Answer:
The pseudocode is as follows:
Input SalesAmount
Input CommissionRate
CommissionEarned= SalesAmount * (CommissionRate/100)
Print CommissionEarned
Explanation:
This gets input for SalesAmount
Input SalesAmount
This gets input for CommissionRate
Input CommissionRate
This calculates the CommissionEarned
CommissionEarned= SalesAmount * (CommissionRate/100)
This prints the calculated CommissionEarned
Print CommissionEarned
WILL MARK BRAINLIST I need help on this asap thanks
Determine the dimensions for T if T = M V^2 A / L^3 where M is a mass, V is a velocity, A is an area, and L is a length.
L / T
M
M L / T^2
M / (L T^2)
No dimensions
Explanation:
ask your dad please and use your brain
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping station is 140 m and that of the exit point is 150 m. The required terminal head is 10 m. Estimate the pipe diameter and pumping head using the explicit design procedure g
Answer:
[tex]D=0.41m[/tex]
Explanation:
From the question we are told that:
Discharge rate [tex]V_r=0.35 m3/s[/tex]
Distance [tex]d=4km[/tex]
Elevation of the pumping station [tex]h_p= 140 m[/tex]
Elevation of the Exit point [tex]h_e= 150 m[/tex]
Generally the Steady Flow Energy Equation SFEE is mathematically given by
[tex]h_p=h_e+h[/tex]
With
[tex]P_1-P_2[/tex]
And
[tex]V_1=V-2[/tex]
Therefore
[tex]h=140-150[/tex]
[tex]h=10[/tex]
Generally h is give as
[tex]h=\frac{0.5LV^2}{2gD}[/tex]
[tex]h=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
Therefore
[tex]10=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
[tex]D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}[/tex]
[tex]D=0.41m[/tex]
Determine the transfer function H(s) for a high pass filter with the following characteristics:
1. a cutoff frequency of 100 kHz
2. a stopband attenuation rate of 40 dB/decade
3. a nominal passband gain of 20 dB, which drops to 14 dB at the cutoff frequency
Write the formula for H(s) that satisfies these requirements.
Answer:
H(s) = 10 / [ 1 + s / (200*10^3π ) ]^2
Explanation:
Characteristics of the high pass filter
Cutoff frequency = 100 kHz
stopband attenuation rate = 40 dB/decade
nominal passband gain = 20dB = 20logK = 20
Formula for H(s) satisfying the requirements above
given that the stopband attenuation = 40 dB/decade the formula for H(s) that will satisfy the requirements is a second order filter
H(s) = K / ( 1 + s/Wo ) ^2 ----- ( 1 )
Wo = 2πf = 2π ( 100 * 10^3 ) = 200 * 10^3 π
K = 10
back to equation ( 1 )
H(s) = 10 / [ 1 + s / (200*10^3π ) ]^2
Explain why veracity, value, and visualization can also be said to apply to relational databases as well as Big Data.
Answer:
Veracity, Value and Visualization are not only the characteristics of Big Data but are also the characteristics of relational databases. Veracity of data is issue with smallest data stores this is the reason that it is important in relation...
In a true Brayton cycle, the pressure ratio is 9. Air input temperature to the cycle 300 K pressure is 100 kPa. The maximum temperature in the cycle is 1300 K. Compressor and turbine their yields are equal to each other. Net work obtained from the cycle is 225 kJ / kg. Accordingly, the cycle find the overall yield. The specific temperatures are variable.
Answer:
i did not known answer but anobody help you
Using 1.5 V batteries, a switch, and three lamps, devise a circuit to apply 4.5 V across eitherone lamp, two lamps in series, or three lamps in series with a single-control switch. Draw theschematic.
Answer: the attached picture is the answer.
Explanation:
Assuming:
the switch position connect to 1, hence 4.5V exist at across lamp1
the switch position connects to 2 hence 4.5 V exist across lamp 1 and lamp 2
the switch position connects to 3, hence, 4.5 V exist across lamp 1, lamp 2 and lamp 3.
In low speed subsonic wind tunnels, the value of test section velocity can be controlled by adjusting the pressure difference between the inlet and test-section for a fixed ratio of inlet-to-test section cross-sectional area.
a. True
b. false
Answer:
Hence the given statement is false.
Explanation:
For low-speed subsonic wind tunnels, the air density remains nearly constant decreasing the cross-section area cause the flow to extend velocity, and reduce pressure. Similarly increasing the world cause to decrease and therefore the pressure to extend.
The speed within the test section is decided by the planning of the tunnel.
