Answers
Related Questions
14. The top plate of the bearing partition
I
a. laps the plate of the exterior wall.
b. is a single member.
c. butts the top plate of the exterior wall.
d. is applied after the ceiling joists are
installed.
Answers
Answer:
d. is applied after the ceiling joists are
installed.
Create a 6-bit full subtractor that uses the Borrow method to subtract two 6-bit binary numbers. You can use the proper basic sub-circuit.
Answers
The boundary work is positive during an expansion process.
a.
False
b.
True
Answers
Answer:
True
Explanation:
During expansion process, the boundary work is positive while in case of contraction, the boundary work is negative. During expansion process, the work is done by the system while in case of compression process, work in done on the system
Hence, the given statement is true
This is the top part of the picture. I need someone to put the correct letter by the word !
Answers
Answer:
J = power steering pump
A = hydraulic hoses
D = pitman arm
G = rack and pinion assembly
B = idler arm
C = center link
F = tie rod
E = tie rod end
H = rack and pinion boot
I = outer tie rod end
Explanation:
If it took a 30m capacity tank mediated by 2cm waterproof water faucet for 10 hours, calculate the flow speed (exit) water from the
Answers
Complete question:
If it took a 30m³ capacity tank mediated by 2cm waterproof water faucet for 10 hours, calculate the flow speed (exit) of water from the tank.
Answer:
the flow rate of the water from the tank is 0.05 m³/min
Explanation:
Given;
volume of water in the tank, v = 30 m³
length of the waterproof faucet, L = 2cm = 0.02 m
duration of water flow through the tank, t = 10 hours
The flow rate of the water from the tank is calculated as;
[tex]Q = \frac{V}{t} = \frac{30 \ m^3}{10\ h \ \times \ 60 \min} = 0.05 \ m^3/ \min[/tex]
Therefore, the flow rate of the water from the tank is 0.05 m³/min
Two consecutive, first order reactions (with reaction rate constant k1 and k2) take place in a perfectly mixed, isothermal continuous reactor (CSTR) A (k1) → B (k2) → C Volumetric flow rates (F) and densities are constant. The volume of the tank (V) is constant. The reactor operate at steady state and at constant temperature. The inlet stream to the reactor contains only A with CA,in = 10 kmol/m3. If k1 = 2 min-1, k2 = 3 min-1, and τ = V/F.= 0.5 min, determine the concentration of C in the stream leaving the reactor.
Answers
Answer:
3 kmol/m^3
Explanation:
Determine the concentration of C in the stream leaving the reactor
Given that the CTSR reaction ; A (k1) → B (k2) → C
K1 = 2 min^-1 , K2 = 3 min^-1 , time constant ; τ = V/F.= 0.5 min also n1 = n2
attached below is the detailed solution
concentration of C leaving the reactor= 3 kmol/mol^3
Given ; Ca = 5 kmol/m^3 , Cb = 2 kmol/m^3 ( from the attached calculations ) Cc = 3 kmol/m^3
banana with an average mass of 0.15 kg and average specific heat of 3.35 kJ/kg · °C is cooled from 20°C to 5°C. The amount of heat transferred from the banana is
a.
62.1 kJ
b.
7.5 kJ
c.
None of these
d.
6.5 kJ
e.
0.85 kJ
f.
17.7 kJ
Answers
Answer:
The amount of heat transferred from the banana is (-)7.54 KJ
Explanation:
As we know,
[tex]Q = m*c*\Delta T[/tex]
Q = Amount of heat transferred
m = mass of banana
[tex]T_2 = 5[/tex] degree Celsius
[tex]T_1 = 20[/tex] degree Celsius
The amount of heat transferred from the banana =
[tex]0.15 * 3.35 * (5 -20)\\-7.54[/tex]KJ (negative sign represents reduction in heat energy)
A Class III two-lane highway is on level terrain, has a measured free-flow speed of 45 mi/h, and has 100% no-passing zones. During the peak hour, the analysis direction flow rate is 150 veh/h, the opposing direction flow rate is 100 veh/h, and the PHF-0.95. There are 5% large trucks and 10% recreational vehicles. Determine the level of service.
Answers
Answer:
LOS = A
Explanation:
Given all the parameters the level of service as seen from the attached graph
is LOS = A
To determine the LOS from the attached graph
calculate the trial value of Vp
Vp = V / PHF
= (100 + 150) / 0.95 = 263 pc/h
since the trial value of Vp = ( 0 to 600 ) pc/h . hence E.T = 1.7 , ER = 1
next we will calculate the flow rate
flow rate = 1 / [ ( 1 + PT(ET - 1 ) + PR ( ER - 1 ) ]
Fhr = 1 / 1.035 = 0.966 ≈ 1
next calculate the real value of Vp
Vp = V / ( PHF * N * Fhr * Fp )
= ( 100 + 150 ) / ( 0.95 * 2 * 1 * 1 )
Vp ≈ 126 pc/h/In
Next calculate the density
D = Vp / S = 126 / ( 45 * 1.61 ) = 1.74 pc/km/In
Illinois furniture , Inc produces all types of coffee furniture the executive secretary is a chair that has been designed using ergonomics to provide comfort during long work hours the chair sells for $130 there are 480 minutes available during the day and the average daily demand has been 50 chairs there are eight tasks
Answers
This question is incomplete, the complete question is;
Illinois furniture , Inc produces all types of office furniture. The "Executive Secretary" is a chair that has been designed using ergonomics to provide comfort during long work hours the chair sells for $130. There are 480 minutes available during the day and the average daily demand has been 50 chairs. There are eight tasks.
TASK PERFORMANCE TIME( MIN ) TASK MUST FOLLOW TASK LISTED
A 4 ---------
B 7 ----------
C 6 A,B
D 5 C
E 6 D
F 7 E
G 8 E
H 8 F,G
1) What is the cycle time for operation?
2) What is the theoretical minimum number of workstation?
Answer:
1) the cycle time for operation is 9.6 min
2) the theoretical minimum number of workstation is 5
Explanation:
Given the data in the question;
production time per day = 480 minutes
average daily demand = 50
Given the data in the question;
1) cycle time for operation
this is simply referred to as the total time for the process from start to finish.
cycle time = production time per day / units demand per day
we substitute
cycle time = 480 min / 50
cycle time = 9.6 min
Therefore, the cycle time for operation is 9.6 min
2) theoretical minimum number of workstation.
theoretical minimum number of workstation = total task time / cycle time
Total task time = ( 4 + 7 + 6 + 5 + 6 + 7 + 8 + 6 ) = 49 min
∴ theoretical minimum number of workstation = 49 min / 9.6 min
theoretical minimum number of workstation = 5.104 ≈ 5
Therefore, the theoretical minimum number of workstation is 5
Problem 9.11 A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 92 MPa m (klein.)) and a yield strength of 900 MPa (65270 psi). The flaw size resolution limit of the flaw detection apparatus is 3 mm (0.1181 in.). (a) If the design stress is one-half of the yield strength and the value of Y is 1.15, what is the critical flaw length
Answers
Answer:
the critical flaw length is 10.06 mm
Explanation:
Given the data in the question;
plane strain fracture toughness [tex]K_{tc[/tex] = 92 Mpa√m
yield strength σ[tex]_y[/tex] = 900 Mpa
design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa
Y = 1.15
we know that;
Critical crack length [tex]a_c[/tex] = 1/π( [tex]K_{tc[/tex] / Yσ )²
we substitute
[tex]a_c[/tex] = 1/π( 92 Mpa√m / (1.15 × 450 Mpa )²
[tex]a_c[/tex] = 1/π( 92 Mpa√m / (517.5 Mpa )²
[tex]a_c[/tex] = 1/π( 0.177777 )²
[tex]a_c[/tex] = 1/π( 0.03160466 )
[tex]a_c[/tex] = 0.01006 m = 10.06 mm
Therefore, the critical flaw length is 10.06 mm
{ [tex]a_c[/tex] = ( 10.06 mm ) > 3 mm
The critical flow is subject to detection