At 800K, a plot of ln[cyclobutane] vs t gives a straight line with a slope of -1.6 s-1. Calculate the time needed for the concentration of cyclobutane to fall to 1/16 of its initial value.
Answer:
hmmm.........
Explanation:
Which of the following is an example of someone who claims that the media has a shooting blanks effect?
A. "Along with parents, peers, and teachers, the media socializes children about how boys and girls are supposed to behave."
B. "My kid saw a cigarette ad in a magazine and now he's smoking. It's the magazine's fault!"
C. "The media doesn't affect me at all because I'm smart enough to know the difference between right and wrong."
D. "There is no definitive evidence that the media affects our behavior"
Answer:
the answer would be d its d
Answer:
Pretty sure the answer is "C"
Explanation:
"The media doesn't affect me at all because I'm smart enough to know the difference between right and wrong."
A rectangular channel 3-m-wide carries 12 m^3/s at a depth of 90cm. Is the flow subcritical or supercritical? For the same flowrate, what depth will five critical flow?
Answer:
Super critical
1.2 m
Explanation:
Q = Flow rate = [tex]12\ \text{m}^3/\text{s}[/tex]
w = Width = 3 m
d = Depth = 90 cm = 0.9 m
A = Area = wd
v = Velocity
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
[tex]Q=Av\\\Rightarrow v=\dfrac{Q}{wd}\\\Rightarrow v=\dfrac{12}{3\times 0.9}\\\Rightarrow v=4.44\ \text{m/s}[/tex]
Froude number is given by
[tex]Fr=\dfrac{v}{\sqrt{gd}}\\\Rightarrow Fr=\dfrac{4.44}{\sqrt{9.81\times 0.9}}\\\Rightarrow F_r=1.5[/tex]
Since [tex]F_r>1[/tex] the flow is super critical.
Flow is critical when [tex]Fr=1[/tex]
Depth is given by
[tex]d=(\dfrac{Q^2}{gw^2})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{12^2}{9.81\times 3^2})^{\dfrac{1}{3}}\\\Rightarrow d=1.2\ \text{m}[/tex]
The depth of the channel will be 1.2 m for critical flow.
Copy bits 3..0 in $s1 to 6..3 in $s2. Bits 6..3 in $s2 are already set to 0. Registers$s0 0..01111$s1 0..0101$s3 0
Answer:
Following are the solution to this question:
Explanation:
To copy 3.0 bits in 50 dollars or run at 50 dollars, it takes just 3.0 bits as well as other bits but masks, and 50 dollars.
Instead of shifting the $ 50 by 3 bits to 6...3 bits of [tex]\$ \ 50, \$ \ 50,0*0000 0003,[/tex] This procedure instead took place at $53 and $50
AND [tex]\$ \ 50,\$ \ 50,0*0000 000f[/tex], take 3..0 bits
SLL [tex]\$ \ 50, \$ \ 50,0*0000 0003,[/tex]Shifts the bits to 6..3
O R [tex]\$ \ 53,\$ \ 53,\$ \ 50 ,[/tex] coping to [tex]\$ \ 53[/tex]
A roadway with a rough-asphalt pavement has a cross slope of 2%, a longitudinal slope of 2.5%, a curb height of 8 cm, and a 90-cm-wide concrete gutter. If the flow rate in the gutter is 0.07 m/s, determine the size (W XL, in mm) and interception capacity (m/s) of a reticuline grate that should be used to intercept as much of the flow as possible.
a. Reticuline grate size?
b. Interception capacity?
Answer:
b
Explanation:
Laminar flow normally persists on a smooth flat plate until a critical Reynolds number value is reached. However, the flow can be tripped to a turbulent state by adding roughness to the leading edge of the plate. For a particular situation, experimental results show that the local heat transfer coefficients for laminar and turbulent conditions are
h_lam(x)= 1.74 W/m^1.5. Kx^-0.5
h_turb(x)= 3.98 W/m^1.8 Kx^-0.2
Calculate the average heat transfer coefficients for laminar and turbulent conditions for plates of length L = 0.1 m and 1 m.
