Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 24 C and exits at 1 bar. If the refrigerant undergoes a throttling process, what is the quality of the refrigerant exiting the expansion valve.

Answer:

Only Technician A

Explanation:

Continuously variable transmission (CVT) does not shift gears because it simply doesn't have gears. It instead uses 2 pulleys which change constantly and continuously in size, and they are linked by a belt. Continuously variable transmission (CVT) is automatic transmission as mentioned in the question above, largely due to the power the belt continuously transmits, making the vehicle's engine work in the optimum power range.

While Technician B saying, that a continuously variable transmission has a maximum of six distinct gear ranges, which is not correct because continuously variable transmission (CVT) works with an unlimited number of gear ratios within a fixed range.

Option C (making lessons learned a regular part of meetings) is the correct approach.

Aside from this, none of the choices are viable methods to learning lessons or gaining knowledge from the past. As a result, the methodology outlined above is the appropriate one.

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Answer:

report on a fight you have witnessed

Answer:

Mark as brainlist pls hello

to move or walk freely and easly

Answer:

Hello bro

Explanation:

I think i can help you something but can you translate it on english plzz

Answer:

Crowdstrike.com.

Explanation:

This question can't be answered correctly without the reading. However, according to crowdstrike.com, "CrowdStrike Named to CNBC's Disruptor 50 List for Second Consecutive Year. Sunnyvale, CA – May 22, 2018 – CrowdStrike® Inc., the leader in cloud-delivered endpoint protection, today announced that it was named to the 2018 CNBC Disruptor 50 list for the second consecutive year."

Hope This Helps!

Answer:

65.61%

Explanation:

we have the following information to answer this question

diameter of the solid steel shaft = 40 mm

outer diametr of the hollow shaft = 40mm

inner diametr pf the hollow shaft = 36mm

[tex]percentage reduction in torque transmission = \frac{Tsolid-Thollow}{Tsolid} *100[/tex]

= (40³ - (40⁴-36⁴)/40)/40³ * 100

= (40³ - 22009.6)/40³ * 100

= 41990.4/64000 * 100

= 0.6561 x 100

= 65.61%

percentage reduction in torque transmission = 65.61%

Solution :

Let [tex]$x(t) = \frac{\sin (20 \pi t)}{\pi t}$[/tex]

[tex]$T_s = 20$[/tex] ms, so [tex]$f_s=\frac{1}{T_s}[/tex]

                           [tex]$=\frac{1}{20}$[/tex]

                           = 0.05 kHz

[tex]$f_s=50 $[/tex] Hz , ws = [tex]$2 \pi f_s = 100 \pi$[/tex]  rad/s

We know that,

FT → [tex]$\frac{\sin (20 \pi \omega)}{\pi \omega}$[/tex]

The sampled signal is :

[tex]$XS(\omega) = \frac{1}{T_s} \sum_{k=- \infty}^{\infty}X (\omega-k\omega S)[/tex]

So, [tex]$XS(\omega) = \frac{1}{20 \times 10^{-3}} \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]

     [tex]$XS(\omega) = 50 \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]

Answer:

wiwhwnwhwwbbwbwiwuwhwhehehewhehehheheheehehehehhehehwh

Explanation:

jwhwhwhwhwhwwhhahwhahahwh

Answer:

Increasing following distance to Four-seconds when encountering other motorists who follow too closely  is an example of appropriate implementation of the IPDE defensive driving strategy for the  maintenance of an appropriate Safety Cushion.

Explanation:

Maintaining the required safety cushion by utilizing the IPDE defensive driving strategy to manage the nine to fifteen space driving zones involves continuous scanning.  Therefore, motorists should be able to identify objects and hazards in the driving scene, line of sight, and path of travel.  They should predict points of driving conflicts.  They should determine appropriate and safe driving actions to take, when, and where.  Finally, action is required to ensure that conflicts are avoided.

Answer:

- the Mach number is 0.24.

- Compressibility becomes effective when Mach number is greater than 0.3, the Mach number of the race cars is less than 0.3, hence, compressibility will not affect their aerodynamics.

Explanation:

Given the data in the question;

Average speed V = 185 miles per hour = ( 185 /2.237 ) m/s = 82.7 m/s

From Almanac, we can find that Indianapolis is at 220 m altitude.

So from table, at that altitude, the standard speed of sound will be 339.4 m/s .

Mach number of the race car will be;

Mach Number = Velocity / sound speed

we substitute

Mach Number = ( 82.7 m/s ) / ( 339.4 m/s )

Mach Number = 0.24

Therefore the Mach number is 0.24.

We know that, compressibility becomes effective when the Mach number is greater than 0.3.

Since the Mach number of the race cars is less than 0.3, compressibility will not affect their aerodynamics.

Answer:

The correct answer is "57%".

Explanation:

Given:

Temperature:

[tex]T_a = 298 \ K[/tex]

[tex]T_3 = 2000 \ K[/tex]

Compression ratio (r),

= [tex]8.2[/tex]

For Otto cycle, the thermal efficiency will be:

⇒ [tex]\eta =1-\frac{1}{(r)^{\nu-1}}[/tex]

By substituting the values, we get

       [tex]=1-\frac{1}{(8.2)^{1.4-1}}[/tex]

       [tex]=1-\frac{1}{(8.2)^{0.4}}[/tex]

       [tex]=57[/tex] (%)  

Answer:

uhhhhhhh lemme think

Explanation:

Answer:

Phase diagrams represent the relationship between temperature and the composition of phases present at equilibrium. An isomorphous system is one in which the solid has the same structure for all compositions. The phase diagram shown is the diagram for Cu-Ni, which is an isomorphous alloy system.

