In a CNC machining operation, the has to be moved from point (5, 4) to point(7, 2)along a circular path with center at (7,2). Before starting operation, the tool is at (5, 4).The correct G and M code for this motion is

Answer:

Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible

Answer: Biometrics

Explanation:

Dual factor authentication refers to an electronic authentication method whereby a user will only be granted an access to an application or a website after the user has successfully been able to present two pieces of evidence which then grants access to the application or website.

Since the technician wants to implement a dual factor authentication system, the biometrics should be implemented during the authorization stage.

Biometrics refers to the body measurements and the calculations that are related to the characteristics of humans. Biometric authentication is used as a form of identification.

A project is meant to achieve clearly defined goals within present constraints

The correct option for the source of work in a project which is NOT a source of authorized work to be performed is option (a)

a. Scope creep

The reason for the selecting the above option is as follows;

In project management, alongside work planned in a work package, authorized work to be performed can come from;

A scope creep, however, is generally is due to the inclusion of the end

users of the project at a time that is behind schedule, insufficient analysis of

the requirements, and project complexity underestimation. and consists of

changes made after a project has commenced

A scope creep therefore, changes the amount of work to be done which

therefore alters the project timeline, and due to project, budgetary, and

end user constraints, CANNOT be a source of authorized work to be

performed

The correct option for the source that is NOT a authorized work to be performed is option a. Scope creep

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Answer:

Amount of gas still in cylinder = 28 pound

Explanation:

Given:

Amount of gas in cylinder = 50 pound

Amount of gas used in Ms. Jones system = 13 pound

Amount of gas used in client system = 9 pound

Find:

Amount of gas still in cylinder

Computation:

Amount of gas still in cylinder = Amount of gas in cylinder - Amount of gas used in Ms. Jones system - Amount of gas used in client system

Amount of gas still in cylinder = 50 - 13 - 9

Amount of gas still in cylinder = 28 pound

Answer:

not understand language

Answer:

80% of the people that were killed weren't wearing a safety flotation device ( in correct terminology Personal Flotation Device, or PFD )

Explanation:

Hence they drowned due to the lack of safety.

The condition present in over 80% of paddling fatalities is that the victims were said to be not wearing a personal flotation device (PFD).

The act of paddling is known to be a sport or something one can do for fun or for games.

Conclusively, a lot of paddling fatalities 1995-2000 was due to the fact that the people who were engage in the sport or act were not putting on  a personal flotation device (PFD) and this can lead to death or drowning.

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Answer: Create lessons learned at the end of the project.

Explanation:

Lessons learned are the experiences that are gotten from a project which should be taken into account for the future projects. Lesson learned are created at the end of the project.

The main objective of the lessons learned is that they show both the positive experience and the negative experience of a project and this will help the future projects that will be undertaken.

Answer:

Explanation:

Be bop

Explanation:

Find a minimal set of cycles that covers all vertices and edges of the circuit graph. For each cycle, define a "mesh" current, and write the Kirchhoff's Voltage Law (KVL) equation with respect to each of the edges in the cycle. Where an edge is part of more than one cycle, all current(s) defined for the edge will contribute to the voltage there.

This will give as many equations as there are mesh currents. Solve the resulting system of equations. The (signed) sum of the mesh currents through any edge is the current in that circuit branch.

__

Example

Consider the attached circuit. It shows mesh currents I1, I2, and I3 in graph cycles with those numbers. The KVL equations are ...

  mesh 1: I1(R3 +R2 +R1) -I2·R1 -I3·R2 = Vi (the voltage across the current source)

  mesh 2: -I1·R1 +I2(r1 +1/(sC)) -I3(1/(sC)) = Vs

  mesh 3: -I1·R2 -I2(1/(sC)) +I3(R2 +sL +1/(sC)) = 0

You will note that the matrix of equation coefficients is symmetric.

__

In this example, you will end with I1 as a function of Vi. If I1 is a given source value, that relation can be used to find Vi.

Answer:

The main sections included in the bill of quantities are Form of Tender, Information, Requirements, Pricing schedule, Provisional sums, and Day works.

Answer:

348 g

Explanation:

The heat gained by the ice equals the heat lost by the hot tea.

Also, since we require a very, very small amount of ice and some water, to still have some ice, the temperature of the hot tea has to decrease to the melting point of ice which is 0°C. So, the final temperature of the mixture is 0 °C.

So, mc(T - T') + mL = -MC(T - T") where m = mass of ice, c =  specific heat of ice = 2090 J/kgK., T = final temperature of mixture = 0 °C, T' = initial temperature of ice = -12 °C, L = heat of fusion of H2O = 3.33 × 10⁵ J/kg, M = mass of hot tea = ρV where ρ = density of tea = 1.00 g/mL and V = volume of hot tea = 350 mL. So, M = ρV = 1.00 g/mL × 350 mL = 350 g = 0.350 kg, C = specific heat of tea = 4190 J/kgK and T" = initial temperature of tea = 85 °C.