Answer:
At L = 0.1 m
h⁻_lam = 11.004K W/m^1.5
h⁻_turb = 7.8848K W/m^1.8
At L = 1 m
h⁻_lam = 3.48K W/m^1.5
h⁻_turb = 4.975K W/m^1.8
Explanation:
Given that;
h_lam(x)= 1.74 W/m^1.5. Kx^-0.5
h_turb(x)= 3.98 W/m^1.8 Kx^-0.2
conditions for plates of length L = 0.1 m and 1 m
Now
Average heat transfer coefficient is expressed as;
h⁻ = 1/L ₀∫^L hxdx
so for Laminar flow
h_lam(x)= 1.74 . Kx^-0.5 W/m^1.5
from the expression
h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5 dx
= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L
= 1.74k/L = [ (x^0.5)/0.5)]⁰^L
= 1.74K × L^0.5 / L × 0.5
h⁻_lam= 3.48KL^-0.5
For turbulent flow
h_turb(x)= 3.98. Kx^-0.2 W/m^1.8
form the expression
1/L ₀∫^L 3.98 . Kx^-0.2 dx
= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L
= (3.98K/L) × (L^0.8 / 0.8)
h⁻_turb = 4.975KL^-0.2
Now at L = 0.1 m
h⁻_lam = 3.48KL^-0.5 = 3.48K(0.1)^-0.5 W/m^1.5
h⁻_lam = 11.004K W/m^1.5
h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2
h⁻_turb = 7.8848K W/m^1.8
At L = 1 m
h⁻_lam = 3.48KL^-0.5 = 3.48K(1)^-0.5 W/m^1.5
h⁻_lam = 3.48K W/m^1.5
h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2
h⁻_turb = 4.975K W/m^1.8
Therefore
At L = 0.1 m
h⁻_lam = 11.004K W/m^1.5
h⁻_turb = 7.8848K W/m^1.8
At L = 1 m
h⁻_lam = 3.48K W/m^1.5
h⁻_turb = 4.975K W/m^1.8
Technician A says vehicles with electronic throttle control do not need a separate cruise control module, stepper motor, or cable to control engine speed. Technician B says a faulty brake light switch may cause the cruise control to not operate. Who is correct?
Answer: its A
Explanation:
People tend to self-disclose to others that are in age, social status, religion, and personality.
Answer:people tend to do this when they are in a different environment they lose something or just have something going on in their life
Explanation:
A three-phase motor rated 25 hp, 480 V, operates with a power factor of 0.74 lagging and supplies the rated load. The motor efficiency is 96%. Calculate the motor input power, reactive power and current.
Answer:
the motor input power is 19.42 KW
the Reactive power is 17.65 KVAR
Current is 31.56 A
Explanation:
Given that;
V = 480V
h.p = 25 hp
p.f = 0.74 lagging
n_motor = 96%
so output = 25hp
and we know that;
1hp = 746 watt
watt = hp × 1hp
so output in watt = 25 × 746 = 18650 Watt = 18.65 KW
n_motor = (output / input) × 100
96 = 1865 / Input
96Input = 1865
Input = 1865 / 96
Input = 19.42 KW
Therefore the motor input power is 19.42 KW
P = √( 3 × V × I × cos∅)
19.42 = √( 3 ×480 × I × 0.74)
I = 31.56 A
Therefore Current is 31.56 A
Q = √( 3 × V × I × sin∅)
we know that
cos∅ = 0.74
so ∅ = cos⁻¹(0.74) = 42.26
so we substitute
Q = √( 3 × 480 × 31.56 × sin(42.26))
= 17.65 KVAR
Therefore the Reactive power is 17.65 KVAR
A settling tank has an influent rate of 0.6 mgd. It is 12 ft deep and has a surface area of 8000 ft². What is the hydraulic retention time?
Answer: hydraulic retention time,τ=28.67 hours
Explanation:
The hydraulic retention time τ (tau), is given as The volume of the settling tank(V) divided by the influent flowrate(Q)
τ =V/Q
But Volume is not known and is given as
Volume = surface area x depth of the tank
= 8000 ft² X 12 ft
= 96,000 ft³
Also, the influent flow rate is in mgd ( million gallons per day), we change it to ft³/sec so as to be in same unit with the volume in ft³
1 million gallons/day = 1.5472286365101 cubic feet/second
0.6mgd = 1.5472286365101 cubic feet/second x 0.6
=0.93cubic feet/second
τ =V/Q
96,000 ft³/0.93 ft³/sec
τ=103,225.8 secs
changing to hours
103,225.8 /3600 =28.67 hours
The hydraulic retention time =28.67 hours
In a CS amplifier, the resistance of the signal source Rsig = 100 kQ, amplifier input resistance (which is due to the biasing network) Rin = 100kQ, Cgs = 1 pF, Cgd = 0.2 pF, gm = 5 mA/V, ro = 25 kΩ, and RL = 20 kΩ. Determine the expected 3-dB cutoff frequency.
Answer:
406.140 KHz
Explanation:
Given data:
Rsig = 100 kΩ
Rin = 100kΩ
Cgs = 1 pF,
Cgd = 0.2 pF, and etc.
Determine the expected 3-dB cutoff frequency
first find the CM miller capacitance
CM = ( 1 + gm*ro || RL )( Cgd )
= ( 1 + 5*10^-3 * 25 || 20 ) ( 0.2 )
= ( 11.311 ) pF
now we apply open time constant method to determine the cutoff frequency
Th = 1 / Fh
hence : Fh = 1 / Th = [tex]\frac{1}{(Rsig +Rin) (Cm + Cgs )}[/tex]
= [tex]\frac{1}{( 200*10^3 ) ( 12.311 * 10^{-12} )}[/tex] = 406.140 KHz
Indicate similarities between a nucleus and a liquid droplet; why small droplets are stable and very big droplets are not?
Answer:
There are several similarities between the nucleus and a liquid droplet.
Explanation:
A droplet of liquid simply is is very small or tiny drop of liquid. It is also considered as a tiny column of liquid that is surrounded by surfaces that have zero shear stress.
A nucleus on the other hand is an assembly between protons and neutrons. The latter is electrically charged whilst the former is positively charged. The number of protons present in an element is very crucial to the qualities of an element.
The main similarities between a nucleus and a liquid droplet are:
1. a nucleus consists of a large amount of neutrons and protons in the same volume as would a liquid which contains large numbers of molecules in the same volume;
2. both the nucleus and the droplet are similar for their homogeneity in electric charge and density;
3. the molecules exert the same amount for forces towards one another as would the nuclear forces in the nucleons.
4. both of them cannot be compressed
5. both molecules and nucleus are can be subject to nuclear fission which simply mean the breaking apart into smaller units (in the case of the nucleus) or the breaking apart into smaller droplets in the case of the liquid molecule.
6. There are two types of phenomena which occurs in both the liquid droplet and the nucleus which are similar to one another. They are:
Evaporation (in the case of the liquid molecule) and reaction emission (in the case of the nucleus). In evaporation, particles are lost, in Atomic transmutation, particles are lost as well.
B) the forces which determine the stability of droplets are surface tension and gravitation. The smaller the area, the stronger the surface tension available to keep the drops from going out of shape.
Cheers
A vortex tube receives 0.3 m^3 /min of air at 600 kPa and 300 K. The discharge from the cold end of the tube is 0.6 kg/min at 245 K and 100 kPa. The discharge from the hot end is at 325 K and 100 kPa. Determine the irreversibility.
Answer:
Irreversibility = 5.361 kW
Explanation:
From the given information:
By applying ideal gas equation at entry:
PV = mRT
600 × 0.3 = m × 0.287 × 300 (where R = 0.287 kJ/kg)
180 = m × 86.1
m = 180/86.1
m = 2.0905 kg/min
At the hot end, using the same ideal gas equation:
PV = mRT
100 × V = 1.4905 × 0.287 × 325
V = 139.026/100
V = 1.3903 m³/ min
This implies that: The total entropy change = Entropy of the universe
So,
[tex]m\bigg [ c_p \ In \dfrac{T_2}{T_o}-R \ In \dfrac{P_2}{P_o} \bigg] + m_2 \bigg [ c_p \ In \dfrac{T_2}{T_o}- R \ In\dfrac{P_2}{P_o} \bigg][/tex]
[tex]= 0.6\bigg [ 1.004 \ In \dfrac{245}{300}-0.287 \ In \dfrac{100}{600} \bigg] +1.4905\bigg [1.004 \ In \dfrac{325}{300}- 0.287\ In\dfrac{100}{600} \bigg][/tex]
= 0.6[-0.2033 + 0.5142] + 1.4905 [0.08036 + 0.5142]
= 1.0727 kJ/min.K
= 0.01787 kw/K
Irreversibility = [tex]T_o [ \Delta S][/tex]
Irreversibility = 300 × 0.01787
Irreversibility = 5.361 kW
Help this is very hard and I don't get it
Answer:
yes it is very hard you should find a reccomended doctor to aid in your situation. But in the meantime how about you give me that lil brainliest thingy :p
A fluid has a mass of 5 kg and occupies a volume of 1 m3 at a pressure of 150 kPa. If the internal energy is 25000 kJ/kg, what is the total enthalpy?
Answer:
155 KJ
Explanation:
The total enthalpy is given by
ΔH=ΔU + PV
Where;
ΔH = enthalpy
ΔU = internal energy = 25000 kJ/kg/ 5 kg = 5000 KJ
P = 150 kPa = 150,000 Pa
V = 1 m3
ΔH= 5000 + (150,000 * 1)
ΔH= 155 KJ
please what is dif
ference between building technology and building engineering.
Answer:
Building technology is building technology such as coding an app or a website
Building engineering is making computers or cars or phones
Explanation:
A 8-core machine has 4 times the performance of a single-core machine of the same frequency. Performance is proportional to frequency. Voltage decreases proportionally to frequency. To achieve the same performance, how much (in percentage) dynamic power would the 8-core system save?
Answer: The 8-core machine saves 87.5% of the dynamic power.
Explanation:
Let Fold = f , Vold = V , Cold = Capacitance
so
Old Dynamic power = Cold × (Vold × Vold) × f
therefore for the 8-core machine
Fnew / Fold = 1/4
Fnew = Fold/4
we were told that Voltage decreases proportional to frequency,
so
Vnew / Vold = 1/4
Vnew = V / 4
So New Capacitance will be;
Cnew = Cold
Thus, New Dynamic power = 8 × Cnew × ( Vnew × Vnew ) × Fnew
= 8 × Cold × (Vold × Vold/16) × ( f/4 )
= 8 × ( Cold ) × ( Vold × Vold ) × ( f ) / 64
= (Old Dynamic Power) / 8
therefore
Old Dynamic Power / New Dynamic Power = 8
Thus, Percentage of power saved will be;
Percentage power saved = 100 × ( Old Dynamic Power - New Dynamic Power ) / Old Dynamic Power
= 100 × (8-1) / 8
= 87.5 %
Therefore The 8-core machine saves 87.5% of the dynamic power.
A stream leaving a sewage pond (containing 80 mg/L of sewage) moves as a plug with a velocity of 40 m/hr. A concentration of 50 mg/L is measured 5,000 m downstream. What is the 1st order decay rate constant in the stream?
Answer:Decay rate constant,k = 0.00376/hr
Explanation:
IsT Order Rate of reaction is given as
In At/ Ao = -Kt
where [A]t is the final concentration at time t and [A]o is the inital concentration at time 0, and k is the first-order rate constant.
Initial concentration = 80 mg/L
Final concentration = 50 mg/L
Velocity = 40 m/hr
Distance= 5000 m
Time taken = Distance / Time
5000m / 40m/hr = 125 hr
In At/ Ao = -Kt
In 50/80 = -Kt
-0.47 = -kt
- K= -0.47 / 125
k = 0.00376
Decay rate constant,k = 0.00376/hr
Calculate the number of vacancies per cubic meter for some metal, M, at 783°C. The energy for vacancy formation is 0.95 eV/atom, while the density and atomic weight for this metal are 6.10 g/cm^3 (at 783°C) and 43.41 g/mol, respectively.
Answer:
Following are the solution to this question:
Explanation:
The number of vacancies by the cubic meter is determined.
[tex]N_V =N exp(\frac{Q_v}{kT})[/tex]
[tex]= \frac{N_A \rho}{A} exp (\frac{Q_v}{kT})[/tex]
[tex]= \frac{6.022 \times 10^{23} \times 6.10}{43.41} \exp(\frac{-0.95}{8.62\times 10^{-5} \times (783+273)})\\\\= \frac{36.7342 \times 10^{23}}{43.41} \exp(\frac{-0.95}{0.0313626})\\\\= 0.846215158 \times 10^{23} \exp(-30.290856)\\\\[/tex]
[tex]=1.57 \times 10^{25} \ cm^{-3}[/tex]
An insulated 40 ft^3 rigid tank contains air at 50 psia and 580°R. A valve connected to the tank is now opened, and air is allowed to escape until the pressure inside drops to 25 psia. The air temperature during this process is kept constant by an electric resistance heater placed in the tank. Determine the electrical work done during this process in Btu assuming variable specific heats.
Answer:
The answer is "[tex]\bold{W_{in} = 645.434573 \ Btu}[/tex]".
Explanation:
Its air enthalpy is obtained only at a given temperature by A-17E.
The solution from the carbon cycle is acquired:
[tex]\to \Delta U= W_{in}-m_{out}h_{out}=0\\\\\to W_{in} = (m_1 - m_2)h[/tex]
[tex]=\frac{Vh}{RT}(P_1- P_2)\\\\= \frac{40 \times 138.66}{0.3704\times 580}(50-25) Btu\\\\= \frac{5,546.4}{214.832}(25) Btu\\\\= 25.8173829(25) Btu\\\\ =645.434573 Btu[/tex]
[tex]W_{in} = 645.434573 \ Btu[/tex]
please help me make a lesson plan. the topic is Zigzag line. and heres the format.
A. Objective
B. Subject matter
C. Learning activities.
D. Assessment.
E. Reinforcement
Explanation:
D. B. C. A. E. Is this a good idea
What is the amount of pearlite formed during the equilibrium cooling of a 1055 steel from 1000°C to room temperature?
Answer: 98.5% of pearlite was formed during the equilibrium cooling
Explanation:
First we calculate the fraction of pro-eutectoid phase which forms for equilibrium cooling of the 1085 steel from 1000°C at room temperature;
we know that in 1085 steel, last two digits denotes the carbon percentage
so 1085 steel contains 0.85% carbon.
Now from the diagram, carbon percentage is greater than the eutectoid com[psition
i.e 0.85 > 0.76
it is a hyper eutectoid steel
so
fraction of pro eutectoid phase W_Fe₃C = (0.85 - 0.76) / ( 6.7 - 0.76)
= 0.09 / 5.94 = 0.015 = 1.5%
Now, the amount of pearlite formed during the equilibrium cooling of the 1055 steel from 1000°C to room temperature will be;
pearlite (C') = (1 - W_Fe₃C)
= 1 - 0.015
= 0.985 = 98.5%
Therefore 98.5% of pearlite was formed during the equilibrium cooling
Anyone help me please ?
Answer:
I can help but I need to know what it looking for
The voltage v= 12cos(60t + 45o) is applied to a 0.1 H inductor. Calculate the inductor's Impedance Z = j XL in ohms.
Answer:
6 Ω
Explanation:
given data :
Voltage ( v ) = 12cos ( 60t + 45° )
L = 0.1 H
calculate the inductor's impedance Z
Z = jXL
= jx = 60
= L = 0.1
hence Z = 60 * 0.1 = 6 Ω
A battery with a nominal voltage of 200-V with a resistance of 10 milliohms to be charged at a constant current of 20 amps from a 3-phase semi-converter with a 220-V (line-to-line) Y-connected 60 Hs supply. Determine:
a. The firing angle of the thyristors for the charging process.
b. The displacement power factor and the supply power factor.
