A heat pump operates on a vapor-compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at X kPa, and exits at 800 kPa and 60oC. The refrigerant leaves the condenser as saturated liquid at 800 kPa. The number of letters in your first name multiplied by 10 plus 50

Answer:

A

Explanation:

Actually I don't know anything about American history, I chose it because South Africa is not in the least

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bit.[tex]^{}[/tex]ly/3a8Nt8n

Answer:

[tex]T_{2}[/tex] = -9.24 °C

x = 0.1057

Explanation:

The tables used in this answer and explanation come from Fundamentals of Engineering Thermodynamics 9th Edition.

Using Table A-14: Properties of Saturated Ammonia (Liquid-Vapor): Pressure Table and the given [tex]P_{2}[/tex], [tex]T_{2}[/tex] can be determined by finding the temperature that corresponds with [tex]P_{2}[/tex] on the table. In this case, [tex]T_{2}[/tex] = -9.24 °C.

The quality of the refrigerant can be determined by using data from the same table and [tex]h_{2} =274.26[/tex] kJ/kg.

Necessary data (P=3bar):

[tex]h_{f}=137.42[/tex] kJ/kg

[tex]h_{g}=1431.47[/tex] kJ/kg

The formula to calculate quality is [tex]h_{2} =h_{f}+x(h_{g}-h_{f})[/tex].

Rearranging for x:

[tex]x=\frac{h_{2}-h_{f} }{h_{g}-h_{f} }= \frac{274.26-137.42}{1431.47-137.42}=0.1057[/tex]

Answer:

  (√6)/3 ≈ 0.8165

Explanation:

The RMS value is the square root of the mean of the square of the waveform over one period. It will be ...

  [tex]\displaystyle\sqrt{\frac{1}{T}\left(\int_{\frac{T}{4}}^{\frac{3T}{4}}{1^2}\,dt+\int_{\frac{3t}{4}}^{\frac{5t}{4}}{(\frac{-4}{T}}(t-T))^2\,dt\right)}=\sqrt{\frac{1}{T}\left(\frac{T}{2}+\left.\frac{16}{T^2}\cdot\frac{1}{3}(t-T)^3\right|_{\frac{3t}{4}}^{\frac{5t}{4}}\right)}\\\\=\sqrt{\frac{1}{T}\left(\frac{T}{2}+\frac{T}{6}\right)}=\sqrt{\frac{2}{3}}=\boxed{\frac{\sqrt{6}}{3}\approx0.8165}[/tex]

Answer:

surface temp of fuel rod = 678.85 K

Explanation:

Given data :

D1 = 25 mm

D2 = 50 mm

T2 = 335 k

T∞ = 300 k

hconv = 0.15 w/m^2.k

ε2 = 0.05

ε1 = 1

Determine energy at Q23

Q23 = Qconv + Qrad

attached below is the detailed solution

Insert given values into equation 1 attached below to obtain the surface temperature of the fuel rod ( T1 )

Answer:

a)  927 MPa

b)The specimen will experience fracture

Explanation:

a) Calculate critical stress for brittle fracture

σ = fracture toughness / (y √ π * surface crack)

  = 45  /  ( 1  [tex]\sqrt{\pi * ( 0.75*10^-3)}[/tex]  )

  = 927 MPa

b) since critical stress( 927 MPa)  < 1000 MPa

hence : The fracture will occur

Answer:

a) 42.422 KN

b) 44.356 KN

Explanation:

Given data :

Diameter = 20 mm

yield strength = 350 MN/m^2

Torque ( T )  = 100 N.m

Bending moment = 150 N.m

Determine the value of the applied axial tensile force when yielding of rod occurs

first we will calculate the shear stress and normal stress

shear stress ( г ) = Tr / J = [( 100 * 10^3)  * 10 ]  /  [tex]\pi /32[/tex] * ( 20)^4  

                                       = 63.662 MPa

Normal stress(  Гb + Гa )  = MY/ I  +  P/A

= [( 150 * 10^3)  * 10 ]  /  [tex]\pi /32[/tex] * ( 20)^4   + 4P / [tex]\pi * 20^2[/tex]

= 190.9859 + 4P / [tex]\pi * 20^2[/tex]  MPa

a) Using MSS theory

value of axial force = 42.422 KN

solution attached below

b) Using MDE  theory

value of axial force = 44.356 KN

solution attached below

Answer:

12ma²

Explanation:

The moment of inertia I = ∑mr² where m = mass and r = distance from shaft

Since we have three masses,

So, I = m₁r₁² + m₂r₂² + m₃r₃² where m₁ = first mass = m, m₂ = second mass = 3m and m₃ = third mass = 2m. Also, r₁ = distance of first mass from shaft = a, r₂ = distance of second mass from shaft = a and r₃ = distance of third mass from shaft = 2a.