Thus by adjusting the pressure difference won't change the worth of test section velocity.
Answer:
The given statement is false .
HELP PLEASE!! ASAP!!!!
can some answer this 2 questions please as paragraph i want it nowww it is graded what action should be taken to make it safe ? also the first question
Actions violated:
Long hair isn't tied upThe girl isn't wearing a lab coatThe girl isn't wearing safety gogglesExtra: There doesn't seem to be an emergency fire blanket in the safeActions to be taken:
Make sure the girl wears a lab coat or kick her outMake sure the girl wears safety goggles or kick her outMake sure her hair is tied up or kick her outEdit: Use these to write your paragraph.
A stream of oxygen enters a compressor at a rate of 200 SCMH. The oxygen exits at 360 K and 500 bar. Determine the volumetric flowrate exiting the compressor using the compressibility factor equation of state.
Answer:
≈ 0.516 m^3/hr
Explanation:
Inlet of compressor = 200 SCMH
sheer standard conditions = 1 atm and 288.5 K
For oxygen :
critical pressure(Pc) = 49.8 atm
critical temperature Tc = 154.6 K
hence at compressor inlet
Tr = T / Tc = 288.5/154.6 = 1.866
Pr = P / Pc = 1 / 49.8 = 0.0204
Z1 ( from compressibility chart ) = 0.98
at compressor outlet
P2 = 500 bar = 500*0.9869 = 493.45 atm , T2 = 360 k
hence : Pr = P / Pc = 493.45 / 49.8 = 9.91
Tr = T / Tc = 360 / 154.6 = 2.33
Z2 ( from compressibility chart ) ≈ 1
V2( volumetric flow rate ) = V1*(P₁Z₂T₂) / (P₂Z₁T₁)
= 200 ( 1 * 1* 360) / (493.45 *0.98*288.5)
= 0.516 m^3/hr
Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution.
x dx/dy−y=x^2sinx
Answer:
Interval: x∈ ( 0, ∞ )
There are no transient terms
Explanation:
x (dy/dx) – y= x^2sinx
Attached below is the detailed solution of the Given problem
There are no transient terms found in the general solution
Interval: x∈ ( 0, ∞ )
An intersection with a four phase signal has a displayed red time of 35 seconds, a start-up lost time of 2 seconds, a yellow time of 4 seconds, and an all red time of 1 second per phase. The total lost time is typically calculated as ____ seconds per cycle.
Answer:
53 sec / cycle
Explanation:
Displayed red time = 35 seconds
Start up lost time = 2 seconds
Yellow time = 4 seconds
Red time = 1 second
Total lost time L = 2n + r
L = lost time
n = number of phase
R = red time
35+2+4+4*1
= 45
L = 2x4+45
= 53 sec/cycle
The total lost time is typically calculated as 53 seconds per cycle
How do your arm muscles work to lift a mug of coffee to your lips?
A.
The biceps muscle contracts while the triceps muscle relaxes. Only then can the forearm move up to lift the mug.
B.
The muscle tendons in your arms stretch to their limit as a result of the flexibility exercises that lift the mug.
C.
Your muscles rapidly increase the speed at which they twitch to trigger your arm movement.
D.
Your biceps and triceps muscles first relax, and then the arm muscle raises the mug.
Answer:
A. The biceps muscle contracts while the triceps muscle relaxes. Only then can the forearm move up to lift the mug.
Explanation:
Skeletal muscle is also known as voluntary muscle and it can be defined as a type of muscle connected with the skeleton by tendons, so as to form a mechanical system that enables the movement of the limbs and other body parts with respect to another.
Generally, skeletal muscles are only found in vertebrates.
The muscles in the arm work to lift a mug of coffee to your lips when the biceps muscle contracts while the triceps muscle relaxes.
Basically, the only way the forearm move up to lift a mug for example, is through the contraction of the biceps muscle and the relaxation of the triceps muscle.
This ultimately implies that, the elbow joint is straightened or bent due to the actions of the biceps muscle and triceps muscle against each other.
To bend the elbow and raise the forearm, the biceps muscle contracts and the triceps muscle relaxes.To straighten the elbow and lower the forearm, the biceps muscle relaxes and the triceps muscle contracts.Answer:A
Explanation:
got it right on plato
Cho thanh có tiết diện thay đổi chịu tải trọng dọc trục (hình 1).
Biết d1 = 5 cm, d2 = 8 cm, a= 15 cm, b=10cm, P1 =400kN, P2 =200kN, E= 2.104 kN/cm2.
a) Vẽ biểu đồ lực dọc.
b) Kiểm tra bền của thanh AC, [ϭ] =10 (kN/cm2).
c) Xác định chuyển vị theo phương dọc trục của tâm tiết diện C
Answer:
saay in English language
Reinforced concrete is a raw material that has always been available, but it was unappreciated by architects until the nineteenth century.
a. True
b. False
Answer: False
Explanation:
Reinforced concrete is simply a combination of the traditional cement concrete with the steel bars which are the reinforcements.
Reinforced concrete is utilized for construction purpose mostly on a large scale. The reinforced concrete was invented by French gardener Joseph in 1849 therefore, it has always been available and appreciated by architects before the 19th century.
what type of slab and beam used in construction of space neddle
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Represent each of the following units as a combination of primitive
dimensions where M=mass, L=length, T=time. As an example, miles per hour would
correspond to [L/T].
a. kilometer
b. quart
c. pascal
d. watt
e. newton
f. horsepower
Answer:
a. unit of length: [L]
b. unit of volume: [[tex]L^3[/tex]]
c. unit of pressure:[tex]P=\frac{F}{A} \equiv\frac{[MLT^{-2}]}{[L^2]}[/tex] [tex][ML^{-1}T^{-2}][/tex]
d. unit of power: [tex]N.m.s^{-1}\equiv [ML^2T^{-3}][/tex]
e. unit of force: [tex][kg.m/s^2]\equiv [MLT^{-2}][/tex]
f. unit of power: [tex]N.m.s^{-1}\equiv [ML^2T^{-3}][/tex]
Force: [tex]F=m.a=m.\frac{v}{t}=m.\frac{x}{t}\div t[/tex]
Power: [tex]P=\frac{W}{t}=\frac{F.x}{t}[/tex]
where:
F = force
A = area
W = work
t = time
a = acceleration
v = velocity
x = displacement
The propeller shaft of the submarine experiences both torsional and axial loads. Draw Mohr's Circle for a stress element on the outside surface of the solid shaft. Determine the principal stresses, the maximum in-plane shear stress and average normal stress using Mohr's Circle.
Answer: Attached below is the missing detail and Mohr's circle.
i) б1 = 9.6 Ksi
б2 = -10.7 ksi
ii) 10.2 Ksi
iii) -0.51Ksi
Explanation:
First step :
direct compressive stress on shaft
бd = P / π/4 * d^2
= -20 / 0.785 * 5^2 = -1.09 Ksi
shear stress at the outer surface due to torsion
ζ = 16*T / πd^3
= (16 * 250 ) / π * 5^3 = 010.19 Ksi
Calculate the Principal stress, maximum in-plane shear stress and average normal stress
Using Mohr's circle ( attached below )
i) principal stresses:
б1 = 4.8 cm * 2 = 9.6 Ksi
б2 = -5.35 cm * 2 = -10.7 ksi
ii) maximum in-plane shear stress
ζ = radius of Mohr's circle
= 5.1 cm = 10.2 Ksi ( Given that ; 1 cm = 2Ksi )
iii) average normal stress
= 9.6 + ( - 10.7 ) / 2
= -0.51Ksi
where are the field poles mounted on an alternator
Answer:
The magnetic field for this type of alternator is established by a set of stationary field poles mounted on the periphery of the alternator frame. The field flux created by these poles is cut by conductors inserted in slots on the surface of the rotating armature.
You have been assigned the task of reviewing the relief scenarios for a specific chemical reactor in your plant. You are currently reviewing the scenario involving the failure of a nitrogen regulator that provides inert padding to the vapor space of the reactor. Your calculations show that the maximum discharge rate of nitrogen through the existing relief system of the vessel is 0.5 kgls, However, your calculations also show that the flow of nitrogen through the l-in supply pipe will be much greater than this. Thus under the current configuration a failure of the nitrogen regulator will result in an over pressuring of the reactor. One way to solve the problem is to install an orifice plate in the nitrogen line, thus limiting the flow to the maximum of 0.5 kg/s. Determine the orifice diameter (in cm) required to achieve this flow. Assume a nitrogen source supply pressure of 15 bar absolute. The ambient temperature is 25°C and the ambient pressure is 1 atm. 3.
Answer:
[tex]D=0.016m[/tex]
Explanation:
From the question we are told that:
Discharge Rate [tex]F_r=0.5kgls[/tex]
Pressure [tex]P=15Kpa[/tex]
Temperature [tex]T=25=>298K[/tex]
Ambient pressure is 1 atm.
Generally the equation for Density is mathematically given by
[tex]\rho=\frac{PM}{RT}[/tex]
[tex]\rho=\frac{15*10^5*28.0134*10^{-3}}{8.314*298}[/tex]
[tex]\rho=16.958kg/m^2[/tex]
Generally the equation for Flow rate is mathematically given by
[tex]F_r=\mu A\sqrt{Q \rho P(\frac{2}{Q+1})^{\frac{Q+1}{Q-1}}}[/tex]
Where
[tex]Q=Heat coefficient\ ratio\ of\ Nitrogen[/tex]
[tex]Q=1.4[/tex]
[tex]\mu= Discharge\ coefficient[/tex]
[tex]\mu=0.68[/tex]
Therefore
[tex]0.5=0.68 A\sqrt{1.4 16.958 15*10^{5}(\frac{2}{1.4+1})^{\frac{1.4+1}{1.4-1}}}[/tex]
[tex]A=2.129*10^{-4}[/tex]
Where
[tex]A=\frac{\pi}{4}D^2[/tex]
[tex]\frac{\pi}{4}D^2=2.129*10^{-4}[/tex]
[tex]D=0.016m[/tex]
What must you do to become ASE certified as an automotive technician?
Answer:
To become ASE certified, you must pass an ASE test and have relevant hands-on work experience. The amount of work experience required can vary by test, and is specified in detail here. ASE recommends submitting the form after you've registered to take an ASE certification test.
Good luck!
Explanation:
Answer: One theme in White Fang is adapting in order to survive. White Fang finally submits to Gray Beaver. He also copes with fighting other dogs. White Fang changes his behaviors so that he can live.
Explanation: its the sample response
Atmospheric pressure is 101 kPa. Pressure inside a tire is measured using a typical tire pressure gage to be 900 kPa. Find gage pressure and absolute pressure in the tire. ___________________________________________________________________
Answer:
The gage and absolute pressures are 900 and 1001 kilopascals, respectively.
Explanation:
The gage pressure ([tex]P_{g}[/tex]), in kilopascals, is the difference between absolute ([tex]P_{abs}[/tex]) and atmospheric pressures ([tex]P_{atm}[/tex]), measured in kilopascals. If we know that [tex]P_{g} = 900\,kPa[/tex] and [tex]P_{atm} = 101\,kPa[/tex], then the gage and absolute pressures are, respectively:
[tex]P_{g} = 900\,kPa[/tex]
[tex]P_{abs} = P_{atm} + P_{g}[/tex]
[tex]P_{abs} = 101\,kPa + 900\,kPa[/tex]
[tex]P_{abs} = 1001\,kPa[/tex]
The gage and absolute pressures are 900 and 1001 kilopascals, respectively.
A micromechanical resonator is to be designed to have a Q factor of 1000 and a natural frequency of 2 kHz. Determine the system-damping factor and the system bandwidth.
Answer:
Explanation:
Given:
Q factor, =1000
natural frequency, [tex]f_n=2000~Hz[/tex]
Damping factor, [tex]\zeta=?[/tex]
Bandwidth, BW=?
We have the relation:
[tex]Q=\frac{1}{2\zeta}[/tex]
[tex]\zeta=\frac{1}{2Q}[/tex]
[tex]\zeta=\frac{1}{2\times 1000}[/tex]
[tex]\zeta=5\times 10^{-4}[/tex]
Bandwidth:
[tex]BW=\frac{f_n}{Q}[/tex]
[tex]BW=\frac{2000}{1000}[/tex]
[tex]BW=2~Hz[/tex]
Your shifts productivity is Slow because one person is not pulling his share. The rest of the team is Getting upset.
Answer:
you are right but then you ddnt ask a question
A 20-mm-diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when one end is twisted through an angle of 15°, what must be the length of the bar?
Answer:
1.887 m
Explanation:
(15 *pi)/180
= 0.2618 rad
Polar moment
= Pi*d⁴/32
= (22/7*20⁴)/32
= 15707.96
Torque on shaft
= ((22/7)*20³*110)/16
= 172857.14
= 172.8nm
Shear modulus
G = 79.3
L = Gjθ/T
= 79.3x10⁹x(1.571*10^-8)x0.2618/172.8
= 1.887 m
The length of the bar is therefore 1.887 meters
A venturimeter of 400 mm × 200 mm is provided in a vertical pipeline carrying oil of specific gravity 0.82, flow being upward. The difference in elevation of the throat section and entrance section of the venturimeter is 300 mm. The differential U-tube mercury manometer shows a gauge deflection of 300 mm. Calculate: (i) The discharge of oil, and (ii) The pressure difference between the entrance section and the throat section.Take the coefficient of meter as 0.98 and specific gravity of mercury as 13.6
Answer:
the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s
Explanation:
Given:
Diameter of the pipe = 100mm = 0.1m
Contraction ratio = 0.5
thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m
The formula for discharge through a venturimeter is given as:
Where,
is the coefficient of discharge = 0.97 (given)
A₁ = Area of the pipe
A₁ =
A₂ = Area at the throat
A₂ =
g = acceleration due to gravity = 9.8m/s²
Now,
The gauge pressure at throat = Absolute pressure - The atmospheric pressure
⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)
Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m
Substituting the values in the discharge formula we get
or
or
Q = 29.28 ×10⁻³ m³/s
Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s
Hope This Helps :D
A designer needs to select the material for a plate under tensile stress. Assuming that the applied tensile force is 13,000 lb and the area under the stress is 4 square inches, determine which material should be selected to assure safety. Assume safety factor is 2. Material A: Ultimate Tensile stress is 8000 lb/in2Material B: Ultimate Tensile stress is 5500 lb/in2
All of the following are instruments involved in changing a tire EXCEPT:
Answer:
can you please give us the option for this question
In a CNC machining operation, the has to be moved from point (5, 4) to point(7, 2)along a circular path with center at (7,2). Before starting operation, the tool is at (5, 4).The correct G and M code for this motion is
Answer: hello your question is incomplete below is the complete question
answer:
N010 GO2 X7.0 Y2.0 15.0 J2.0 ( option 1 )
Explanation:
Given that the NC machining has to be moved from point ( 5,4 ) to point ( 7,2 ) along a circular path
GO2 = circular interpolation in a clockwise path
G91 = incremental dimension
hence the correct option is :
N010 GO2 X7.0 Y2.0 15.0 J2.0
How much energy does it take to boil water for pasta? For a one-pound box of pasta
you would need four quarts of water, which requires 15.8 kJ of energy for every degree
Celsius (°C) of temperature increase. Your thermometer measures the starting
temperature as 48°F. Water boils at 212°F.
a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?
b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?
c. [2 pts] How much energy is required to heat the four quarts of water from
48°F to 212°F (boiling)?
Answer:
a. 164 °F b. 91.11 °C c. 1439.54 kJ
Explanation:
a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?
Since the starting temperature is 48°F and the final temperature which water boils is 212°F, the number of degrees Fahrenheit we would need to raise the temperature is the difference between the final temperature and the initial temperature.
So, Δ°F = 212 °F - 48 °F = 164 °F
b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?
To find the degree change in Celsius, we convert the initial and final temperature to Celsius.
°C = 5(°F - 32)/9
So, 48 °F in Celsius is
°C₁ = 5(48 - 32)/9
°C₁ = 5(16)/9
°C₁ = 80/9
°C₁ = 8.89 °C
Also, 212 °F in Celsius is
°C₂ = 5(212 - 32)/9
°C₂ = 5(180)/9
°C₂ = 5(20)
°C₂ = 100 °C
So, the number of degrees in Celsius you must raise the temperature is the temperature difference between the final and initial temperatures in Celsius.
So, Δ°C = °C₂ - °C₁ = 100 °C - 8.89 °C = 91.11 °C
c. [2 pts] How much energy is required to heat the four quarts of water from
48°F to 212°F (boiling)?
Since we require 15.8 kJ for every degree Celsius of temperature increase of the four quarts of water, that is 15.8 kJ/°C and it rises by 91.11 °C, then the amount of energy Q required is Q = amount of heat per temperature rise × temperature rise = 15.8 kJ/°C × 91.11 °C = 1439.54 kJ
A 1m3 tank containing air at 25℃ and 500kPa is connected through a valve to
another tank containing 5kg of air at 35℃ and 200kPa. Now the valve is opened,
and the entire system is allowed to reach thermal equilibrium, which is at 20℃
(Take: Ru = 8.314 kJ / kg.K).
Answer:
The right answer is "2.2099 m³".
Explanation:
Given:
Mass,
m = 5 kg
Temperature,
T = 35℃
or,
= 35 + 273
Pressure,
P = 200 kPa
Gas constant,
R = 0.2870 kj/kgK
By using the ideal gas equation,
The volume will be:
⇒ [tex]PV=mRT[/tex]
or,
⇒ [tex]V=\frac{mRT}{P}[/tex]
By substituting the values, we get
[tex]=\frac{5(0.2870)(35+273)}{200}[/tex]
[tex]=\frac{441.98}{200}[/tex]
[tex]=2.2099 \ m^3[/tex]