Hope it help you friend

Answer:

f = 0.04042

Explanation:

temperature = 0°C = 273k

p = 600 Kpa

d = 40 millemeter

e = 10 m

change in  P = 235 N/m²

μ = 2m/s

R = 188.9 Nm/kgk

we solve this using this formula;

P = ρcos*R*T

we put in the values into this equation

600x10³ = ρcos * 188.9 * 273

600000 = ρcos51569.7

ρcos = 600000/51569.7

=11.63

from here we find the head loss due to friction

Δp/pg = feμ²/2D

235/11.63 = f*10*4/2*40x10⁻³

20.21 = 40f/0.08

20.21*0.08 = 40f

1.6168 = 40f

divide through by 40

f = 0.04042

Answer:

a. Covalent modification = Seconds to minutes

b. Allosteric control = Milliseconds

c. Gene expression = Hours

Explanation:

Covalent modifications refer to the addition and/or removal of chemical groups by the action of particular enzymes such as methylases, acetylases, phosphorylases, phosphatases, etc. For example, histones are chromatin-associated proteins covalently modified by enzymes that add methyl groups (histone methylation), acetyl groups (histone acetylation), phosphate groups (histone phosphorylation), etc. Moreover, allosteric control, also known as allosteric regulation, is a type of regulation of the enzyme activity by binding an effector molecule (allosteric modulator) at a different site than the enzyme's active site, thereby triggering a conformational change on the enzyme upon binding of an effector. Finally, gene expression encompasses the cellular processes by which genetic information flows from genes to proteins (i.e., transcription >> translation). In metabolic pathways, enzymes that are able to catalyze irreversible reactions represent sites of control (for example, during glycolysis, pyruvate kinase is an enzyme that catalyzes an irreversible reaction, thereby serving as a control site). In turn, enzymatic activity is modulated by covalent modifications or reversible binding of allosteric effectors. Finally, metabolic pathways are also modulated by gene regulatory mechanisms that control the transcription of specific enzymes required for such pathways. During these processes, the times required for allosteric regulation, covalent modification (e.g., phosphorylation) and transcriptional control can be counted in milliseconds, seconds, and hours, respectively.

Answer:

p q output ¬(p ∧ q)

0 0 1

0 1 1

1 0 1

0 0 0

p q output ¬p ∨ ¬q

0 0 1

0 1 1

1 0 1

0 0 0

Explanation:

We'll create two separate truth tables for both sides of the equation, and see if they match.

The expressions in the question use AND, OR and NOT operators.

Let's start with ¬(p ∧ q)

Now let's move on to the second expression ¬p ∨ ¬q

Therefore we can say the two expressions are equivalent.

Attached  the truth table to verify the first De Morgan's law ¬(p ∧ q) ≡ ¬p ∨ ¬q:

As you can see from the attached truth table, the truth values for ¬(p ∧ q) and ¬p ∨ ¬q are the same for all combinations of p and q, confirming the validity of the first De Morgan's law.

De Morgan's law is a fundamental principle in propositional logic.

It states that the negation of a conjunction (AND) is equivalent to the disjunction (OR) of the negations of the individual propositions.

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#SPJ3

Answer:

Answer to the following question is as follows;

Explanation:

It is critical to communicate well during negotiations and conversation in order to attain your objectives. Communication also becomes essential in business and corporation . Effective communication may help you as well as your employees develop a strong professional relationship, which can boost morale and efficiency.

Answer:

I didn't understand your question or is it a fun fact

Answer:

The objective of a protection scheme is to keep the power system stable by isolating only the components that are under fault, whilst leaving as much of the network as possible still in operation.

Explanation:

The devices that are used to protect the power systems from faults are called protection devices.

Answer:

Abolition of intermediaries (rent collectors under the pre-Independence land revenue system); Tenancy regulation (to improve the contractual terms including the security of tenure); A ceiling on landholdings (to redistributing surplus land to the landless);

Types of Land Reform

Abolition of Intermediaries

The first step taken by the Indian government under land reforms post-independence was passing the Zamindari Abolition Act. The primary reason of a backward agrarian economy was the presence of intermediate entities like, jagirdars and zamindar who primarily focussed on collecting sky-rocketing rents catering to their personal benefits, without paying attention to the disposition of farms and farmers. Abolition of such intermediaries not only improved conditions of farmers by establishing their direct connection with the government but also improved agricultural production.

Regulation of Rents

This was in direct response to the unimaginably high rents which were charged by intermediaries during British rule, which resulted in a never-ending cycle of poverty and misery for tenants. Indian government implemented these regulations to protect farmers and labourers from exploitation by placing a maximum limit on the rent that could be charged for land.

Tenure Security

Legislations were passed in all states of the country to grant tenants with permanent ownership of lands and protection from unlawful evictions on expiry of the lease. This law protects tenants from having to vacate a property immediately after their tenure is over unless ordered by law. Even in that case, ownership can be regained by tenants with the excuse of personal cultivation.