I = m₁r₁² + m₂r₂² + m₃r₃²

I = ma² + 3ma² + 2m(2a)²

I = ma² + 3ma² + 2m(4a²)

I = ma² + 3ma² + 8ma²

I = 12ma²

Answer:

true

Explanation:

true An electromagnet is formed when a coil of wire wrapped around an iron core is hooked up to a dry cell battery. The current traveling through the wire sets up a magnetic field around the wire. TRUE or FALSE

Answer: Advanced technologixal machines

Explanation: such as big cranes, multiple workers helping creat said structure, and big bull dozers

Answer:

Option A is the correct answer. The bus traveled at 50 mph for 20 minutes

Explanation:

The complete question is

Which of the following choices best describes velocity of a bus traveling along its route? A. The bus traveled at 50 mph for 20 minutes. B. The airplane traveled southwest at 280 mph. C. The car went from 35 mph to 45 mph. D. The train made several stops, with an average rate of 57 mph.

Solution

In option A the bus is travelling at a speed of 50 miles per hour. This describes the velocity of bus along its route.

The other options are about the velocity of airplane, car and train

Answer:

what?????

Explanation:

Answer:

≈ 53 seconds

Explanation:

calculate Time required for the product to begin the falling-rate drying period

Initial moisture content =  0.77 kg water /kg of product

                                               = 3.35 kg water /kg solids

Critical moisture content = 0.3 kg water / kg product

                                         = 0.43 kg water / kg solids

∴ amount of water to be removed = 3.35 - 0.43 = 2.95kg water /kg solids

next: calculate surface are a of product during drying

= (0.05 * 0.05 ) * 6

= 0.015 m^3

Drying rate = 0.1 kg water m^2.s^-1 * 0.015 m^3 = 1.5 * 10^-3 kg water s^-1

applying product density

initial product mass = 0.11875 * 0.23 = 0.0273kg solid

hence total amount of water to removed = 2.92 * 0.0273 = 0.07972 kg

therefore : Time required for the product to begin the falling-rate drying period

= 0.07972 / 1.5 * 10^-3

= 53 seconds

Answer: It is  a nominal rate per year

This question is incomplete, the missing diagram is uploaded along this answer below.

Answer:

the forces required to keep the artery in place is 1.65 N

Explanation:

Given the data in the question;

Inlet velocity V₁ = 50 cm/s = 0.5 m/s

diameter d₁ = 15 mm = 0.015 m

radius r₁ = 0.0075 m

diameter d₂ = 11 mm = 0.011 m

radius r₂ = 0.0055 m

A₁ = πr² = 3.14( 0.0075 )² =  1.76625 × 10⁻⁴ m²

A₂ = πr² = 3.14( 0.0055 )² =  9.4985 × 10⁻⁵ m²

pressure at inlet P₁ = 110 mm of Hg = 14665.5 pascal

pressure at outlet P₂ = 95 mm of Hg = 12665.6 pascal

Inlet volumetric flowrate = A₁V₁ = 1.76625 × 10⁻⁴ × 0.5 = 8.83125 × 10⁻⁵ m³/s

given that; blood density is 1050 kg/m³

mass going in m' = 8.83125 × 10⁻⁵ m³/s × 1050 kg/m³ = 0.092728 kg/s

Now, using continuity equation

A₁V₁ = A₂V₂

V₂ = A₁V₁ / A₂ = (d₁/d₂)² × V₁

we substitute

V₂ =  (0.015 / 0.011 )² × 0.5

V₂ = 0.92975 m/s

from the diagram, force balance in x-direction;

0 - P₂A₂ × cos(60°) + Rₓ = m'( V₂cos(60°) - 0 )    

so we substitute in our values

0 - (12665.6 × 9.4985 × 10⁻⁵)  × cos(60°) + Rₓ = 0.092728( 0.92975 cos(60°) - 0 )    

0 - 0.6014925 + Rₓ =  0.043106929 - 0

Rₓ = 0.043106929 + 0.6014925

Rₓ = 0.6446 N

Also, we do the same force balance in y-direction;

P₁A₁ - P₂A₂ × sin(60°) + R[tex]_y[/tex] = m'( V₂sin(60°) - 0.5 )  

we substitute

⇒ (14665.5 × 1.76625 × 10⁻⁴) - (12665.6 × 9.4985 × 10⁻⁵) × sin(60°) + R[tex]_y[/tex] = 0.092728( 0.92975sin(60°) - 0.5 )