Making m subject of the formula, we have

mc(T - T') + mL = -MC(T - T")

m[c(T - T') + L] = -MC(T - T")  

m = -MC(T - T")/[c(T - T') + L]

substituting the values of the variables into the equation, we have

m = -MC(T - T")/[c(T - T') + L]

m = -0.350 kg × 4190 J/kgK(0 °C - 85 °C.)/[2090 J/kgK(0 °C. - (-12 °C)) + 3.33 × 10⁵ J/kg]

m = -1466.5 J/kK(- 85 °C)/[2090 J/kgK(0 °C + 12 °C)) + 3.33 × 10⁵ J/kg]

m = 124652.5 J/[2090 J/kgK(12 °C) + 3.33 × 10⁵ J/kg]

m = 124652.5 J/[25080 J/kg + 3.33 × 10⁵ J/kg]

m = 124652.5 J/358080 J/kg

m = 0.3481 kg

m = 348.1 g

m ≅ 348 g

So, the minimum amount of ice to be added is 348 g.

Answer:

business

Explanation:

because shopping has great income

Answer:

is ur question right

Explanation:

is ur question right

is ur question right

Answer:

  parallel

Explanation:

All components in this circuit are tied in parallel. Each component experiences the same voltage from one terminal to the other. It is a parallel circuit.

Answer:

Explanation:

Two that come to mind:

Answer:

what does it mean

Explanation:

sorry

plz write in English

Answer:

flow chart

Explanation:

it's a flow chart right ? don't know the explanation cause I am not even sure I am right

Answer:

[tex]Q_2 = 32[/tex] mL/s

Explanation:

Given :

The flow is incompressible viscous flow.

The initial flow rate, [tex]Q_1[/tex] = 1 mL/s

Initial diameter, [tex]D_1= D_0[/tex]

Initial length, [tex]L_1=L_0[/tex]

The initial pressure difference to maintain the flow, [tex]P_1=P_0[/tex]

We know for a viscous flow,

[tex]$\Delta P = \frac{32 \mu V L}{D^2}$[/tex]

[tex]$\Delta P = \frac{32 \mu Q L}{\frac{\pi}{4}D^4}$[/tex]

[tex]$Q \propto \Delta P \times D^4$[/tex]

[tex]$\frac{Q_1}{Q_2}= \frac{P_1}{P_2} \times \left( \frac{D_1}{D_2} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{P_0}{2P_0} \times \left( \frac{D_0}{2D_0} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{1}{2} \times \left( \frac{1}{2} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{1}{32}$[/tex]

∴ [tex]Q_2 = 32[/tex] mL/s

The flow rate if the pressure difference was increased to 2P0 and the diameter was increased to 2D0 is; Q2 = 32 mL/s

We are given;

Initial flow rate; Q1 = 1 mL/s

Initial uniform diameter; D0

Initial Length; L0

Initial Pressure difference; P0

Relationship between pressure, flow rate and diameter for vicious flow is given by;

Q1/Q2 = (P1/P2) × (D1/D2)⁴

Where;

Q1 is initial flow rate

Q2 is final flow rate

P1 is initial pressure difference

P2 is final pressure difference

D1 is initial diameter

D2 is final diameter

We are told that the pressure difference was increased to 2P0 and the diameter was increased to 2D0. Thus;

P2 = 2P0

D2 = 2D0

Thus;

1/Q2 = (P0/2P0) × (D0/2D0)⁴

>> 1/Q2 = ½ × (½)⁴

1/Q2 = 1/32

Q2 = 32 mL/s

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Answer:

Hey mate......

Explanation:

This is ur answer.....

Hope it helps!

Brainliest pls!

Follow me :)

Answer:

Non-functional requirements when defined and executed well will help to make the system easy to use and enhance the performance

Explanation:

The claim will be impractical. The reason behind this is provided below throughout the explanation segment.

Thus, the idea of using water instead of petroleum appears unworkable throughout traditional engines.

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Answer:

3.983 MW

Explanation:

Given that:

At the inlet:

Velocity (v₁) = 30 m/s

Enthalpy (h₁) = 3348 kJ/kg

At the outlet:

Velocity (v₂) = 60 m/s

Enthalpy (h₂) = 2550 kJ/kg

Mass flow rate (m) = m₁ = m₂ = 5kg/s

According to the steady flow energy equation:

[tex]Q+ m_1 (h_1 + \dfrac{v_1^2}{2000}+ \dfrac{gz_1}{1000} )= m_2(h_2+\dfrac{v_2^2}{2000}+\dfrac{gz_2}{1000})+W_{shaft}[/tex]

Since the elevation (z) is negligible and flow via the turbine is adiabatic:

Then,

Q = 0 and z₁ = z₂

∴

[tex]W_{shaft} = (mh_1-mh_2) + (\dfrac{mv_1^2-mv_2^2}{2000})[/tex]

[tex]W_{shaft} = ((5*3348) -(5*2550)) + (\dfrac{(5*(30)^2)-(5*(60)^2)}{2000})[/tex]

[tex]W_{shaft} = (16740-12750) + (\dfrac{4500-18000}{2000})[/tex]

[tex]W_{shaft} = (16740-12750) + (-6.75)[/tex]

[tex]W_{shaft} = 3983.25 \ kW[/tex]

[tex]\mathbf{W_{shaft} = 3.983 \ MW}[/tex]

Answer:

I don't understand the question

Answer:

As per my experience Egor Popov is one the best book for mechanics of materials.

Explanation:

hope this helps

Answer:

B

Explanation:

Anxiety is very common especially nowadays but it's especially common in psychological disorders