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soldi70 [24.7K]
3 years ago
12

How many volts of electricity would it take to power up an entire city? Take Tokyo for example. Please explain!

Engineering
1 answer:
jek_recluse [69]3 years ago
3 0

Answer:

maybe a couple million volts

Explanation:

because there are outlets you need to charge your phone maybe you car if you have a tesla and skyscrapers

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Explain the difference between thermoplastics and thermosets giving structure property correlation.
Misha Larkins [42]

Answer:

Explanation:

Thermosetting polymers are infusible and insoluble polymers. The reason for such behavior is that the chains of these materials form a three-dimensional spatial network, intertwining with strong equivalent bonds. The structure thus formed is a conglomerate of interwoven chains giving the appearance and functioning as a macromolecule, which as the temperature rises, simply the chains are more compacted, making the polymer more resistant to the point where it degrades.

Macromolecules are molecules that have a high molecular mass, formed by a large number of atoms. Generally they can be described as the repetition of one or a few minimum units or monomers, forming the polymers. In contrast, a thermoplastic is a material that at relatively high temperatures, becomes deformable or flexible, melts when heated and hardens in a glass transition state when it cools sufficiently. Most thermoplastics are high molecular weight polymers, which have associated chains through weak Van der Waals forces (polyethylene); strong dipole-dipole and hydrogen bond interactions, or even stacked aromatic rings (polystyrene). Thermoplastic polymers differ from thermosetting polymers or thermofixes in that after heating and molding they can overheat and form other objects.

Thermosetting plastics have some advantageous properties over thermoplastics. For example, better resistance to impact, solvents, gas permeation and extreme temperatures. Among the disadvantages are, generally, the difficulty of processing, the need for curing, the brittle nature of the material (fragile) and the lack of reinforcement when subjected to tension. But even so in many ways it surpasses the thermoplastic.

The physical properties of thermoplastics gradually change if they are melted and molded several times (thermal history), these properties are generally diminished by weakening the bonds. The most commonly used are polyethylene (PE), polypropylene (PP), polybutylene (PB), polystyrene (PS), polymethylmethacrylate (PMMA), polyvinylchloride (PVC), ethylene polyterephthalate (PET), Teflon (or polytetrafluoroethylene, PTFE) and nylon (a type of polyamide).

They differ from thermosets or thermofixes (bakelite, vulcanized rubber) in that the latter do not melt when raised at high temperatures, but burn, making it impossible to reshape them.

Many of the known thermoplastics can be the result of the sum of several polymers, such as vinyl, which is a mixture of polyethylene and polypropylene.

When they are cooled, starting from the liquid state and depending on the temperatures to which they are exposed during the solidification process (increase or decrease), solid crystalline or non-crystalline structures may be formed.

This type of polymer is characterized by its structure. It is formed by hydrocarbon chains, like most polymers, and specifically we find linear or branched chains

4 0
3 years ago
A room is kept at −5°C by a vapor-compression refrigeration cycle with R-134a as the refrigerant. Heat is rejected to cooling wa
Fed [463]

Answer:

note:

<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>

Download docx
4 0
3 years ago
Read 2 more answers
Water flows through a pipe and enters a section where the cross sectional area is larger. Viscosity, friction, and gravitational
BaLLatris [955]

Answer:

(A) and (D)

Explanation:

1) P2 is less than P1, that is when P1 increases in pressure, the velocity V1 of the water also increases. Therefore, on the other hand, since P2 is directly proportional to V1, P2 and V2 will be less than P1 and V1 respectively.

2) For P2 greater than P1 and V2 also is greater than V1. Since P2 is directly proportional to V2, hence, when P2 increases in pressure, P1 reduces in pressure. Similarly, velocity, V2 also increases and V1 reduces.

3 0
3 years ago
We can model a certain battery as a voltage source in series with a resistance. The open-circuit voltage of the battery is 10 V
Angelina_Jolie [31]

Answer:

51.4 Ohms

Explanation:

By applying voltage division rule

V_f=v_i\times \frac {R_l}{R_l+R_m} where v is voltage, subscripts i and f represnt initial and final, R is resistance, m is internal and l is external.Substituting 7V for final voltage, 10V for initial voltage and the external resistance as 120 Ohms then

7=10*\frac {120}{120+R_m}\\7R_m+840=1200\\R_m={1200-840}{7}=51.428571\approx 51.4 Ohms

3 0
3 years ago
A piston-cylinder device contains 0.1 m3 of liquid water and 0.9 m² of water vapor in equilibrium at 800 kPa. Heat is transferre
docker41 [41]

Answer:

Initial temperature = 170. 414 °C

Total mass = 94.478 Kg

Final volumen = 33.1181 m^3

Diagram  = see picture.

Explanation:

We can consider this system as a close system, because there is not information about any output or input of water, so the mass in the system is constant.  

The information tells us that the system is in equilibrium with two phases: liquid and steam. When a system is a two phases region (equilibrium) the temperature and pressure keep constant until the change is completed (either condensation or evaporation). Since we know that we are in a two-phase region and we know the pressure of the system, we can check the thermodynamics tables to know the temperature, because there is a unique temperature in which with this pressure (800 kPa) the system can be in two-phases region (reach the equilibrium condition).  

For water in equilibrium at 800 kPa the temperature of saturation is 170.414 °C which is the initial temperature of the system.  

to calculate the total mass of the system, we need to estimate the mass of steam and liquid water and add them. To get these values we use the specific volume for both, liquid and steam for the initial condition. We can get them from the thermodynamics tables.

For the condition of 800 kPa and 170.414 °C using the thermodynamics tables we get:

Vg (Specific Volume of Saturated Steam) = 0.240328 m^3/kg

Vf (Specific Volume of Saturated Liquid) = 0.00111479 m^3/kg

if you divide the volume of liquid and steam provided in the statement by the specific volume of saturated liquid and steam, we can obtain the value of mass of vapor and liquid in the system.

Steam mass = *0.9 m^3 / 0.240328 m^3/kg = 3.74488 Kg

Liquid mass = 0.1 m^3 /0.00111479 m^3/kg = 89.70299 Kg  

Total mass of the system = 3.74488 Kg + 89.70299 Kg = 93,4478 Kg

If we keep the pressure constant increasing the temperature the system will experience a phase-change (see the diagram) going from two-phase region to superheated steam. When we check for properties for the condition of P= 800 kPa and T= 350°C we see that is in the region of superheated steam, so we don’t have liquid water in this condition.  

If we want to get the final volume of the water (steam) in the system, we need to get the specific volume for this condition from the thermodynamics tables.  

Specific Volume of Superheated Steam at 800 kPa and 350°C = 0.354411 m^3/kg

We already know that this a close system so the mass in it keeps constant during the process.

 

If we multiply the mass of the system by the specific volume in the final condition, we can get the final volume for the system.  

Final volume = 93.4478 Kg * 0.354411 m^3/kg = 33.1189 m^3

You can the P-v diagram for this system in the picture.  

For the initial condition you can calculate the quality of the steam (measure of the proportion of steam on the mixture) to see how far the point is from for the condition on all the mix is steam. Is a value between 0 and 1, where 0 is saturated liquid and 1 is saturated steam.  

Quality of steam = mass of steam / total mass of the system

Quality of steam = 3.74488 Kg /93.4478 Kg = 0,040 this value is usually present as a percentage so is 4%.  

Since this a low value we can say that we are very close the saturated liquid point in the diagram.  

6 0
4 years ago
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