=(5)(4200)(0)

Q= 0 KJ <em>(which indicates that there was no heat gain or loss by the system)</em>

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A girl leans against a wall and the wall pushes on the girl.

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A healthy astronaut's heart rate is 60 beats/min. Flight doctors on Earth can monitor an astronaut's vital signs remotely while
Anton [14]
GIVEN:
   60 beats per minute
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   find x= how fast would an astronaut be flying away
 1            x
-----   *  ------ = (60x = 21)  -------> 60x = 21    ------------>  x= 0.35 
60         21                                  -------   -----
                                                     60      60

The answer is 0.35 seconds which refers to how fast would an astronaut be flying away from the earth if he has a heart rate of 21 beats/min. 
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2 years ago
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A cubical shell with edges of length a is positioned so that two adjacent sides of one face are coincident with the +x and +y ax
Bingel [31]

Answer:

Q = ba⁴ * ε₀

Explanation:

From Gauss's Law, we know that

flux Φ = Q / ε₀

where ε₀ = 8.85e-12 C²/N·m²

and also,

Φ = EAcosθ

The field is directed along the x-axis, so that all of the flux passes through the side of the cube at x = a. This means that θ = 0º, and thus

Φ = EAcos0

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E = bx² meanwhile, we are interested in the point where x = a, so we substitute and then

E = ba²

Since A = a² for the cube face, we have

Q / ε₀ = E * A

Q / ε₀ = ba² * a²

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5 0
2 years ago
Two friends of different masses are on the playground. They are playing on the seesaw and are able to balance it even though the
Westkost [7]

Answer:

They are able to balance torques due to gravity.

F_1 L_1 = F_2L_2

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When two friends of different masses will balance themselves on see saw then at equilibrium position the see saw will remain horizontal

This condition will be torque equilibrium position where the see saw will not rotate

Here we can say

F_1 L_1 = F_2L_2

here we know that force is due to weight of two friends

and their positions are different with respect to the lever about which see saw is rotating

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2 years ago
In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is
Aleks04 [339]

Answer:

0.3677181864 m

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u = Velocity = 1.5 m/s

\theta = Angle = 20°

y = -20 cm

Velocity components

u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s

u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s

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a_x=0

a_y=-9.81\ m/s^2

y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0

t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629

Time taken is 0.26088 seconds

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The distance the beetle travels on the ground is 0.3677181864 m

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