HELP PLEASE!! ASAP!!!!
can some answer this 2 questions please as paragraph i want it nowww it is graded what action should be taken to make it safe ? also the first question

Answer:

The answers are in the explanation. The pH is 5.91

Explanation:

The CH3NH2 reacts with HCl as follows:

CH3NH2 + HCl → CH3NH3⁺ + Cl⁻

When 200mL of HCl are added, the moles of CH3NH2 and HCl are reacting completely producing CH3NH3+ and Cl-. That means the species present are:

no H+. All reacted

yes H2O. Because the water is present in the solutions of HCl and CH3NH2

yes Cl-. Is a product of the reaction

Yes CH3NH2. Is consumed in the reaction but comes from the equilibrium of CH3NH3+

yes CH3NH3+. Is the other product of the reaction. MAJOR SPECIES

When 300.00mL of HCl are added, 100mL are in excess:

yes H+. Is in excess: H+ + Cl- = HCl in water. MAJOR SPECIES. Determine the pH of the solution.

yes H2O. Is present because the reactants are diluted.  

yes Cl-. Is a product of reaction and comes from HCl.

Yes CH3NH2. The reactant is over but comes from the equilibrium of CH3NH3+

yes no CH3NH3+. Yes. Is a product and remains despite HCl is in excess.

To find the pH:

At equivalence point the ion that determines pH is CH3NH3+. Its concentration is:

0.100L * (0.200mol/L) = 0.0200 moles / 0.300L = 0.0667M CH3NH3+

The equilibrium of CH3NH3+ is:

Ka = Kw/kb = 1x10-14/4.4x10-4 = 2.273x10-11 = [H+] [CH3NH2] / [CH3NH3+]

As both [H+] [CH3NH2] comes from the same equilibrium:

[H+] =  [CH3NH2] = X

2.273x10-11 = [X] [X] / [0.0667M]

1.5159x10-12 = X²

X = 1.23x10-6M = [H+]

As pH = -log [H+]

pH = 5.91

The pH at the equivalent point for this titration is "5.91".

[tex]CH_3NH_2 = 0.200\ M\\\\ \text{volume} = 100.0\ mL = 0.100\ L\\\\HCl = 0.100\ M\\\\[/tex]

We must now quantify the pH well at the equivalence point.

We know that even at the point of equivalence, moles of acid and moles of the base are equivalent. As such, first, we must calculate the number of moles of the given base.

Calculating the Moles in [tex]CH_3NH_2 = 0.200\ M \times 0.100\ L = 0.0200\ moles[/tex]

Calculating the Moles in [tex]HCl = 0.0200 \ moles[/tex]

Calculating the volume of [tex]HCl[/tex]:

[tex]\to \text{Molarity} = \frac{ \text{moles}}{\text{volume \ (L)}} \\\\\to \text{Volume} = \frac{\text{moles}}{\text{molarity}}\\\\[/tex]

                [tex]= \frac{0.0200 \ moles}{ 0.100\ M}\\\\= 0.200 \ L\\\\= 200 \ mL\\\\[/tex]

Calculating the reaction among the acid and base:

[tex]CH_3NH_2 + HCl \longrightarrow CH_3NH_3^{+} + Cl^-[/tex]

[tex]0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200[/tex]

Therefore the conjugate acid of the bases exists at the standard solution.

Then we must calculate the new molar mass of [tex]CH_3NH_3^+[/tex].

Total volume[tex]= 100 + 200 = 300\ mL = 0.300\ L[/tex]

[tex][CH_3NH_3^+] = \frac{0.0200\ mole}{ 0.300\ L}= 0.0667\ M[/tex]

Using the ICE table

[tex]CH_3NH_3^+ + H_2O \longrightarrow CH_3NH_2 + H_3O^+[/tex]

[tex]I \ \ \ \ \ \ \ \ \ \ 0.0667 \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ 0\\\\C\ \ \ \ \ \ \ \ -x\ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ +x\\\\E \ \ \ \ \ \ \ \ \ \ \ \ 0.0667-x \ \ \ \ \ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ \ \ \ \+x\\\\\to Ka = \frac{[CH_3NH_2] [H_3O^+] }{[CH_3NH_3^+]}[/tex]

Calculating [tex]K_a[/tex] from [tex]K_b[/tex]

[tex]\to K_a \times K_b = 1\times 10^{-14}\\\\\to K_a = \frac{1\times 10^{-14}}{4.4\times 10^{-4}} = 2.27\times 10^{-11}\\\\[/tex]

                           [tex]= 2.27\times 10^{-11} \\\\= x\times \frac{x}{(0.0667-x)}[/tex]

The x in the 0.0667-x can be ignored since the Ka value is just too small and it also does not follow the five percent criteria.

[tex]\to 2.27 \times 10^{-11} \times 0.0667 = x_2\\\\\to x_2 = 1.515\times 10^{-12}\\\\\to x = 1.23\times 10^{-6}\ M\\\\\to [H_3O^+] = x = 1.23\times 10^{-6}\ M\\\\[/tex]

We have the formula to calculate pH.

[tex]\to pH = - \log [H_3O^+] = - \log 1.23\times 10^{-6}\ M= 5.91[/tex]

The pH at the equivalent point for this titration is "5.91".

Find out more information about the pH here:

brainly.com/question/15289741

Answer:

[tex]v_2=549.2 m/s\\[/tex]

Explanation:

Given:

[tex]P_1=2500kPa\\T_1=1500 k\\V_1=5 m/s\\P_2=100 kPa\\T_2=1400 k\\A_2=10 cm^2[/tex]

Solution:

For [tex]Co_2[/tex] y=1.4

Since Nozzle is adiababic

So,

[tex]h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}\\\frac{v_2^2}{2}=(h_2-h_2)+\frac{r^2}{2}\\v_2^2=2(h_1-h_2)+v_1^2\\v_2=\sqrt{2(h_1-h_2)+v_1^2}[/tex]

Now,

[tex]h_1-h_2=Cp_1T_1-CP_2T_2\\h_1-h_2=(1989-1838.2)*10^3\\ =150.8 * 10^3\\Cp for co_2\\C_{p1}=1.326 kj/kg\\C_{p2}=1.313 kj/kg\\v_2=\sqrt{301600+25}\\ =549.2 m/s[/tex]

Answer:

Veracity, Value and Visualization are not only the characteristics of Big Data but are also the characteristics of relational databases. Veracity of data is issue with smallest data stores this is the reason that it is important in relation...

Answer:

Explanation:

Tech Social Media giant FB is one of those companies. Not long ago the ceo was brought to court to accusations that his company was selling user data. Turns out this is true and they are selling their users private data to companies all over the word. Once the news turned to something else, people focused on something new but the company still continues to sell it's users data the same as before. This is completely unethical as the information belongs to the user and they are not getting anything while the corporation is profiting.

Answer:

a) 1.9%

b) T2s = 505.5 k = 232.5°C

Explanation:

P1 = 95 kPa

T1 = 27°C  = 300 k

P2 = 600 kPa

T1 = 277°c  = 550 k

Table used : Table ( A - 17 ) Ideal gas properties of air

a) determining the isentropic efficiency of the compressor

Л = ( h2s - h1 ) / ( h2a -  h1 ) ---- ( 1 )

where ; h1 = 300.19 kJ/kg , T1 = 300 K , h2a = 554.74 kJ/kg , T2 = 550 k

To get h2s we have to calculate the the value of Pr2 using Pr1(relative pressure)

 Pr2 = P2/P1 * Pr = ( 600 / 95 ) * 1.306  hence; h2s = 500.72 kJ/kg

back to equation1

Л = 0.019 = 1.9%

b) Calculate the exit temperature of the air if compressor is reversible

if compressor is reversible the corresponding exit temperature

T2s = 505.5 k = 232.5°C

given that h2s = 500.72 kJ/kg

Answer:

Interval:  x∈ ( 0, ∞ )

There are no transient terms

Explanation:

x (dy/dx) – y= x^2sinx

Attached below is the detailed solution of the Given problem

There are no transient terms found in the general solution

Interval:  x∈ ( 0, ∞ )

Answer:

a)  Vr = - a^2/r cosθ  + aß / r

    Vθ = 1/r [ -a^2/r * sinθ ]

b) attached below

Explanation:

potential function

Ø= a^2 /r  cosØ + aßlnr ----- ( 1 )

a = radius ,  ß = constant

a) Expressions for Vr and Vθ

Vr =  dØ / dr  ----- ( 2 )

hence expression : Vr = - a^2/r cosθ  + aß / r

Vθ = 1/r dØ / dθ ------ ( 3 )

back to equation 1

dØ / dr = - a^2/r sinθ + 0  --- ( 4 )

Resolving equations 3 and 4

Vθ = 1/r [ -a^2/r * sinθ ]

b) expression for stream function

attached below

Answer:

[tex]\eta_{turbine} = 0.603 = 60.3\%[/tex]

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in  vapor state:

h₂ = [tex]h_{g\ at\ 125KPa}[/tex] = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than [tex]s_g[/tex] and greater than [tex]s_f[/tex] at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

[tex]x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88[/tex]

Now, we will find [tex]h_{2s}[/tex](enthalpy at the outlet for the isentropic process):

[tex]h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg[/tex]

Now, the isentropic efficiency of the turbine can be given as follows:

[tex]\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%[/tex]

Answer: Attached below is the missing detail and Mohr's circle.

i) б1 =  9.6 Ksi

б2 = -10.7 ksi

ii) 10.2 Ksi

iii)  -0.51Ksi

Explanation:

First step :

direct compressive stress on shaft

бd = P / π/4 * d^2

      = -20 / 0.785 * 5^2  = -1.09 Ksi

shear stress at the outer surface due to torsion

ζ = 16*T / πd^3

  = (16 * 250 ) / π * 5^3  = 010.19 Ksi

Calculate the Principal stress, maximum in-plane shear stress and average normal stress

Using Mohr's circle ( attached below )

i) principal stresses:

б1 = 4.8 cm * 2 = 9.6 Ksi

б2 = -5.35 cm * 2 = -10.7 ksi

ii) maximum in-plane shear stress

ζ  = radius of Mohr's circle

   = 5.1 cm = 10.2 Ksi   ( Given that ; 1 cm = 2Ksi )

iii) average normal stress

 = 9.6 + ( - 10.7 ) / 2

  = -0.51Ksi

Answer:

jsow

hfhcffnbxhdhdhdhdhdhdddhdhdgdhdhdhdhdhdhhhdhdjsksmalalaksjdhfgrgubfghhhhhhh

Explanation:

j

The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.

Reducing such a need to move in between multiple tabs, the split-screen has been valuable for increasing wages. In the several instances running a two or more desktop system will allow different programs to run throughout multiple devices. That works with the same process on both PC and laptop monitors.

Just display them side by side, instead of the switching among both the apps that has been used frequently. In this phase, an app that the snap to either left or right occupies a third of the display, and yet another app holds the two-thirds remaining. It refers to Split-Screen Mode.

Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization.

Therefore, The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.

Learn more about average power on:

https://brainly.com/question/14831024

#SPJ2

Answer:

[tex]T_o=82.1sec[/tex]

Explanation:

From the question we are told that:

Lost Time [tex]t=12secs[/tex]

Sum of critical flow ratios [tex]X=0.72[/tex]

Generally the Webster Method's equation for Optimum cycle time is is mathematically given by

 [tex]T_o=\frac{1.5t+5}{1-x}[/tex]

 [tex]T_o=\frac{1.5*12+5}{1-0.72}[/tex]

 [tex]T_o=82.1sec[/tex]

Answer:

1709.07 ft^3/s

Explanation:

Annual peak streamflow = Log10(Q [ft^3/s] )

mean = 1.835

standard deviation = 0.65

Probability of levee been overtopped in the next 15 years = 1/5

Determine the design flow ins ft^3/s

P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2

                         ∴  T = 67.72 years

Q₁₅ = 1 - 0.2 = 0.8

Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )

K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )

    = 2.1504

back to equation 1

Zt = 1.835 + ( 2.1504 * 0.65 )  = 3.23276

hence:

Log₁₀ ( Qt(ft^3/s) ) = Zt  = 3.23276

hence ; Qt = 10^3.23276

                  = 1709.07 ft^3/s

Answer:

The code is as follows:

<form name = "myForm">

       <div>

           <input type="radio" name="vehicle" value="D0" id="D0"/>

           <label for="D0">Car</label>

       </div>

       <div>

           <input type="radio" name="vehicle" value="D1" id="D1"/>

           <label for="D1">Bike</label>

       </div>

   </form>

Explanation:

This defines the first button

           <input type="radio" name="vehicle" value="D0" id="D0"/>

           <label for="D0">Car</label>

This defines the second button

           <input type="radio" name="vehicle" value="D1" id="D1"/>

           <label for="D1">Bike</label>

The code is self-explanatory, as it follows all the required details in the question

Answer:

8.6 ft³/s

Explanation:

The force due to the water jet F = mv where m = mass flow rate = ρQ where ρ = density of water = 62.4 lbm/ft³ and Q = volume flow rate. v = velocity of water jet = 30 ft/s

So, F = mv

F = ρQv

making Q subject of the formula, we have

Q = F/ρv

Since F = force due to water jet = force needed to hold the plate against the water stream = 500 lbf = 500 × 1 lbf = 500 × 32.2 lbmft/s² = 16100 lbmft/s²

Since

Q = F/ρv

Substituting the values of the variables into the equation for Q, we have

Q = F/ρv

Q = 16100 lbmft/s²/(62.4 lbm/ft³ × 30 ft/s)

Q = 16100 lbmft/s²/1872 lbm/ft²s

Q = 8.6 ft³/s

So, the volume flow rate is 8.6 ft³/s.

Complete Question

Complete Question is attached below.

Answer:

[tex]V'=5m/s[/tex]

Explanation:

From the question we are told that:

Diameter [tex]d=0.10m[/tex]

Power [tex]P=4.0kW[/tex]

Head loss [tex]\mu=10m[/tex]

 [tex]\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu[/tex]

 [tex]\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10[/tex]

 [tex]H_m=(\frac{200*10^3}{1000*9.8}-10)[/tex]

 [tex]H_m=10.39m[/tex]

Generally the equation for Power is mathematically given by

 [tex]P=\rho gQH_m[/tex]

Therefore

 [tex]Q=\frac{P}{\rho g H_m}[/tex]

 [tex]Q=\frac{4*10^4}{1000*9.81*10.9}[/tex]

 [tex]Q=0.03935m^3/sec[/tex]

Since

 [tex]Q=AV'[/tex]

Where

 [tex]A=\pi r^2\\A=3.142 (0.05)^2[/tex]

 [tex]A=7.85*10^{-3}[/tex]

Therefore

 [tex]V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}[/tex]

 [tex]V'=5m/s[/tex]

Answer:

Explanation:

B. you would grab the plug closest to the outlet

If you don't have enough experience, it's always best to leave socket changing to the experts. If you make a mistake, you might inflict harm and potentially endanger yourself and other people. Read on if you're interested in learning how to change a socket safely. Thus, option D is correct.

Grip the plug, not the electrical cable, when taking an electrical plug out of its socket. Before handling an electrical switch, socket, or outlet, hands must be fully dry.

Reduce the extra so that it rests only on top of the existing plasterboard. If necessary, push it back a little by using your finger. Fill the dent with ready-mixed filler or powdered filler, whichever you want, and bring it flush with the surrounding wall. Allow to dry, then sand off any excess.

Therefore, It is acceptable to grasp the electrical cord when removing an electrical plug from its socket

Learn more about electrical plug here:

https://brainly.com/question/28932892

#SPJ5

Answer:

a.  164 °F b. 91.11 °C c. 1439.54 kJ

Explanation:

a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?

Since the starting temperature is 48°F and the final temperature which water boils is 212°F, the number of degrees Fahrenheit we would need to raise the temperature is the difference between the final temperature and the initial temperature.

So, Δ°F = 212 °F - 48 °F = 164 °F

b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?

To find the degree change in Celsius, we convert the initial and final temperature to Celsius.

°C = 5(°F - 32)/9

So, 48 °F in Celsius is

°C₁ = 5(48 - 32)/9

°C₁ = 5(16)/9

°C₁ = 80/9

°C₁ = 8.89 °C

Also, 212 °F in Celsius is

°C₂ = 5(212 - 32)/9

°C₂ = 5(180)/9

°C₂ = 5(20)

°C₂ = 100 °C

So, the number of degrees in Celsius you must raise the temperature is the temperature difference between the final and initial temperatures in Celsius.

So, Δ°C = °C₂ - °C₁ = 100 °C - 8.89 °C = 91.11 °C

c. [2 pts] How much energy is required to heat the four quarts of water from

48°F to 212°F (boiling)?

Since we require 15.8 kJ for every degree Celsius of temperature increase of the four quarts of water, that is 15.8 kJ/°C and it rises by 91.11 °C, then the amount of energy Q required is Q = amount of heat per temperature rise × temperature rise =  15.8 kJ/°C × 91.11 °C = 1439.54 kJ

Solution :

The spring is expanded by 2 times of the block when it moves down an inclined by x times.

Here, [tex]$x_1$[/tex] = 39 mm

        [tex]x_2[/tex] = 225 mm

a). From the work energy principal,

   Work forces = kinetic energy

[tex]$(mg \sin 50^\circ)\times \frac{99}{1000}-(\mu_k mg \cos 50^\circ) \times \frac{99}{1000} -\frac{1}{2}k(0.225^2 - 0.039^2)=\frac{1}{2}m(V^2_2-0.08^2)$[/tex]

[tex]$(112.6 \times 0.099)-(14.17 \times 0.099)-4.91= 7.5(V^2_2-0.08^2)$[/tex]

[tex]$9.75= 7.5(V^2_2-0.08^2)$[/tex]

[tex]$1.3= V^2_2-0.08^2$[/tex]

[tex]$V_2=1.14\ m/s$[/tex]

b). calculating the distance travelled by the block before it comes to rest.

Substitute the value of [tex]V_2[/tex] in (1),

[tex]$-(\mu_kmg \cos 50^\circ)x + (mg \sin 50^\circ)x-\frac{1}{2}k\left( ( 2x+0.039)^2 - 0.039^2\right)= -\frac{1}{2}m(0.08)^2$[/tex]

[tex]$-14.17x+112.6x - 100(4x^2+0.156x)=-0.048$[/tex]

[tex]$98.43x - 100(4x^2+0.156x)+0.048=0$[/tex]

[tex]$98.43x - 400x^2-15.6x+0.048=0$[/tex]

[tex]$82.83x - 400x^2+0.048=0$[/tex]

[tex]$ 400x^2- 82.83x-0.048=0$[/tex]

x = 0.20 m

Answer:

Tensile stress = 0.1855Kpsi

Total deformation = 0.0012243 in

Unit strain =  1.855 *10^-5   or  18.55μ

Change in the rod diameter = 5.02 * 10^ -6 in

Explanation:

Data given: D= 0.82 in

                   L = 5.5 ft * 12 = 66 in

load (p) = 3000 (Ibf) /32.174 = 93.243 Ibm

Area = (π/4) D² = (π/4) 0.82²  = 0.502655 in²

∴ Tensile stress Rt = P/A = 93.243/0.502655 = 185.50099 pound/in²

                           Rt = 0.1855 Kpsi

∴ Total deformation = PL / AE = Rt * L/ Eal

                                 = 0.1855 * 10³  * 66 / 10000 * 10³

                                 = 0.0012243 in

∴the unit strains = total deformation / L = 0.0012243/ 66

                          =0.00001855 = 1.855 *10^-5

                         = 18.55μ

∴ Change in rod   Δd/ d = μ ΔL/L

                           = (0.33) 1.855 *10^-5 * 0.82

                           = 5.02 * 10^ -6 in

jrjrkeekkekekkwkkakkllalllalallalllalalaallalalaalalalalalallallallllallalalallaaallalallllllallllllllalaalalalaaaaalalaaaaaaalgjgiejxpwunfifjruritiririirieoeowowowowowowowowooeowowowoeeoeowowowowowowowoowowwowowowoozoeisiaokseekxidjdkdjfidjfjdjfjfjrifjrifjdirjdjrjfjrjfjrjfjrfuejwwuxmaneanfjkaosndjxieneamalhaqzeeshanvhorahfuensiwjakaksjdhfhfnfhfndjxnxmakaalalalwlwlwwow

Answer:

The answer is "15 N".

Explanation:

Please find the complete question in the attached file.

In frame B:

For just slipping:

[tex]\to \frac{P}{2} \cos \theta =mg \sin \theta\\\\\to P=2 mg \tan \theta \\\\[/tex]

        [tex]=2 \times 1 \times g \times \tan 37^{\circ}\\\\ =2 \times 10 \times \frac{3}{4}\\\\ =15 \ N[/tex]

Answer:

Each of the metal specimens HAS an indentation near the center to ensure that the fracture point would occur in this region. Tension tests WERE CONDUCTED as follows. Two pieces of reflective tape WERE PLACED approximately 1 inch apart in the center of the specimen where the indentation 4 WAS LOCATED. The width and the thickness of the specimen at this location WAS MEASURED using a Vernier caliper. Then the specimen WAS SECURED in the MTS Load Frame. A laser extensometer WAS PLACED into position to measure the deformation of the specimen. The laser extensometer WAS USED to measure the original distance between the pieces of reflective tape. The MTS WAS SET to elongate the specimen one tenth of an inch every minute.

Answer:

1. Drawing grid.

2. Orthogonal.

3. Geometries.

4. CTRL+N.

5. Thirteen (13).

Explanation:

CAD is an acronym for computer aided design and it is typically used for designing the graphical representation of a building plan. An example of a computer aided design (CAD) software is AutoCAD.

Some of the features of an AutoCAD software are;

1. Drawing grid: is a regular pattern of dots displayed on the screen of an AutoCAD software, which acts as a visual aid and it's also used to define the extent of a drawing.

2. Ortho is short or an abbreviation for orthogonal, which means either vertical or horizontal.

3. Tangent is a point where two geometries meet at just a single point.

4. If you want to create a new drawing, simply press CTRL+N for the short cut key.

5. There are thirteen object snaps (Osnap) that can help you perform your task on AutoCAD easily. The 13 object snaps (Osnap) are; Endpoint, Midpoint, Apparent intersect, Intersection, Quadrant, Extension, Tangent, Center, Insert, Perpendicular, Node, Parallel, and Nearest.

Answer: hello your question is incomplete below is the complete question

answer:

N010 GO2 X7.0 Y2.0 15.0 J2.0  ( option 1 )

Explanation:

Given that the NC machining has to be moved from point ( 5,4 ) to point ( 7,2 ) along a circular path

GO2 = circular interpolation in a clockwise path

G91 = incremental dimension

hence the correct option is :

N010 GO2 X7.0 Y2.0 15.0 J2.0  

Answer:

the current consumed is 3.3 A

Explanation:

Given;

resistance, R = 30 ohms

inductance, L = 200 mH

Voltage supply, V = 230 V

frequency of the coil, f = 50 Hz

impedance, Z = 69.6 Ohms

The current consumed is calculated as;

[tex]I = \frac{V}{Z} \\\\I = \frac{230}{69.6} \\\\I = 3.3 \ A[/tex]

Therefore, the current consumed is 3.3 A