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Serga [27]
3 years ago
11

When placed outside and exposed to the atmosphere, an ice cube melts, forming a thin film of liquid on the ground. Overnight, th

e liquid freezes, returning to the initial temperature of the ice cube. The water making up the cube undergoes (a) a thermodynamic cycle, (b) a reversible process, (c) an irreversible process, (d) none of the above.
Engineering
1 answer:
ludmilkaskok [199]3 years ago
8 0

Answer:

Option A and C are correct.

Explanation:

First of all, a thermodynamic circle is closed and gotten when our starting and ending points are the same. Now, from the question, we can see that the water starts and ends on the same temperature and pressure. Thus, we can say that a thermodynamic circle has occurred. So, option A is correct.

Now, to achieve a reversible process, there should be no exergy losses could cause rise in exergy entropy. There are two possible causes which we cannot eliminate as both are useful. First is exergy friction while the second cause is heat exchange at different temperatures of heat reservoirs and our cube. This second cause is confirmed because our cube while melting is on lower temperature than it's surrounding. This causes rise in entropy in overall system, thus making the process irreversible. So option C is correct.

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As soldiers get bigger and packs get heavier, a parachutist and load can weigh as much as 490 lbf. The standard 28-ft parachute
vovangra [49]

Answer:

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Explanation:

8 0
3 years ago
A 15-ft beam weighing 570 lb is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground
7nadin3 [17]

Answer:

I. Tension (cable A) ≈ 6939 lbf

II. Tension (cable B) ≈ 17199 lbf

Explanation:

Let's begin by listing out the data that we were given:

mass of beam (m) = 570 lb, deceleration (cable A) = -20 ft/s², deceleration (cable B) = -2 ft/s²,

g = 32.17405 ft/s²

The tension on an object is given by the product of mass of the object by gravitational force plus/minus the product of mass by acceleration.

Mathematically represented thus:

T = mg + ma

where:

T = tension, m = mass, g = gravitational force,

a = acceleration

I. For Cable A, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-20)]

T = 18339.2085 - 11400 = 6939.2085

T ≈ 6939 lbf

II. For Cable B, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-2)]

T = 18339.2085 - 1140 = 17199.2085

T ≈ 17199 lbf

4 0
4 years ago
Technician A says the compressor is the dividing line of the refrigeration system, low- to high-side. Technician B says the expa
Fiesta28 [93]

The statements regarding the refrigeration system as expressed by the technicians are correct. Hence, the correct statement is Option c.

<h3>What is a refrigeration system?</h3>

Refrigeration systems are a crucial process in the industry and home applications as they carry out cooling or keep the room temperature at a preferred value.

The missing information in the question is:

A.technician A only

B.technician B only

C.both A and B

D.neither A or B

Hence, The statements regarding the refrigeration system as expressed by the technicians are correct.  the correct statement is Option c.

Learn more about Refrigeration systems:

brainly.com/question/12135611

#SPJ1

7 0
2 years ago
What can be used to measure the alcohol content in gasoline? A. Graduated cylinder B. Electronic tester C. Scan tool D. Either a
Dovator [93]

Answer:

GRADUATED CYLINDER

Explanation:

4 0
3 years ago
Determine the resolution of a manometer required to measure the velocity of air at 50 m/s using a pitot-static tube and a manome
oksano4ka [1.4K]

Answer:

a)  Δh = 2 cm,  b) Δh = 0.4 cm

Explanation:

Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.

            P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used  in such a way that the height of point 2 of is the same of point 1

           y₁ = y₂

subtitute

           P₁ = P₂ + ½ ρ v₂²

           P₁ -P₂ = ½ ρ v²

where ρ is the density of fluid  

now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface

           ΔP =ρ_{Hg} g h - ρ g h

           ΔP =  (ρ_{Hg} - ρ) g h

as the static pressure we can equalize the equations

          ΔP = P₁ - P₂

         (ρ_{Hg} - ρ) g h = ½ ρ v²

         v = \sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}

in this expression the densities are constant

        v = A  √h

       A =\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }

 

They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³

we look for the constant

        A = \sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }

        A = 454.55

we substitute

       v = 454.55 √h

to calculate the uncertainty or error of the velocity

         h = \frac{1}{454.55^2} \ v^2

       Δh = \frac{dh}{dv}   Δv

       \frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}

Suppose we have a height reading of h = 20 cm = 0.20 m

             

a) uncertainty 2.5 m / s ( 0.05)

        \frac{\delta v}{v} = 0.05

       \frac{\Delta h}{h} = 2 0.05  

       Δh = 0.1 h

       Δh = 0.1  20 cm

       Δh = 2 cm

b) uncertainty 0.5 m / s ( Δv/v= 0.01)

        \frac{\Delta h}{h} =  2 0.01

        Δh = 0.02 h

        Δh = 0.02 20

        Δh = 0.1 20 cm

        Δh = 0.4 cm = 4 mm

5 0
3 years ago
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