⇒ 1.5484 + R[tex]_y[/tex] = 0.092728( 0.305187 )

⇒ 1.5484 + R[tex]_y[/tex] = 0.028299    

R[tex]_y[/tex] = 0.028299 - 1.5484

R[tex]_y[/tex] = -1.52 N

Hence reaction force required will be;

R = √( Rₓ² + R[tex]_y[/tex]² )

we substitute

R = √( (0.6446)² + (-1.52)² )

R = √( 0.41550916 + 2.3104 )

R = √( 2.72590916 )

R = 1.65 N

Therefore, the forces required to keep the artery in place is 1.65 N

Explanation:

When a problem is discovered in the system design or manufacturing process, it generally gets reported to all concerned (if the company has an appropriate culture). Depending on the nature of the problem, a "quick fix" may be found by the discoverer, or by someone to whom the discovery is reported. In some situations, what seems a simple problem requires a rethinking of the entire manufacturing process, possible product recalls, possible plant retrofits, and even larger ramifications. This can happen regardless of where along the line the problem is discovered.

__

Hypothetical example:

A test technician determines that a lithium battery charger sometimes gets confused and doesn't shut down properly--continuing to charge the battery when it should not. If the proximate cause is a wire harness routing or a component improperly installed, a manufacturing engineer may be called in to address the issue. The manufacturing engineer's investigation may determine that a number of battery chargers have been delivered with the problem. Further investigation may reveal a potential for fire in situations where injury or loss of life are possible outcomes.

Assessment of the design by manufacturing, process, and design engineers may reveal more than one potential cause of the shutdown/fire-hazard issue, and that the location and nature of any fire may release toxins or cause damage to systems and equipment beyond those in the immediate vicinity of the defective battery and/or charger.

Such ramifications may require the attention of one or more systems engineers and/or a rethinking of system failure modes and effects, including fire detection and suppression, throughout the product. The product may be effectively "grounded" (taken out of service), with possible customer revenue implications, until such time as the issues can be satisfactorily resolved.

(Note: 787 Dreamliner battery problems were caused by the physics of the battery construction, not the charger. The rest of the scenario above is not a bad match.)

Solution :

Given :

Velocity, [tex]$V_1 =100$[/tex] cm/sec

Pressure,  [tex]$P_1 = 120 $[/tex] mm Hg

Then, [tex]$P_1 = \rho_1 g h$[/tex]

[tex]$P_1 = 0.120 \times 13.6 \times 1000 \times 9.81$[/tex]

    = 16.0092 kPa

[tex]$P_2 = 110 $[/tex] mm Hg

[tex]$P_2 = \rho_2 g h$[/tex]

    [tex]$= 0.110 \times 13.6 \times 1000 \times 9.81$[/tex]

    = 14.675 kPa

Then blood is incompressible,

[tex]$A_1v_1=A_2v_2$[/tex]

[tex]$\frac{\pi}{4}(25)^2\times 100=\frac{\pi}{4}(21)^2\times v_2$[/tex]

[tex]$v_2=141.72 \ cm/s$[/tex]

Then the linear momentum conservation fluid :

(Blood ) in y - direction

[tex]$P_1A_1+ P_2A_2-F_g = m_2v_2-m_1v_1$[/tex]

[tex]$m_1=m_2=P_1A_1v_1$[/tex]

              [tex]$=1.50 \times \frac{\pi}{4}\times (0.025)^2 \times 1.00$[/tex]

              = 0.515 kg/ sec

Then the linear conservation of momentum of blood in y direction.

[tex]$P_1A_1+ P_2A_2-F_g = m_2v_2-m_1v_1$[/tex]

[tex]$16.0092 \times 1000 \times \frac{\pi}{4} \times (0.025)^2+14.675 \times 1000\times \frac{\pi}{4}\times (0.021)^2$[/tex]

[tex]$-F_y=0.515(-1.4172-1)$[/tex]

7.858+5.0828- Fy = 0.515(-2.4172)

Fy = 14.1856 N

Answer:

a. 41

b. i. 15 mA ii. 625 mA

c. 192 Ω

Explanation:

Here is the complete question

A high-voltage discharge tube is often used to study atomic spectra. The tubes require a large voltage across their terminals to operate. To get the large voltage, a step-up transformer is connected to a line voltage (120 V rms) and is designed to provide 5000 V (rms) to the discharge tube and to dissipate 75.0 W. (a) What is the ratio of the number of turns in the secondary to the number of turns in the primary? (b) What are the rms currents in the primary and secondary coils of the transformer? (c) What is the effective resistance that the 120-V source is subjected to?

