Answer:
Air is a gas
Explanation:
i think. beavuse it cant be a liqued or a solid. i dont think a plasma. i would answer gas
2. In a certain Village in eastern Uganda, residents use on average 1.5 Kg per inhabitant of
wood per day in a traditional stove of 12% efficiency. If the village has 200 in habitants,
Find,
i.
The
total mass of wood consumed daily.
ii.
The Energy demand of the village in KWh.
iii.
The actual energy consumption of the village in KWh.
iv.
If an investor would like to provide this village with a new energy converter of 30%
efficiency, how many Kilograms of wood will be required per day
probably alot lol
Explanation:
it probably is
Suppose we are given three boxes, Box A contains 20 light bulbs, of which 10 are defective, Box B contains 15 light bulbs, of which 7 are defective and Box C contains 10 light bulbs, of which 5 are defective. We select a box at random and then draw a light bulb from that box at random. (a) What is the probability that the bulb is defective? (b) What is the probability that the bulb is good?
Answer:
0.49
0.51
Explanation:
Probability that bulb is defective :
Let :
b1 = box 1 ; b2 = box 2 ; b3 = box 3
d = defective
P(defective bulb) = (p(b1) * (d|b1)) + (p(b2) * p(d|b2)) + (p(b3) * p(d|b3))
P(defective bulb) = (1/3 * 10/20) + (1/3 * 7/15) + (1/3 * 5/10))
P(defective bulb) = 10/60 + 7/45 + 5/30
P(defective bulb) = 1/6 + 7/45 + 1/6 = 0.4888
= 0.49
P(bulb is good) = 1 - P(defective bulb) = 1 - 0.49 = 0.51
FOR BRAINLIST ITS A DCP
The Battle of Sabine Pass
The Battle of Galveston
The Battle of Palmito Ranch
The Battle of Vicksburg
Answer:
I think The Battle of Sabine Pass
What can be used to measure the alcohol content in gasoline? A. Graduated cylinder B. Electronic tester C. Scan tool D. Either a graduated cylinder or an electronic tester
Answer:
GRADUATED CYLINDER
Explanation:
A 0.06-m3 rigid tank initially contains refrigerant- 134a at 0.8 MPa and 100 percent quality. The tank is connected by a valve to a supply line that carries refrigerant- 134a at 1.2 MPa and 36°C. Now the valve is opened, and the refrigerant is allowed to enter the tank. The valve is closed when it is observed that the tank contains saturated liquid at 1.2 MPa. Determine (a) the mass of the refrigerant that has
entered the tank and (b) the amount of heat transfer.
Answer:
a) 0.50613
b) 22.639 kJ
Explanation:
From table A-11 , we will make use of the properties of Refrigerant R-13a at 24°C
first step : calculate the volume of R-13a ( values gotten from table A-11 )
V = m1 * v1 = 5 * 0.0008261 = 0.00413 m^3
next : calculate final specific volume ( v2 )
v2 = V / m2 = 0.00413 / 0.25 ≈ 0.01652 m^3/kg
a) Calculate the mass of refrigerant that entered the tank
v2 = Vf + x2 * Vfg
v2 = Vf + [ x2 * ( Vg - Vf ) ] ----- ( 1 )
where: Vf = 0.0008261 m^3/kg, V2 = 0.01652 m^3/kg , Vg = 0.031834 m^3/kg ( insert values into equation 1 above )
x2 = ( 0.01652 - 0.0008261 ) / 0.031834
= 0.50613 ( mass of refrigerant that entered tank )
b) Calculate the amount of heat transfer
Final specific internal energy = u2 = Uf + ( x2 + Ufg ) ----- ( 2 )
uf = 84.44 kj/kg , x2 = 0.50613 , Ufg = 158.65 Kj/kg
therefore U2 = 164.737 Kj/kg
The mass balance ( me ) = m1 - m2 --- ( 3 )
energy balance( Qin ) = ( m2 * u2 ) - ( m1 * u1 ) + ( m1 - m2 ) * he
therefore Qin = 41.184 - 422.2 + 403.655 = 22.639 kJ
Rainfall rates for successive 20-min period of a 140min storm are 1.5, 1.5, 6.0, 4.0, 1.0, 0.8, and 3.2 in/hr, totaling 6.0in. Determine the rainfall excess during successive 20-min periods by the NRCS method. The soil in the basin belongs to group A. It is an agriculture row crop land with contoured pattern in good hydrologic condition. The soil is in average condition before the storm (moisture condition II).
Water vapor at 5 bar, 3208C enters a turbine operating at steady state with a volumetric flow rate of 0.65 m3/s and expands adiabatically to an exit state of 1 bar, 1608C. Kinetic and potential energy effects are negligible. Determine for the turbine (a) the power developed, in kW, (b) the rate of entropy production, in kW/K, and (c) the isentropic turbine efficiency.
