In the High Low Logic Index low levels are bearish and high levels are bullish, generally True False

here's your answer..

Answer:

1700 W

Explanation:

The heat transfer rate P = kA(T - T')/d where k = thermal conductivity of wall = 1.7 W/m-K, A = area of wall = 0.5 m × 1.2 m = 0.6 m², T = temperature of inner surface = 1400 K, T = temperature of outer surface = 1150 K and d = thickness of wall = 0.15 m

So, P = kA(T - T')/d

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

P = 1.7 W/m-K × 0.6 m²(1400 K - 1150 K)/0.15 m

P = 1.7 W/m-K × 0.6 m² × 250 K/0.15 m

P = 255 Wm/0.15 m

P = 1700 W

So, the heat transfer rate through the wall is 1700 W

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

Answer:

the speed Vc of the current and its direction measured clockwise from the north is 0.71 m/s and 231.02° respectively

Explanation:

Given the data in the question and as illustrated in the diagram below;

The absolute velocity of the ship Vs is 6 Knots due east

so we convert to meter per seconds

Vs = 6 Knots × [tex]\frac{0.51444 m/s}{1 Knots}[/tex] = 3.0866 m/s

Next we determine the relative velocity of the ship Vs/c

Vs/c = AB / t

given that distance between A to B = 10 nautical miles which requires 2 hours

so we substitute

Vs/c = 10 nautical miles / 2 hrs

Vs/c = [10 nautical miles × [tex]\frac{1852 m}{1 nautical-miles}[/tex] ] / [ 2 hrs × [tex]\frac{3600s}{1hr}[/tex] ]

Vs/c = 18520 / 7200

Vs/c = 2.572 m/s

Now, from the second diagram below, { showing the relative velocity polygon }

Now, using COSINE RULE, we calculate the velocity current.

Vc = √( V²s + V²s/c - 2VsSs/ccos10 )

we substitute

Vc = √( (3.0866)² + (2.572)² - (2 × 3.0866 × 2.572 × cos10 ) )

Vc = √( (3.0866)² + (2.572)² - (2 × 3.0866 × 2.572 × 0.9848 ) )

Vc = √( 9.527099 + 6.615184 - 15.6361 )

Vc = √0.506183

Vc = 0.71 m/s

Next, we use the SINE RULE to calculate the direction;

Vc/sin10 = Vs/c / sinθ

we substitute

0.71 / sin10 = 2.572 / sinθ

0.71 / 0.173648 = 2.572 / sinθ

4.0887 = 2.572 / sinθ

sinθ  = 2.572 / 4.0887

sinθ = 0.62905

θ = sin⁻¹( 0.62905 )

θ = 38.98°

So, angle measured clock-wise will be;

θ = 270° - 38.98°

θ = 231.02°

Therefore, the speed Vc of the current and its direction measured clockwise from the north is 0.71 m/s and 231.02° respectively

Answer:

I dont kno

Explanation:

Im so sorry

Solution :

Cost

Destination           Destination         Destination                     Maximum supply

Origin 1                       5                          7                                           600

Origin 2                     10                         10                                          800

                         15, for > 200            15, for > 200

         Demand          500                       700

Variables

Destination       1          2

Origin 1             [tex]$X_1$[/tex]        [tex]$$X_2[/tex]

Origin 2            [tex]$X_3$[/tex]        [tex]$$X_4[/tex]

Constraints   :   [tex]$X_1$[/tex], [tex]$$X_2[/tex], [tex]$X_3$[/tex], [tex]$$X_4[/tex]  ≥ 0

Supply : [tex]$X_1$[/tex] + [tex]$$X_2[/tex]  ≤ 600

              [tex]$X_3$[/tex] + [tex]$$X_4[/tex] ≤ 800

Demand : [tex]$X_1$[/tex] + [tex]$$X_3[/tex]  ≥ 500

              [tex]$X_2$[/tex] + [tex]$$X_4[/tex] ≥ 700

Objective function :

Min z = [tex]$5X_1+7X_2+10X_3+10X_4, \ (if \ X_3, X_4 \leq 200)$[/tex]

[tex]$=5X_1+7X_2+(10\times 200)+(X_3-200)15+(10 \times 200)+(X_4-200 )\times 15 , \ \ (\text{else})$[/tex]

Costs :

                  Destination 1       Destination  2

Origin 1         5                             7

Origin 2        10                           10

                     15                            15

Variables :

[tex]$X_1$[/tex]        [tex]$$X_2[/tex]

300    300  

200   400

[tex]$X_3$[/tex]      [tex]$$X_4[/tex]

Objective function : Min z = 10600

Constraints:

Supply    600 ≤ 600

                600 ≤ 800

Demand   500 ≥ 500

                 700 ≥ 500

Therefore, the total cost is 10,600.

Answer:

Resistance of the lamp (r) = 156.52 ohm (Approx.)

Explanation:

Given:

Total power p = 140 W

Resistance of the coffee maker = 300 Ohm

Voltage v = 120 V

Find:

Resistance of the lamp (r)

Computation:

We know that

p = v² / R

SO,

Total power p = [Voltage²/Resistance of the lamp (r)] + [Voltage²/Resistance of the coffee maker]

140 = [120² / r] + [120²/300]

140 = 120²[1/r + 1/300]

140 = 14,400 [1/r + 1/300]

Resistance of the lamp (r) = 156.52 ohm (Approx.)

Answer:

the rate of flow of heat through the roof is 45616.858 BTU/hr

Explanation:

Given the data in the question;

pin thickness [tex]t_p[/tex] = 2 in

fiber glass thickness [tex]t_f[/tex] = 4 in

Asphalt shingles thickness [tex]t_a[/tex] = 0.1 in

R/inch for pine = 1.28

R/inch for fiberglass = 3.0

R/inch for Shingles = 4.0

Temperature inside the house [tex]T_{inside[/tex] = 700 F

Temperature outside the house [tex]T_{outside[/tex] = 300 F

area of the roof A = 500 ft²

we calculate the total Resistance;

R = ( 2 × 1.28 ) + ( 4 × 3.0 ) + ( 0.1 × 4.0 )

R = 2.56 + 12 + 0.4

R = 14.96

Now, we determine the rate of heat flow;

dQ/dt = ΔT(A) / R

⇒ ( [tex]T_{inside[/tex] - [tex]T_{outside[/tex] )A / R

we substitute

⇒ (( 700 - 300 ) × 500 ) / 14.96

⇒ ( 400 × 500 ) / 14.96

⇒ 200000 / 14.96

⇒ 13368.98 watt

we know that 1 watt = 3.412142 BTU/hr

⇒ ( 13368.98 × 3.412142 ) BTU/hr

⇒ 45616.858 BTU/hr

Therefore, the rate of flow of heat through the roof is 45616.858 BTU/hr

Complete Question

Analyze the boundary work done during the process having a rigid tank contains air at 500 kPa and 150°C. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65°C and 400 kPa, respectively.

Determine the boundary work done during this process and heat Lose

Answer:

a)  [tex]W=0[/tex]

b)  [tex]dQ=-61.03KJ/kg[/tex]

Explanation:

From the question we are told that:

Pressure of air [tex]P_1=500kpa[/tex]

Temperature of Air [tex]T_2=150°C[/tex]

Pressure drop [tex]P_2=400kpa[/tex]

Temperature of drop [tex]T_2=65 \textdegree C[/tex]

Generally the Constant Volume Process  is mathematically given by

 [tex]V_1=V_2=V[/tex]

Therefore

a)

Generally the equation for  boundary work w is mathematically given by

 [tex]W=pdv[/tex]

 [tex]W=P(V_2-V_1)[/tex]

 [tex]W=P(V_V)[/tex]

 [tex]W=0KJ[/tex]

b)

Generally the equation for Heat Change is mathematically given by

 [tex]dQ=dU+dW[/tex]

 [tex]dQ=dU[/tex]

 [tex]dQ=C_v(T_2-T_1)[/tex]

Where

   C_v=Specific Heat capacity of Air

  [tex]C_v=0.718 kJ/kg K[/tex]

 [tex]dQ=0.718(338-423)[/tex]

 [tex]dQ=-61.03KJ/kg[/tex]

Explanation:

Answer ⬇️

Social and ethical issues in engineering, ethical principles of engineering, professional code of ethics, some specific social problems in engineering practice: privacy and data protection, corruption, user orientation, digital divide, human rights, access to basic services.

➡️dhruv73143(⌐■-■)

Answer:

The units on the rate of heat transfer are Joule/second, also known as a Watt.

Explanation:

Heat flow is calculated using the rock thermal conductivity multiplied by the temperature gradient. The standard units are mW/m2 = milli Watts per meter squared. Thus, think of a flat plane 1 meter by 1 meter and how much energy is transferred through that plane is the amount of heat flow.

hope it helps .

The rate of heat transfer is measured in Joules per second, also known as Watts.

Heat transfer is a thermal engineering discipline that deals with the generation, use, conversion, and exchange of thermal energy between physical systems.

Heat transfer mechanisms include thermal conduction, thermal convection, thermal radiation, and energy transfer via phase changes.

The rate of heat transfer through a unit thickness of material per unit area per unit temperature difference is defined as thermal conductivity. Thermal conductivity varies with temperature and is measured experimentally.

Heat is typically transferred in a combination of these three types and occurs at random. Heat transfer rate is measured in Joules per second, also known as Watts.

Thus, Joules per second or watts is the unit of rate of heat transfer.

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Answer:

35.92 kpsi

Explanation:

Given data:

diameter of the steel bar d = 0.875 in

Area A = πd^2/4 = π(0.875)^2/4

length L = 15.0 ft

Load P = 21.6 kip

Modulus of elesticity E = 29×10^6 Psi

Assume we are asked to determine axial stress in the bar which is given as

[tex]\sigma = Load, P/ Area, A[/tex]

[tex]\sigma = 4P/\pi d^2[/tex]

substitute the value

[tex]\sigma = \frac{4\times 21.6}{\pi \times (0.875)^2} \\=35.92\ kpsi[/tex]

Answer:

Explanation:

The missing diagram attached to the question is shown in the attached file below:

The very first thing we need to do in other to solve this question is to determine the mass of both the tractor and the mass of the gravel

For tractor, the mass is:

[tex]m_1 = \dfrac{2400 \ lb }{32.2 \ ft/s^2}[/tex]

[tex]m_1 = 74.53 \ lb.s^2/ft[/tex]

For gravel, the mass is:

[tex]m_2 = \dfrac{900 \ lb}{32.2 \ft/s^2}[/tex]

[tex]m_2 = 27.95 \ lb.s^2/ft[/tex]

From the diagram, let's consider the force along the horizontal components and vertical components;

So,

[tex]\sum F_x = ma_x \\ \\ 2F = (m_1+m_2) a \\ \\ F = \dfrac{1}{2}(74.53 4 + 27.950)lb.s^2/ft(2 \ ft/s^2) \\ \\ F = 102.484 \ lb[/tex]

[tex]\sum F_y = 0 \\ \\ 2N_A+2N_B - 2400 -900 = 0 \\ \\ N_A +N_B = 1650 \ lb[/tex]

Consider the algebraic sum of moments in the plane of A, with counter-clockwise moments being positive.

[tex]\sum M_A = I_o \alpha + \sum ma (d) \\ \\ = -2400 (20) + 2N_B (60) -900(110) = 0 - (74.534)(2)(20) - (27.950)(2)(40)[/tex]

[tex]=-48000 + 2N_B (60) -99000 = -2981.36-2236 \\ \\ = + 2N_B (60) = -2981.36-2236+48000+99000 \\ \\ = + 2N_B (60) = 141782.64 \\ \\ N_B = \dfrac{141782.64}{120} \\ \\ N_B = 1181.522 \ lb[/tex]

Replacing the value of 1181.522 lb for [tex]N_B[/tex] in equation (1)

[tex]N_A[/tex] + 1181.522 lb = 1650 lb

[tex]N_A[/tex] = (1650 - 1181.522)lb

[tex]N_A[/tex] = 468.478 lb

The net reaction on each of the rear wheels now is:)

[tex]F_R = \sqrt{N_A^2 +F^2}[/tex]

[tex]F_R = \sqrt{(468.478)^2 + (102.484)^2}[/tex]

[tex]\mathbf{F_R =479.6 \ lb}[/tex]

Now, we can determine the angle at the end of the rear wheels at which the resultant reaction force is being made in line with the horizontal

[tex]\theta = tan ^{-1}( \dfrac{468.478 }{102.484})[/tex]

[tex]\theta = 77.7^0[/tex]

Finally, the net reaction on each of the front wheels is:

[tex]F_B = N_B[/tex]

[tex]F_B =[/tex] 1182 lb  

Answer:

A. S0 = 1, S1 = 0, S2 = 0

lines need to send data for the fifth bit in an 8 bit system

Answer:

Dark shadow

Explanation:

Shadow is nothing but space when the light is blocked by an opaque object. It is just that part where light does not reach. When you stand in the sun, you are able to see your shadow behind you. ... This is because our body is opaque and does not allow the light to pass through it

Mark brainliest

Answer:

a). 139498.24 kg

b). 281.85 ohm

c). 10.2 ohm

Explanation:

Given :

Diameter, d = 22 m

Linear strain, [tex]$\epsilon$[/tex] = 3%

                        = 0.03

Young's modulus, E = 30 GPa

Gauge factor, k = 6.9

Gauge resistance, R = 340 Ω

a). Maximum truck weight

σ = Eε

σ = [tex]$0.03 \times 30 \times 10^9$[/tex]

[tex]$\frac{P}{A} =0.03 \times 30 \times 10^9$[/tex]

[tex]$P = 0.03 \times 30 \times 10^9\times \frac{\pi}{4}\times (0.022)^2$[/tex]

 = 342119.44 N

For the four sensors,

Maximum weight = 4 x P

                            =  4 x 342119.44

                            = 1368477.76 N

Therefore, weight in kg is [tex]$m=\frac{W}{g}=\frac{1368477.76}{9.81}$[/tex]

                     m = 139498.24 kg

b). Change in resistance

[tex]k=\frac{\Delta R/R}{\Delta L/L}[/tex]

[tex]$\Delta R = k. \epsilon R$[/tex]    , since [tex]$\epsilon= \Delta L/ L$[/tex]

[tex]$\Delta R = 6.9 \times 0.03 \times 340$[/tex]

[tex]$\Delta R = 70.38 $[/tex] Ω

For 4 resistance of the sensors,

[tex]$\Delta R = 70.38 \times 4 = 281.52$[/tex] Ω

c). [tex]$k=\frac{\Delta R/R}{\epsilon}$[/tex]

If linear strain,

[tex]$\frac{\Delta R}{R} \approx \frac{\Delta L}{L}$[/tex]  , where k = 1

[tex]$\Delta R = \frac{\Delta L}{L} \times R$[/tex]

[tex]$\Delta R = 0.03 \times 340$[/tex]

[tex]$\Delta R = 10.2 $[/tex] Ω

Answer:

Following are the responses to the given question:

Explanation:

Effective red duration is applied each cycle r=30 second D/D/1 queuing

In total, its approach delay is 83.33 sec vehicle per cycle

Flow rate(s) of saturated = 1,000 vehicles each hour

Total vehicle delay per cycle[tex]= \frac{v \times 30^2}{2(1-\frac{v}{0.2778})}[/tex]

[tex]\to \frac{v\times 30^2}{2(1-\frac{v}{0.2778})}= 83.33\\\\\to 900v=166.66-599.928v\\\\\to v=0.111 \frac{veh}{sec}\\\\[/tex]

The flow rate for such total approach is 0.111 per second.

The overall flow velocity of the approach is 400 cars per hour

The approach capacity refers to the number of arrivals per cycle.