Answer:
a) ( ∝ ) = 69.6548
b) supply power factor = 0.6709
displacement power factor = 0.8208
Explanation:
Given data:
Nominal voltage ( E ) = 200-V
resistance (r) = 10 milliohms
constant current ( I ) = 20 amps
Phase ; 3-phase
semi-converter with 220-v ( line-to-line ) , 220√2 ( phase voltage )
frequency ; 60 Hz
a) determine the firing angle of thyristors
Vo = E + I*r
= 200 + 20*10*10^-3
= 200.2 v
attached below is the remaining part of the solution
firing angle of thyristors for charging process ( ∝ ) = 69.6548
b) determine displacement power factor and supply power factor
attached below is the detailed solution
Displacement power factor ( Dpf ) = cos ( ∝ /2 ) = 0.8208
displacement power factor = g * Dpf
= 0.81747 * 0.8208 = 0.6709
Products exit a combustor at a rate of 100 kg/sec, and the air-fuel ratio is 9. Determine the air flow rate. a. 9 kg/sec b. 90 kg/sec c. 100 kg/sec d. 10 kg/sec
Answer: the air flow rate a is 90 kg/sec; Option b) 90 kg/sec is the correct answer
Explanation:
Given that;
product of combustor flow rate m = 100 kg/s
air-fuel = 9
Airflow rate = ?
⇒We know that in the combustor, air fuel are mixed and then ignited,
⇒air fuel products are exited at the combustor
let air and fuel be a and b respectively
⇒ a + b = 100 kg/sec ----- let this be equation 1
now
⇒ air / fuel = 9
a / b = 9
a = 9b -----------let this be equation 2
now input a = 9b in equation 1
9b + b = 100 kg/sec
10b = 100 kg/sec
b = 10 kg/sec
we know that
a = 9b
so a = 9 × 10 = 90 kg/sec
Therefore the air flow rate a is 90 kg/sec
How many snaps points does an object have?
Answer:
what do you mean by that ? snap points ?
A laboratory furnace wall is constructed of 0.2 m thick fireclay brick having a thermal conductivity of 1.82 W/m-K. The wall is covered on the outer surface with insulation of thermal conductivity of 0.095 W/m-K. The furnace inner brick surface is at 950 K and the outer surface of the insulation material is at 300 K. The maximum allowable heat transfer rate through the wall of the furnace is 830 W/m^2. Determine how thick in cm the insulation material must be.
Answer:
The appropriate solution will be "6.4 cm".
Explanation:
The given values are:
Length,
l = 0.2 m
Thermal conductivity,
K₁ = 1.82 W/m-K
K₂ = 0.095 W/m-K
Temperature,
T = 950 K
T = 300 K
Heat transfer rate,
Q = 830 W/m²
Now,
⇒ [tex]Q = \frac{\Delta T}{\frac{L_1}{K_1 A} +\frac{L_2}{K_2 A} }=\frac{A \Delta T}{\frac{L_1}{K_1 } +\frac{L_2}{K_2 } }[/tex]
⇒ [tex]\frac{Q}{A} =\frac{\Delta T}{\frac{L_1}{K_1} +\frac{L_2}{K_2} }[/tex]
On substituting the above given values in the equation, we get
⇒ [tex]830=\frac{(980-300)}{\frac{0.2}{1.82} +\frac{x}{0.095} }[/tex]
On applying cross-multiplication, we get
⇒ [tex]\frac{0.2}{1.82} +\frac{x}{0.095} =\frac{950-300}{830}[/tex]
⇒ [tex]\frac{0.2}{1.82} +\frac{x}{0.095} =\frac{650}{830}[/tex]
⇒ [tex]x =0.639 \ m[/tex]
⇒ [tex]x=6.345 \ i.e., 6.4 \ m[/tex]
Which of following are coding languages used in controlling a robot? *
A. Scratch
B. B/B--
C. C/C++
D. Robot Z
Answer:
C/C++
Explanation:
C/C++
Series aiding is a term sometimes used to describe voltage sources of the same polarity in series. If a 5 V and a 9 V source are connected in this manner, what is the total voltage?
Answer:Total Voltage = 14V
Explanation: it is possible that a circuit can contain more than one source of electromotive force which can cause flow of current in the same or opposite direction . When the connection to voltage sources allows for current from the voltage sources to flow in same direction,it is termed Series aiding Thus, the Total/effective voltage in a series aiding circuit is computed as the sum of series aiding voltages .
Here we have the series aiding voltages to be 5V and 9V ,
therefore,
Total Voltage = 5V + 9V
= 14V