Answer:
A) 371.28 kW
b) 0.1547 Kw/K
c) 85%
Explanation:
pressure (p1) = 5 bar
exit pressure ( p2 ) = 1 bar
Initial Temperature ( T1 ) = 320°C
Final temp ( T2 ) = 160°C
Volume ( V ) = 0.65 m^3/s
A) Calculate power developed ( kW )
P = m( h1 - h2 ) = 1.2 ( 3105.6 - 2796.2 ) = 371.28 kW
B) Calculate the rate of entropy production
Δs = m ( S2 - S1 ) = 1.2 ( 7.6597 - 7.5308 ) = 0.1547 Kw/K
c) Calculate the isentropic turbine efficiency
For an isentropic condition : S2s = S1
therefore at state , value of h2 at isentropic condition
attached below is the remaining part of the solution
Note : values of [ h1, h2, s1, s2 , v1 and m ] are gotten from the steam tables at state 1 and state 2
is a baby duck swimming in a lake a learned behavior
Answer:
True because some ducks can't swim but have to learn it
FOR BRAINLIST HELP PLEASE IS A DCP
A- Causes of the 13t Amendment
B- Reasons for Women's Suffrage
C- Reasons for the Freedmen's Bureau
D- Causes of the Plantation System
Answer:
C
Explanation: Freedmens Bureau provided resources for southerners and newly freed slaves
Nitrogen (N2) enters an insulated compressor operating at steady state at 1 bar, 378C with a mass flow rate of 1000 kg/h and exits at 10 bar. Kinetic and potential energy effects are negligible. The nitrogen can be modeled as an ideal gas with k 5 1.391. (a) Determine the minimum theoretical power input required, in kW, and the corresponding exit temperature, in 8C. (b) If the exit temperature is 3978C, determine the power input, in kW, and the isentropic compressor efficiency.
Answer:
A)
i) 592.2 k
ii) - 80 kw
B)
i) 105.86 kw
ii) 78%
Explanation:
Note : Nitrogen is modelled as an ideal gas hence R - value = 0.287
A) Determine the minimum theoretical power input required and exit temp
i) Exit temperature :
[tex]\frac{T_{2s} }{T_{1} } = (\frac{P2}{P1} )^{\frac{k-1}{k} }[/tex]
∴ [tex]T_{2s}[/tex] = ( 37 + 273 ) * [tex](\frac{10}{1} )^{\frac{1.391-1}{1.391} }[/tex] = 592.2 k
ii) Theoretical power input :
W = [tex]\frac{-n}{n-1} mR(T_{2} - T_{1} )[/tex]
where : n = 1.391 , m = 1000/3600 , T2= 592.2 , T1 = 310 , R = 0.287
W = - 80 kW ( i.e. power supplied to the system )
B) Determine power input and Isentropic compressor efficiency
Given Temperature = 3978C
i) power input to compressor
W = m* [tex]\frac{1}{M}[/tex] ( h2 - h1 )
h2 = 19685 kJ/ kmol ( value gotten from Nitrogen table at temp = 670k )
h1 = 9014 kj/kmol ( value gotten from Nitrogen table at temp = 310 k )
m = 1000/3600 , M = 28
input values into equation above
W = 105.86 kw
ii) compressor efficiency
П = ideal work output / actual work output
= ( h2s - h1 ) / ( h2 - h1 ) = ( T2s - T1 ) / ( T2 - T1 )
= ( 592.2 - 310 ) / ( 670 - 310 )
= 0.784 ≈ 78%
The Aluminum Electrical Conductor Handbook lists a dc resistance of 0.01558 ohm per 1000 ft at 208C and a 60-Hz resistance of 0.0956 ohm per mile at 508C for the all-aluminum Marigold conductor, which has 61 strands and whose size is 1113 kcmil. Assuming an increase in resistance of 2% for spiraling, calculate and verify the dc resistance. Then calculate the dc resistance at 508C, and determine the percentage increase due to skin effect.
Answer:
a) 0.01558 Ω per 1000 feet
b) 0.0923 Ω per mile
c) 3.57%
Explanation:
a) Calculate and verify the DC resistance
Dc resistance = R = р [tex]\frac{l}{A}[/tex]
for aluminum at 20°C
р = 17 Ωcmil/ft
hence R = 17 * 1000 / ( 113000 ) = 0.01527 Ω per 1000 feet
there when there is an increase in resistance of 2% spiraling
R = 1.02 * 0.01527 = 0.01558 Ω per 1000 feet
b) Calculate the DC resistance at 50°C
R2 = R1 ( [tex]\frac{T+t2}{T+ t1}[/tex] )
where ; R1 = 0.01558 , T = 228 , t2 = 50, t1 = 20 ( input values into equation above )
hence R2 = ( 0.01746 / 0.189 ) Ω per mile = 0.0923 Ω per mile
c ) Determine the percentage increase due to skin effect
AC resistance = 0.0956 ohm per mile
Hence; Increase in skin effect
= ( 0.0956 -0.0923 ) / 0.0923
= 0.0357 ≈ 3.57%
Air expands through a turbine from 10 bar, 900 K to 1 bar, 500 K. The inlet velocity is small compared to the exit velocity of 100 m/s. The turbine operates at steady state and develops a power output of 3200 kW. Heat transfer between the turbine and its surroundings and potential energy effects are negligible. Calculate the mass flow rate of air, in kg/s, and the exit area, in m2 .