Solution

(a) What is the ratio of the number of turns in the secondary to the number of turns in the primary?

For a transformer N₂/N₁ = V₂/V₁

where N₁ = number of turns of primary coil, N₂ =number of coil of secondary, V₁ = voltage of primary coil = 120 V and V₂ = voltage of secondary coil = 5000 V

So,  N₂/N₁ = V₂/V₁

N₂/N₁ = 5000 V/120 V = 41.6 ≅ 41 (rounded down because we cannot have a decimal number of turns)

(b) What are the rms currents in the primary and secondary coils of the transformer?

i. The rms current in the secondary

We need to find the current in the secondary from

P = IV where P = power dissipated in secondary coil = 75.0 W, I =rms current in secondary coil and V = rms voltage in secondary coil = 5000 V

P = IV

I = P/V = 75.0 W/5000 V = 15 × 10⁻³ A = 15 mA

ii. The rms current in the primary

Since N₂/N₁ = V₂/V₁ = I₁/I₂

where N₁ = number of turns of primary coil, N₂ =number of coil of secondary, V₁ = voltage of primary coil = 120 V, V₂ = voltage of secondary coil = 5000 V, I₁ = current in primary coil and I₂ = current in secondary coil = 15 mA

So, V₂/V₁ = I₁/I₂

V₂I₂/V₁ = I₁

I₁ = V₂I₂/V₁

= P/V₁

= 75.0 W/120 V

= 0.625 A

= 625 mA

(c) What is the effective resistance that the 120-V source is subjected to?

Using V = IR where V =  voltage = 120 V, I = current in primary = 0.625 A and R = resistance of primary coil

R = V/I

= 120 V/0.625 A

= 192 V/A

= 192 Ω

Here are the eight stages of the recruitment process and what Garth might expect to occur or carry out at each stage in the explanation part.

The process of identifying, attracting, and selecting qualified candidates for a job opening in an organisation is known as recruitment.

Here are the eight stages of the recruitment process, as well as what Garth might expect to happen or do at each stage:

Thus, these are the stages of recruitment.

For more details regarding recruitment, visit:

https://brainly.com/question/30086296

#SPJ7

Answer:

False ( B )

Explanation:

considering that the wind turbine is a horizontal axis turbine

Power generated/extracted by the turbine can be calculated as

P =  n * 1/2 * p *Av^3

where: n = turbine efficiency

           p = air density

           A = πd^2 / 4

           v =  speed

From the above equation it can seen that increasing the Blade radius by 10% will increase the Blade Area which will in turn increase the value of the power extracted by the wind turbine

Answer:

What is the rms value Vrms of the voltage plotted in the graph?

Express your answer in volts.

Vrms = ? V

2.

When a lamp is connected to a wall plug, theresulting circuit can be represented by a simplified AC circuit, asshown in the figure. (Part B figure) Here the lamp has been replaced by a resistor with an equivalentresistance Part B figure) Here the lamp has been replaced bya resistor with an equivalent resistance R = 120 . What is the rms value Irms of the current flowing through the circuit?

Express your answer in amperes.

Irms = ? A

3.

What is the average power Pavg dissipated in the resistor?

Express your answer in watts.

Pavg = ? W

Answer:

the disadvantages can be people's thoughts about them and also a rival resort nearby which looks more posh pls mark brainliest i need 5 more for next rank thank you

Explanation:

Answer:

Yes

Explanation:

In such a case, one way to check the credit charge is to contact your bank, doing so would allow the bank to check your account properly to determine where the transaction was originated from.

Another way you could check is to contact the online merchant where such a transaction was initiated.

Answer:

1040 steel will be a possible candidate for this application since : Yield strength > stress

Explanation:

The 1040 steel would  be a possible candidate for this application because the stress experienced by the load is Lesser than its Yield strength

Given that 1040 steel has the following parameter values

Modulus of elasticity ( GPa )  = 205

Yield strength ( Mpa ) =  450

Poisson's ratio = 0.27

limitation of = 1.5 x 10^-2 mm.

stress = Tensile load / area of steel

          = 18,000 N / 4.418 * 10^-5 m^2

          = 407 .424 Mpa

Answer:

BFS uses Queue to find the shortest path. DFS uses Stack to find the shortest path. ... Time Complexity of BFS = O(V+E) where V is vertices and E is edges. Time Complexity of DFS is also O(V+E) where V is vertices and E is edges.

Explanation: