identify the part of the direct current motor​

Answer:

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

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

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:

[tex]Hr=4.2*10^7\ btu/hr[/tex]

Explanation:

From the question we are told that:

Water flow Rate [tex]R=4.5slug/s=144.78ib/sec[/tex]

Initial Temperature [tex]T_1=60 \textdegree F[/tex]

Final Temperature  [tex]T_2=140 \textdegree F[/tex]

Let

Specific heat of water [tex]\gamma= 1[/tex]

And

 [tex]\triangle T= 140-60[/tex]

 [tex]\triangle T= 80\ Deg.F[/tex]

Generally the equation for Heat transfer rate of water  [tex]H_r[/tex] is mathematically given by

Heat transfer rate to water= mass flow rate* specific heat* change in temperature

 [tex]H_r=R* \gamma*\triangle T[/tex]

 [tex]H_r=144.78*80*1[/tex]

 [tex]H_r=11582.4\ btu/sec[/tex]

Therefore

 [tex]H_r=11582.4\ btu/sec*3600[/tex]

 [tex]Hr=4.2*10^7\ btu/hr[/tex]

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:

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:

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:

Resistance and voltage drop will still continue to rise, although at a slower pace than on the desired axis.

Explanation:

Pulling the strain gauge in the long axis causes the wires to elongate/thin, the effect of this is that it will lead to an increase in resistance and voltage drop (V = I*R).

As a result of the resultant effect, resistance and voltage drop will still continue to rise, although at a slower pace than on the desired axis, such as the long axis.

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:

Yes

Explanation:

Answer:

true

Explanation:

hehehe

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:

hope it's helpful please like and Follow me

Answer:

Geography is the science that studies and describes the surface of the Earth in its physical, current and natural aspect, or as a place inhabited by humanity.

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:

100 kWh

Explanation:

Since the freezer has a rating of 500 watts and runs for 200 hours in a month, the energy consumption can be gotten by getting the product of the rating of the freezer in kilowatts and the amount of time the fridge is on per month.

The rating of the freezer = 500 watts = 0.5 kW, time = 200 h

Energy consumption = rating * time = 0.5 kW * 200 h

Energy consumption = 100 kWh

Therefore the refrigerator uses 100 kWh per month

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:

a. 6.48 kW b. 1678.34 kg c. 777.92 W d. 201.42 kg

Explanation:

(a) the rate of heat transfer to the iced water in the tank

The rate of heat transfer to the outer surface of the spherical tank is P = P₁ + P₂ where P₁ = rate of heat transfer to the outer surface by radiation and P₂ = rate of heat transfer to the outer surface by convection through air

P₁ = εσAT⁴ where ε = emissivity of outer surface ε= 1, σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m-K⁴, A = area of outer surface of spherical tank = 4πR² where R = outer radius of spherical tank = inner radius + thickness = inner diameter/2 + 5 cm = 2 m/2 + 0.05 m = 1 m + 0.05 m = 1.05 m and T = temperature of surroundings = 20 °C = 273 + 20 = 293 K.

P₁ = εσAT⁴

P₁ = 1 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.05 m)² × (293 K)⁴

P₁ = 1 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.1025 m²) × 7370050801 K⁴

P₁ = 184285909263.7647π × 10⁻⁸ W

P₁ = 578951258703.16 × 10⁻⁸ W

P₁ = 5789.51 W

P₂ = hA(T - T₁) where h = coefficient of thermal convection of air = 2.5 W/m²-K, A = outer surface area of spherical tank = 4πR², T = temperature of surroundings = 20 °C = 273 + 20 = 293 K and T₁ = temperature of outer surface of spherical tank = 0 °C = 273 + 0 = 273 K.  

P₂ = hA(T - T₁)

P₂ = 2.5 W/m²-K × 4π(1.05 m)² × (293 K - 273 K)

P₂ = 2.5 W/m²-K × 4π(1.1025 m²) × 20 K

P₂ = 220.5π W

P₂ = 692.72 W

So, P = P₁ + P₂ = 5789.51 W + 692.72 W = 6482.23 W

Since we are neglecting the thermal resistance of the spherical tank, the rate of heat absorption of the outer surface equals the rate of heat absorption in the inner surface. The rate of heat absorption at the inner surface equals the rate of heat transfer to the iced water.

So, rate of heat transfer to the iced water = P = 6482.23 W = 6.48223 kW 6.48 kW

(b) the amount of ice at 0°C that melts during a 24-h period. The heat of fusion of water is 333.7 kJ/kg.