Environmentally friendly time ratio to cycle length:

[tex],\frac{g}{C} \ is = \frac{400}{1000}=0.4\\\\r= c-g\\\\30\ sec =C - 0.4 C\\\\C=50 \ sec[/tex]

Answer:

Yes

Explanation:

Answer:

true

Explanation:

hehehe

Answer:

the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

Explanation:

Given the data in the question;

From the first law of thermodynamics;

dQ = dU + dW ------ let this be equation 1

where dQ is the heat transfer, dU is internal energy and dW is the work done.

from the question, the process is isothermal ( internally reversible process )

Thus, the change in internal energy is 0

dU = 0

given that; Air is compressed by a 40-kW compressor from P1 to P2

since it is compressed, dW = -40 kW

we substitute into equation 1

dQ = 0 + ( -40 kW )

dQ = -40 kW

Now, change in entropy of air is;

ΔS[tex]_{air[/tex] = dQ / T

given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K

so we substitute

ΔS[tex]_{air[/tex] =  -40 kW / 298.15 K

ΔS[tex]_{air[/tex] =  -0.13416 ≈ -0.1342 kW/K

Therefore, the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

Answer:

the theoretical fracture strength of the brittle material is 5.02 × 10⁶  psi

Explanation:

Given the data in the question;

Length of surface crack α = 0.25 mm

tip radius ρ[tex]_t[/tex] = 1.2 × 10⁻³ mm

applied stress σ₀ = 1200 MPa

the theoretical fracture strength of a brittle material = ?

To determine the the theoretical fracture strength or maximum stress at crack tip, we use the following formula;

σ[tex]_m[/tex] = 2σ₀[tex]([/tex] α / ρ[tex]_t[/tex] [tex])^{\frac{1}{2}[/tex]

where α is the Length of surface crack,

ρ[tex]_t[/tex] is the tip radius,

and σ₀ is the applied stress.

so we substitute

σ[tex]_m[/tex] = (2 × 1200 MPa)[tex]([/tex] 0.25 mm / ( 1.2 × 10⁻³ mm ) [tex])^{\frac{1}{2}[/tex]

σ[tex]_m[/tex] = 2400 MPa × [tex]([/tex] 208.3333 [tex])^{\frac{1}{2}[/tex]

σ[tex]_m[/tex] = 2400 MPa × 14.43375

σ[tex]_m[/tex] = 34641 MPa

σ[tex]_m[/tex] = ( 34641 × 145 )psi

σ[tex]_m[/tex] = 5.02 × 10⁶  psi

Therefore, the theoretical fracture strength of the brittle material is 5.02 × 10⁶  psi

Answer:

In an electric motor, the stator provides a magnetic field that drives the rotating armature; in a generator, the stator converts the rotating magnetic field to electric current. In fluid powered devices, the stator guides the flow of fluid to or from the rotating part of the system.

Answer:

The answer is "shear".

Explanation:

Every transformation could be shown by some kind of conventional matrix.

Its standard matrices of shape k here could be any real value enabling shear transformation to parallel to the y axis.

[tex]\left[\begin{array}{cc}1&0\\k&1\\\end{array}\right][/tex]

This coefficient matrix A is the standard matrix during transformation. With the use of a the transformation could be entered into T(x)=Ax

And standard transfer function is standard.

[tex]\left[\begin{array}{cc}1&0\\6&1\\\end{array}\right][/tex]

The matrix in the form of the shear matrix.

Answer:

the wire that can carry much current is AWG 24

AWG 24 is the  wire can carry a higher current. The diameter of the wire increases with decreasing gauge. The electrical resistance to the signals decreases with increasing wire diameter. Hence, option D is correct.

Any current above 10 mA has the potential to deliver an unpleasant to severe shock, while 200 mA and above is considered lethal. Despite the fact that currents above 200 mA might cause serious burns and unconsciousness, they usually do not cause death if the sufferer receives timely medical attention.

Higher voltage and lower current are frequently more effective combinations. The amount of current that a wire is rated to handle is frequently indicated by its capacity rating. People can use the same cable to drive a load twice as large by doubling the voltage and maintaining the same current.

Thus, option D is correct.

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Answer:

Opened Push-button Switch (i.e. a PTM Switch)

Explanation:

Tha's just another symbol for a switch, but this one specifies that the switch is a push-button type of switch.

Since it's not touching and completing the line, the state of the switch is initially open.

Answer:

V = 1.5062 m/s

Explanation:

look to the photos