Answer:
False
Explanation:
Flow back to surface
Words and numbers can be
printed using many different
or type styles.
Problem 1. Network-Flow Programming (25pt) A given merchandise must be transported at a minimum total cost between two origins (supply) and two destinations (demand). Destination 1 and 2 demand 500 and 700 units of merchandise, respectively. At the origins, the available amounts of merchandise are 600 and 800 units. USPS charges $5 per unit from origin 1 to demand 1, and $7 per unit from origin 1 to demand 2. From origin 2 to demand 1 and 2, USPS charges the same unit cost, $10 per unit, however, after 200 units, the unit cost of transportation increases by 50% (only from origin 2 to demand 1 and 2).
a) Formulate this as a network-flow problem in terms of objective function and constraint(s) and solve using Excel Solver.
b) How many units of merchandise should be shipped on each route and what is total cost?
Solution :
Cost
Destination Destination Destination Maximum supply
Origin 1 5 7 600
Origin 2 10 10 800
15, for > 200 15, for > 200
Demand 500 700
Variables
Destination 1 2
Origin 1 [tex]$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.
g Consider the following observations on shear strength (MPa) of a joint bonded in a particular manner. 22.6 40.4 16.4 72.4 36.6 109.8 30.0 4.4 33.1 66.7 81.5 (a) What are the values of the fourths
The object in ....................... shadow is not seen completely
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
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.
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]
A 2-m-internal-diameter spherical tank made of 0.5-cm-thick stainless steel (k = 15 W/m·K) is used to store iced water at 0°C in a room at 20°C. The walls of the room are also at 20°C. The outer surface of the tank is black (emissivity ε = 1), and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. Assuming the entire steel tank to be at 0°C and thus the thermal resistance of the tank to be negligible, determine
(a) the rate of heat transfer to the iced water in the tank and
(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. Now, consider a 2-m internal diameter double- walled spherical tank configuration is used instead to store iced water at 0°C in a room at 20°C. Each wall is 0.5 cm thick, and the 1.5-cm-thick air space between the two walls of the tank is evacuated in order to minimize heat transfer. The surfaces surrounding the evacuated space are polished so that each surface has an emissivity of 0.15. The temperature of the outer wall of the tank is measured to be 20°C. Assuming the inner wall of the steel tank to be at 0°C, determine
(c) the rate of heat transfer to the iced water in the tank for this double-walled tank configuration and
(d) the amount of ice at 0°C that melts during a 24-h period for this double-walled tank configuration.
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
whats is the purpose of the stator winding
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.
7
Which wire can carry a higher current?
ered
tof
Select one:
A. AWG 6
B. AWG 18
C. AWG 12
D. AWG 24
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.
What is meant by high current?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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If the same type of thermoplastic polymer is being tensile tested and the strain rate is increased, it will: g
Answer:
It would break I think need to try it out
Explanation:
True or false all workers who do class 1 asbestos work must be part of a medical surveillance program
Answer:
Yes
Explanation:
Answer:
true
Explanation:
hehehe
what is Geography? pliz help
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.
1. (15) A truck scale is made of a platform and four compression force sensors, one at each corner of the platform. The sensor itself is a short steel cylinder, 22 mm in diameter. A single stain gauge is pre-stressed to 3% strain and bonded on the outer surface of the cylinder. The strain gauges have a nominal resistance (before pre-stressing) of 340 Ohms and a gauge factor of 6.9. The steel used for the cylinders has a modulus of elasticity of 30 GPa. Calculate: a. The maximum truck weight that the scale can measure. b. The change in resistance of the sensors for maximum weight. c. The sensitivity of the scale assuming the response of the strain gauges is linear.
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
Discuss typical advantages and disadvantages of an irrigation system?
A small ship capable of making a speed of 6 knots through still water maintains a heading due east while being set to the south by an ocean current. The actual course of the boat is from A to B, a dis- tance of 10 nautical miles that requires exactly 2 hours. Determine the speed vC of the current and its direction measured clockwise from the north.
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
A step-up transformer has 20 primary turns and 400 secondary turns. If the primary current is 30 A, what is the secondary current
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.
What is secondary current ?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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Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length 0.25 mm (0.01 in) and tip radius of curvature of 1.2 x 10-3 mm (4.7 x 10-5 in.) when a stress of 1200 MPa (174,000 psi) is applied.
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
4 An approach to a pretimed signal has 30 seconds of effective red, and D/D/1 queuing holds. The total delay at the approach is 83.33 veh-s/cycle and the saturation flow rate is 1000 veh/h. If the capacity of the approach equals the number of arrivals per cycle, determine the approach flow rate and cycle length.
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]
The roof of a house has three layers: (1) 2 inch thick pine, (2) 4 inches of fiberglass insulation and (3) 0.1 inch thick Asphalt shingles. Calculate the rate of flow of heat (BTU/hour) through the roof. The temperature inside the house is 700F and the temperature outside is 300F. The R/inch for pine is 1.28, the R/inch for fiberglass is 3.0 and the R/inch for Shingles is 4.0. Take the area of the roof to be 500 square feet. The Unit of R is mixed: ft2 . 0F.h/BTU
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
A 2400-lb rear-wheel drive tractor carrying 900 lb of gravel starts from rest and accelerates forward at 3ft/s2. Determine the reaction at each of the two (a) rear wheels A, (b) front wheels B.
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
Find the velocity and rate of flow of water through a rectangular channel of 6m wide and 3m deep when it's running full. The channel is having a bed slope of 1 in 2000. Take Chezy's coefficient, C=55
Answer:
V = 1.5062 m/s
Explanation:
look to the photos
I ran across this symbol in some Electrical wiring documents and I am unaware of what this means. Any help?
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.
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What would the Select lines need to be to send data for the fifth bit in an 8-bit system (S0 being the MSB and S2 being the LSB)?
A. S0 = 1, S1 = 0, S2 = 0
B. S0 = 0, S1 = 0, S2 = 0
C. S0 = 0, S1 = 1, S2 = 0
D. S0 = 0, S1 = 1, S2 = 1
Answer:
A. S0 = 1, S1 = 0, S2 = 0
lines need to send data for the fifth bit in an 8 bit system
A frost free, 17 cu. ft. refrigerator-freezer uses energy at a rate of 500. watts when you hear the compressor running. If the fridge runs for 200. hours per month, how many kilowatt-hours of energy does the refrigerator use each month
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
1.8 A water flow of 4.5 slug/s at 60 F enters the condenser of steam turbine and leaves at 140 F. Determine the heat transfer rate (Btu/hr)
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]
What happens to resistance in the strain gauge and voltage drop from a connected Wheatstone bridge if you were to pull the strain gauge along the long axis
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.
Air is compressed by a 40-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process as a result of heat transfer to the surrounding medium at 20°C. Determine the rate of entropy change of the air. State the assumptions made in solving this problem
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.
As an engineer which types of ethical issues or problem you can face in industrial environment.
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(⌐■-■)
Provide two programming examples in which multithreading provides better performance than a single-threaded solution. Provide one example where singlethreaded solution performs better than multi-threaded solution
Answer:
I dont kno
Explanation:
Im so sorry
Consider the following example: The 28-day compressive strength should be 4,000 psi. The slump should be between 3 and 4 in. and the maximum aggregate size should not exceed 1 in. The coarse and fine aggregates in the storage bins are wet. The properties of the materials are as follows:________.
Cement : Type I, specific gravity = 3.15
Coarse Aggregate: Bulk specific gravity (SSD) = 2.70; absorption
capacity = 1.1%; dry-rodded unit weight = 105 lb./ft.3
surface moisture = 1%
Fine Aggregate: Bulk specific gravity (SSD) = 2.67; absorption
capacity = 1.3%; fineness modulus = 2.70;
surface moisture = 1.5%