Since the amount of heat, Q = Pt where P = heat transfer rate to iced water = 6482.23 W and t = time = 24 h = 24 h × 60 min/h × 60 s/min = 86400 s.

Also, Q = the latent heat required to melt the ice at 0 °C = mL where m = mass of ice melted and L = latent heat of fusion of ice = 333.7 kJ/kg

So, Pt = mL

m = Pt/L

= 6482.23 W × 86400 s/333.7 × 10³ J/kg

= 560064672/333.7 × 10³

= 1678.34 kg

(c) the rate of heat transfer to the iced water in the tank for this double-walled tank configuration

Since P is the rate of heat transfer to the outer surface, this is also the rate of heat transfer to the outer 0.5 cm thick wall = P₃ = 6482.23 W

P₃ = kA(T - T₃)/d where k = thermal conductivity of outer wall = 15 W/m²-K

A = surface area of outer wall = 4πR'² where R' = radius of outer wall = radius of inner wall + thickness of inner wall + thickness of vacuum + thickness of outer wall = 2.0 m/2 + 0.5 cm + 1.5 cm + 0.5 cm = 1 m + 2.5 cm = 1 m + 0.025 m = 1.025 m, T = temperature of surroundings = 20 °C = 273 + 20 = 293 K, T₃ = temperature of inner surface of outer wall of spherical tank and d = thickness of outer surface of tank = 0.5 cm = 0.05 m

P₃ = kA(T - T₃)/d

making T₃ subject of the formula, we have

P₃d = kA(T - T₃)

P₃d/kA = (T - T₃)

T₃ = T - P₃d/kA

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

T₃ = 293 K - 6482.23 W × 0.05 m/[15 W/m-K × 4π(1.025 m)²]

T₃ = 293 K - 324.1115 Wm/[15 W/m-K × 4π(1.050625 m²)]

T₃ = 293 K - 324.1115 Wm/[63.0375π W/m-K)]

T₃ = 293 K - 324.1115 Wm/[198.0381 W/m-K)]

T₃ = 293 K - 1.64 K

T₃ = 291.36 K

Since the 1.5 cm thick air space is evacuated, all the heat gets to the inner 0.5 cm thick wall by radiation.

So P = εσAT₃⁴

P₄ = εσAT₃⁴ where ε = emissivity of outer surface ε = 0.15, σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m-K⁴, A = area of inner surface of outer wall of spherical tank = 4πR"² where R" = outer radius of inner thick wall of spherical tank = inner radius + thickness of inner wall = inner diameter/2 + 0.5 cm = 2 m/2 + 0.005 m = 1 m + 0.005 m = 1.005 m and T = temperature of outer wall = 291.36 K.

P₄ = 0.15 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.005 m)² × (291.36 K)⁴

P₄ = 0.15 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.010025 m²) × 7206422389.51 K⁴

P₄ = 24762024365.028π × 10⁻⁸ W

P₄ = 77792193833.18 × 10⁻⁸ W

P₄ = 777.92 W

Now P₄ is the heat transfer rate to the inner surface which is at temperature T₄

Since T₄ = 0 °C, P₄ is the rate of heat transfer to the iced water

So, rate of heat transfer to the iced water P₄ = 777.92 W

(d) the amount of ice at 0°C that melts during a 24-h period for this double-walled tank configuration

Since the amount of heat, Q = P₄t where P₄ = heat transfer rate to iced water = 777.92 W and t = time = 24 h = 24 h × 60 min/h × 60 s/min = 86400 s.

Also, Q = the latent heat required to melt the ice at 0 °C = mL where m = mass of ice melted and L = latent heat of fusion of ice = 333.7 kJ/kg

So, P₄t = mL

m = P₄t/L

= 777.92 W × 86400 s/333.7 × 10³ J/kg

= 67212288/333.7 × 10³

= 201.42 kg

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  

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]

In an ideal transformer, the ratio of input voltage to output voltage is equal to the ratio of the number of turns in primary coil to number of turns in the secondary coil. Therefore, the secondary current in the given case is 1.5 A.

Secondary current refers to the electric current that flows in the secondary winding of a transformer. A transformer is a device that transfers electrical energy from one circuit to another by means of electromagnetic induction.

It consists of two or more coils of insulated wire, called windings, that are wound around a common magnetic core. In a transformer, an alternating current (AC) flows through the primary winding, which produces a magnetic field that induces a voltage in the secondary winding.

The secondary current in the above given case is 1.5 A.

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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:

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 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

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:

It would break I think need to try it out

Explanation:

Answer:

I dont kno

Explanation:

Im so sorry

Answer:

V = 1.5062 m/s

Explanation:

look to the photos