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zysi [14]
2 years ago
9

A rigid tank A of volume 0.6 m3 contains 5 kg air at 320K and the rigid tank B is 0.4 m3 with air at 600 kPa, 360 K. They are co

nnected to a piston cylinder initially empty with closed valves. The pressure in the cylinder should be 800 kPa to float the piston. Now the valves are slowly opened and the entire process is adiabatic. The internal energy of the mixture at final state is:_____.
a. 229 k/kg.
b. 238 kJ/kg
c. 257 kg
d. cannot be determined.

Physics
1 answer:
neonofarm [45]2 years ago
6 0

Answer:

the internal energy of the mixture at final state = 238kJ/kg

Explanation:

Given

V= 0.6m³

m=5kg

R=0.287kJ/kg.K

T=320 K

from ideal gas equation

PV = nRT

where P is pressure, V is volume, n is number of mole, R is ideal gas constant , T is the temperature.

Recall, mole = mass/molar mass

attached is calculation of the question.

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A 5.0-g marble is released from rest in the deep end of a swimming pool. An underwater video reveals that its terminal speed in
Temka [501]

Answer:

a) a = -g = 9.8 m/s² , b) a = 0 m/s² and c)   t1 = 0.0213 s

Explanation:

a) At the moment the marble is released its velocity is zero, so it has no resistance force, the only force acting is its weight, so the acceleration is the acceleration of gravity

       a = -g = 9.8 m / s²

b) When the marble goes its terminal velocity all forces have been equalized, therefore, the sum of them is zero and consequently if acceleration is also zero

      a = 0 m / s²

c) We have to assume a specific type of resistive force, for liquid in general the resistive force is proportional to the speed of the body.

The expression of this situation is

         v = mg / b (1 -e^{-bt/m} )

For a very long time the exponential is zero, so the terminal velocity is

        v_{T} = mg / b

        b = mg /  v_{T}

        b = 5 10-3 9.8 / 0.3

        b = 0.163

We already have all the data to calculate the time for v = ½ v_{T}

        ½ v_{T} = v_{T} (1 -e^{-bt/m})

        ½ = 1- e (- 0.163 t1 / 5 10-3)

        e (-32.6 t1) = 1-0.5              (by  ln())

       -32.6 t1 = ln 0.5

       t1 = -1 / 32.6 (-0.693)

       t1 = 0.0213 s

3 0
2 years ago
An object is 6.0 cm in front of a converging lens with a focal length of 10 cm.Use ray tracing to determine the location of the
Masja [62]

As we know that

\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}

here we know that

d_0 = 6 cm

f = 10 cm

now from above equation we have

\frac{1}{d_i} + \frac{1}{6} = \frac{1}{10}

d_i = -15 cm

so image will form on left side of lens at a distance of 15 cm

This image will be magnified and virtual image

Ray diagram is attached below here

8 0
2 years ago
Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 1350 N and that c 5 360 mm, determine the moment
riadik2000 [5.3K]

Answer:

291.598 N-m

291.6 N-m

Explanation:

Let's first take a  look at the free bodily diagrammatic representation.

The first diagram will aid us in answering  question (a), so as the second diagram will facilitate effective understanding when solving for question (b).

Let's first determine our angle θ from the diagram

To find angle θ ; we have :

tan θ  = \frac{360+240}{450}

tan θ  = \frac{600}{450}

tan θ  = 1.333

θ  = tan⁻¹ (1.333)

θ  = 53.13°

Now, to determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A.

We have:

M__B}=(Fcos \theta *240)-(Fsin \theta *450)

where Force(F) = Force in the cord AC = 1350 N and θ  = 53.13° ; we have:

M__B}=(1350&cos 53.13^0 *240)-(1350sin 53.13^0 *450)

M__B}= 194400.463-485999.348

M__B}=-291598.885 N-mm\\

M__B}=-291.598 N-m

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A = 291.598 N-m

b) From the second diagram, taking the moment at point B (M__B}),

we have:

M__B}=-Fcos \theta *360 - Fsin \theta * 0

M__B}=-Fcos \theta *360 - 0

M__B}=-Fcos \theta *360

where Force(F) =  1350 N and θ  = 53.13° ; we have:

M__B}= -1350*cos53.13^0*360

M__B}= -291600 N-mm

M__B}= -291.6 N-m

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point C = 291.6 N-m

6 0
2 years ago
A physical change occurs when a material changes shape or size but the composition of the material does not change. True or Fals
Elden [556K]
It is true that a physical change occurs when a material changes shape or size, but the composition of the material does not change. The correct answer is True. 
6 0
2 years ago
The diagram shows movement of thermal energy. At bottom a fire has red curved lines labeled Y with arrowheads pointing upward to
Pepsi [2]

Answer:

X and Z

Explanation:

Conduction occurs through direct physical contact.  Heat transferred from the pot to the handle, and from the handle to the hand, are both examples of conduction.

6 0
2 years ago
Read 2 more answers
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