Answer:
- the mass flow rate of air is 7.53 kg/s
- the exit area is 0.108 m²
Explanation:
Given the data in the question;
lets take a look at the steady state energy equation;
m" = W"[tex]_{cv[/tex] / [ (h₁ - h₂ ) -[tex]\frac{V_2^2}{2}[/tex] ]
Now at;
T₁ = 900K, h₁ = 932.93 k³/kg
T₂ = 500 K, h₂ = 503.02 k³/kg
so we substitute, in our given values
m" = [ 3200 kW × [tex]\frac{1\frac{k^3}{s} }{1kW}[/tex] ] / [ (932.93 - 503.02 )k³/kg -[tex]\frac{100^2\frac{m^2}{s^2} }{2}[/tex]|[tex]\frac{ln}{kg\frac{m}{s^2} }[/tex]||[tex]\frac{1kJ}{10^3N-m}[/tex]| ]
m" = 7.53 kg/s
Therefore, the mass flow rate of air is 7.53 kg/s
now, Exit area A₂ = v₂m" / V₂
we know that; pv = RT
so
A₂ = RT₂m" / P₂V₂
so we substitute
A₂ = {[ [tex](\frac{8.314}{28.97}\frac{k^3}{kg.K})[/tex]×[tex]500 K[/tex]×[tex](7.54 kg/s)[/tex] ] / [(1 bar)(100 m/s )]} |[tex]\frac{1 bar}{10N/m^2}[/tex]||[tex]10^3N.m/1k^3[/tex]
A₂ = 0.108 m²
Therefore, the exit area is 0.108 m²
In this lab, we assumed that the flip-flops did not contribute to the timing constraints of the circuit. Unfortunately, this is not the case. As you saw when you simulated the D flip-flop, the sampling action does not happen instantaneously. In fact, a flip-flop will become unstable if the inputs do not remain stable for a certain amount of time prior to the rising-edge event (setup time) and a certain amount of time after the rising-edge event (hold time). Assume a setup and hold time of 2ns and 1ns, respectively. What would the theoretical maximum clock rate for the synchronous adder be in this scenario
What 2 things can you never eat for breakfast?
i know the answer but lets see if you do
Answer:
you can't eat your school computer or a pencil.
An oil with density 900 kg/m3 and kinematic viscosity 0.0002 m2/s flows upward through an inclined pipe as shown in figure below. The pressure and elevation are known at sections 1 and 2, 10 m apart. Assuming steady laminar flow
Answer:
P=900KG/M3
U=0.0002 M2/S
RE=PV/U
=900*10/0.0002
=45000000
Explanation:
The Reynold number will be 4.5×10⁷. Reynold's number is found as the ratio of the inertial to the viscous force.
What is density?Density is defined as the mass per unit volume. It is an important parameter in order to understand the fluid and its properties. Its unit is kg/m³.
The mass and density relation is given as
mass = density × volume
The ratio of inertial to viscous force is known as Reynold's number.
[tex]\rm R_E= \frac{\rho u L}{\mu} \\\\ \rm R_E=\frac{900 \times 10}{0.0002} \\\\ R_E=45000000[/tex]
Hence, the Reynold number will be 4.5×10⁷.
To learn more about the density refers to the link;
brainly.com/question/952755
#SPJ2
Sarah is a site investigator for a large construction firm. She is considering Miguel, a former geology student with experience as an intern at an architecture firm, for an assistant site investigation position. Which of the following is most relevant to her decision?
Answer: A. whether his geology studies exposed him to principles of geotechnical engineering
Explanation:
The options include:
a. whether his geology studies exposed him to principles of geotechnical engineering
b. the size of the geology program he attended
c. the size of the architecture firm
d. whether the architecture firm was intending to offer Miguel a full-time position
Since Miguel, is a former geology student with experience as an intern at an architecture firm, and Sarah is considering him for an assistant site investigation position, the option that will be relevant for her to make a decision is to know whether his geology studies exposed him to principles of geotechnical engineering.
Geotechnical engineering, is a branch of engineering that makes use of principles of rock mechanics to solve engineering challenges. Since Sarah needs him for an assistant site investigation position, he'll need to investigate souls, rocks and evaluate them.
What forced induction device is more efficient?
A. Turbo Charger
B. Supercharger
C. Centrifugal Supercharger
D. Procharger
Answer:
A
Explanation:
Answer:
a
Explanation: