electrical engineering​

Answer:

10.007

Explanation:

Assuming we have to find out the compression ratio of the engine

Given information

Cubic capacity of the engine, V = 245 cc

Clearance volume, V_c = 27.2 cc

over square-ratio = 1.1

thus,

D/L = 1.1

where,

D is the bore

L is the stroke

Now,

Volume of the engine V =[tex]\frac{\pi}{4} D^2L[/tex]

plugging values we get

245 = [tex]\frac{\pi}{4} D^3/1.1[/tex]

Solving we get D =7 cm

therefore,  L= 7/1.1 =6.36 cm

Now,

the compression ratio is given as:

r =(V+V_c)/V_c

on substituting the values, we get

r = (245+27.2)/27.2 =10.007

Hence, Compression ratio = 10.007

Answer:

I would say Firewall security. Malware is pretty corrupting, but the firewall I think is good for protection if you have a good one.

Explanation:

Answer:

Explanation:

Given that:

[tex]y = \int^t_og'(t-s) f(s) ds \ \text{is solution to } \ my"ky= f(t)[/tex]

where;

[tex]g'(0) = \dfrac{1}{m}[/tex]     and [tex]mg"+kg = 0[/tex]

[tex]\text{Using Leibniz Formula to prove the above equation:}[/tex]

[tex]\dfrac{d}{dt} \int ^{b(t)}_{a(t)} \ f (t,s) \ ds = f(t,b(t) ) * \dfrac{d}{dt}b(t) - f(t,a(t)) *\dfrac{d}{dt}a(t) + \int ^{b(t)}_{a(t)}\dfrac{\partial}{\partial t} f(t,s) \ dt[/tex]

So, [tex]y = \int ^t_0 g' (t-s) f(s) \ ds[/tex]

[tex]\text{By differentiation with respect to t;}[/tex]

[tex]y' = g'(o) f(t) \dfrac{d}{dt}t- 0 + \int^{t}_{0}g'' (t-s) f(s) ds \\ \\ y' = \dfrac{1}{m}f(t) + \int ^t_0 g'' (ts) f(s) \ ds[/tex]

[tex]y'' = \dfrac{1}{m} f'(t) + g"(0) f(t) + \int^t_o g"'(t-s) f(s)ds --- (1)[/tex]

[tex]Since \ \ mg" (t) +kg (t) = 0 \\ \\ \implies g" (t) = -\dfrac{k}{m} g(t) --- (111) \\ \\ put \ t \ =0 \ we \ get;\\g" (0) = - \dfrac{k}{m } g(0) \\ \\ g"(0) = 0 \ \ \ \ ( because \ g(0) =0) \\ \\[/tex]

[tex]Now \ differentiating \ equation (111) \ with \ respect \ to \ t \\ \\ g"'(t) = -\dfrac{k}{m}g(t) \\ \\ replacing \ it \ into \ equation \ (1) \\ \\ y" = \dfrac{1}{m}f' (t) + 0 + \int ^t_o \dfrac{-k}{m}g' (t-s) f(s) \ ds \\ \\ y" = \dfrac{1}{m}f' (t) - \dfrac{k}{m} \int ^t_o g' (t-s) \ f(s) \ ds \\ \\ y" = \dfrac{1}{m}f'(t) - \dfrac{k}{m}y \\ \\ my" = f'(t)-ky \\ \\ \implies \mathbf{ my" +ky = f'(t)}[/tex]

Where are your answer options?

Answer:

Implement a hazard communication program

Explanation: i took the quiz

Answer:

a) the inductance of the coil is 6 mH

b) the emf generated in the coil is 18 mV  

Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l =  35.0 cm = 0.35 m

a) the inductance of the coil (in mH)

inductance of solenoid

L = N²μA / l

A = πd²/4  

so

L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)

L =  0.00841549 / 1.4

L = 6 × 10⁻³ H    

L = 6 × 10⁻³ × 1000 mH

L = 6 mH

Therefore, the inductance of the coil is 6 mH

b)

Emf ( ∈ ) = L di/dt

given that; di/dt = 3.00 A/sec

{∴ di = 3 - 0 = 3 and dt = 1 sec}

Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV  

Therefore, the emf generated in the coil is 18 mV  

Answer:

it is indeed C

Explanation:

Answer:

c

Explanation:

Answer:

Explanation:

This is Answer....

Answer: hello attached below is the properly written chain reaction  to your question

answer :

d[NO] / dt = [tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] + K_{3f}[N][OH][/tex]

d[N] / dt = [tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] - K_{3f}[N][OH][/tex]

Explanation:

write out expressions for  d[NO] / dt and d[N] / dt

Given :

properly written chain reaction ( attached below)

Expression for d[NO] / dt can be written as

[tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] + K_{3f}[N][OH][/tex]

Expression for d[N] / dt can be written as

[tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] - K_{3f}[N][OH][/tex]

Answer:

El =50 degre

Explanation:

Yan ang answer

Answer: hello your question lacks the required diagram attached below is the diagram

answer :  29528.1  N/m^2

Explanation:

Given data :

dimensions of tank :

Length = 5-m

Width = 4-m

Depth = 2.5-m

acceleration of tank = 2m/s^2

Determine the maximum gage pressure in the tank

Pa ( pressure at point A )  = s*g*h1

    = 10^3 * 9.81 * 3.01

    = 29528.1  N/m^2

attached below is the remaining part of the solution

Answer:

The application of regulations in locations are very important.

Explanation:

The application of regulations in locations are very important in order to gain more benefit from it because people choose those places that are well regulated and having more facilities. If the location has baths, showers, electric floor heaters and other necessities so the people prefer the place over another and increase of clients occurs which give more benefits to the place owners.

Answer:

c   Waste stack

Explanation:

Answer:

C Electrical Connections

Explanation:

In reading says . However, electrical

connections aren’t shown in construction drawings.

Answer:

Explanation:

From the information given:

[tex]E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa \\ \\ d = 25 \ mm \ \\ \\ D = 45 \ mm \ \\ \\ L = 761 \ mm \\ \\ P = -88 kN[/tex]

The total load is distributed across both the rod and tube:

[tex]P = P_1+P_2 --- (1)[/tex]

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

[tex]\delta_1=\delta_2[/tex]

[tex]\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}[/tex]

[tex]\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}[/tex]

[tex]P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})[/tex]

[tex]P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}[/tex]

[tex]P_2 = 6.6212 \ P_1[/tex]

Replace [tex]P_2[/tex] into equation (1)

[tex]P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\ -88 = 7.6212 \ P_1 \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\ P_1 = -11.547 \ kN[/tex]

Finally, to determine the normal stress in aluminum rod:

[tex]\sigma _1 = \dfrac{P_1}{A_1} \\ \\ \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}[/tex]

[tex]\sigma_1 = - 23.523 \ MPa}[/tex]

Thus, the normal stress = 23.523 MPa in compression.

Answer:

a

Explanation:

Answer:

1.933 KN-M

Explanation:

Determine the largest permissible bending moment when the composite bar is bent  horizontally

Given data :

modulus of elasticity of steel = 200 GPa

modulus of elasticity of aluminum = 75 GPa

Allowable stress for steel = 220 MPa

Allowable stress for Aluminum = 100 MPa

a = 10 mm

First step

determine moment of resistance when steel reaches its max permissible stress

next : determine moment of resistance when Aluminum reaches its max permissible stress

Finally Largest permissible bending moment of the composite Bar = 1.933 KN-M

attached below is a detailed solution

Answer:

Attached below is the derived next state equations

Explanation:

Attached below is the derived next state equations

used for the solution of the given problem.

Answer:

a. 2.30

b. decreases with increasing velocity.

c. 0.179 kg/s.

Explanation:

Without mincing let's dive straight into the solution to the question above.

                                                         [a].

The improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate can be determined by calculating turbulent flow:

The turbulent flow over the plate= 10 × 0.5/ 20.92 × 10⁻6 = 2.39 × 10⁵.

While the turbulent flow correlation = 0.037( 2.39 × 10⁵)^[tex]\frac{4}{5}[/tex] (0.7)^[tex]\frac{1}{3}[/tex] = 659.6.

Array of slot noozle = [10 × (2  × 0.004)]/ 20.92  × 10^-6] = 3824.

where A = 4/56 =0.714.

And Ar = [ 60 + 4 (40/2  × 4) - 2 ]^2 ]-1/2 = 0.1021.

N = 2/3 (0.1021)^3/4 [ 2  ×  3824/ ( 0.0714 / 0.1021) + (.1021/0.0714)] (0.700)^0.42 =24.3.

h = 24.3  ×  0.030/0.004 = 91.1 W/m^2k.

Therefore; 659.6  × 0.030/0.5 = 39.0 W/m²k.

The turbulent flow = 0.5 × 39.6 × 0.5( 140 -15) = 1237.5 W.

The slot noozle = 91.1  ×  0.5  ×  0.5 [ 140 -15] = 2846.87W.

The improvement in cooling rate = 2846.87/ 1237.5 = 2.30.

                                                     [b].

2.3 [ (2^2/3)/ 2^4/5] = 2.1

Thus, it decreases with increasing velocity

                                                      [c].

The  air mass rate requirement for the slotted nozzle arrangement = 9 × 0.995 (0.5 × 0.004)10 = 0.179 kg/s.

Answer:

a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s

b) The pressure difference across the pump is approximately 293.118 kPa

Explanation:

The efficiency of the pump = 78%

The power of the pump = 5 -kW

The height of the pool above the underground water, h = 30 m

The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

[tex]P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}[/tex]

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

[tex]\eta_t[/tex] = The efficiency of the pump = 78%

[tex]\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}[/tex]

[tex]Q_{max}[/tex] = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump [tex]Q_{max}[/tex] ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

Answer:

a) 75%

b) 82%

Explanation:

Assumptions:

[tex]\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}[/tex]

Properties: The density of water [tex]\delta = 1000 kg/m^3[/tex]

Conversions:

[tex]165 \ ft \ to \ meters = 50 m \\ \\7000 \ lbm/s \ to \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \ to \ kilowatt = 1166 kw \\ \\[/tex]

Analysis:

Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:

[tex]e_{mech_{in}} - e_{mech_{out}} = gh - 0[/tex]

Then;

[tex]gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}[/tex]

gh = 0.491 kJ/kg

[tex]\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg[/tex]

= 1559 kW

Therefore; the overall efficiency is:

[tex]\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}[/tex]

[tex]= \dfrac{1166 \ kW}{1559 \ kW}[/tex]

= 0.75

= 75%

b) mechanical efficiency of the turbine:

[tex]\eta_{turbine- generator} = \eta_{turbine}\times \eta_{generator}[/tex]

thus;

[tex]\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%[/tex]

def digit_sum(number):

 sumOfDigits = 0

 while number > 0:

     sumOfDigits += number % 10

     number = number // 10

 return sumOfDigits

or you can use,

In [2]: digit_sum(10)

Out[2]: 1

In [3]: digit_sum(434)

Out[3]: 11

or you can use,

digits = "12345678901234567890"

digit_sum = sum(map(int, digits))

print("The equation is: ", '+'.join(digits))

print("The sum is:", digit_sum)

Have a great day <3

Answer:

T = 438.87 N.m

Explanation:

The power required to raise the 4961 N load in 10 meters for 2 minutes is:

[tex]P = \dfrac{4961*10}{2*60}\\ \\ P = 413.42 Nm/sec[/tex]

P = Torque  × W

[tex]413.42 = T \times \dfrac{2 * \pi*9}{60}[/tex]

[tex]413.42 = T \times0.942 \\ \\ T = \dfrac{413.42}{0.942}[/tex]

[tex]T = \dfrac{413.42}{0.942}[/tex]

T = 438.87 N.m

Answer:

The pavement of the carpool lanes is marked with the diamond symbol. These carpool lanes are reserved for buses and vehicles with a minimum of two or three people and that includes the driver

Explanation:

hope this helped

Answer:

Explanation:

Crystalline melting temperature (Tm) is termed as the temperature required for a  crystalline polymer to change to a fluid or glasslike crystalline spaces of a semi-crystalline polymer liquefy (expanded sub-atomic movement).  

Crystallization of polymers is an interaction related with incomplete arrangement of their atomic and molecular chains. These chains crease together and structure requested districts called lamellae, which form bigger spheroidal designs named spherulites. Polymers can solidify after cooling from melting, mechanical extending, or dissolvable dissipation. Crystallization influences the optical, mechanical, and synthetic chemical properties of the polymer.  

For a crystalline polymer, a required polymer chain is present in or goes along a few crystalline and amorphous zones. The crystalline zones are comprised of intermolecular & intramolecular arrangements or deliberate and thus firmly stuffed plan of atoms or chain fragments, and an absence of it brings about the development of amorphous zones.  

The mechanical property boundary, for example, shear modulus expansions in the temperature of perception for polymer material framework.  

The temperature reaction of direct linear polymers might be seen as partitioned into three particularly separate fragments:  

1. Above Tm: The polymer stays as fluid whose consistency & viscosity would rely upon atomic molecular weight and temperature.  

2. Between Tm and Tg: This area may go between close to 100% crystalline & 100% amorphous chain atomic bunches relying upon the polymer underlying consistency. The amorphous part carries on similar to supercooled fluid in this section. The generally actual conduct of the polymer in this moderate portion is similar to an elastic rubber.  

3.Below Tg: The polymer material saw as glass is hard and inflexible, showing and emanating a predetermined coefficient of thermal extension. The glass is more like a crystalline strong than the fluid in personal conduct standard regarding mechanical property boundaries. In regard to the molecular atomic request, in any case, the glass all the more intently takes after the fluid. There is little contrast between the direct linear and cross-connected polymer beneath Tg.

Answer:

Die of intoxication by water first

Explanation:

We assume that the weight of the man is 154.35 pounds which is 70 kg

LD50 water = 90g per kg

Maximum concentration = 90x70

= 6300grams

Convert grams to liters

6300/100

= 6.3 litres

From here we get amount of kool aid

6.3 x 1.97x10^-4

= 1.24x10^-3

= 1.24grams

1.24 grams is below 420 kool aid is lower than LD50 with about 6 grams for 1 kg (6x70kg = 420). So 420 is lethal dose. But 1.24 is less than this so the man has to die of water intoxication first.

Answer:

Explanation:

The missing part of the question is attached in the diagram below, the second diagram shows the schematic view of the stress-strain curve and the plane stress.

From the given information:

The elastic modulus is:

[tex]E = \dfrac{\sigma}{\varepsilon} \\ \\ E = \dfrac{150 \ MPa}{0.0217} \\ \\ E = 69.124 \ GPa[/tex]

Hence, suppose 0.2% offset cuts the stress-strain curve at a designated point A from the image attached below, then the yield strength relating to the stress axis from the curve will be [tex]\sigma_y[/tex] = 270 MPa.

The shear yield strength by using von Mises criteria is estimated as;

[tex]\tau_1 = \dfrac{\sqrt{2}}{3}\sigma_y \\ \\ \tau_1 = \dfrac{\sqrt{2}}{3}*270 \\ \\ \tau_1 = 127.28 \ MPa[/tex]

The shear yield strength by using Tresca criteria is:

[tex]\tau_2 = \dfrac{1}{2}\sigma_y \\ \\ \tau_2= \dfrac{1}{2}*270 \\ \\ \tau_2 = 135 \ MPa[/tex]