3. According to the drag equation the velocity of an object moving through a fluid can be modeled by the equation -- -ky- where k is a constant.
(a) Find the general solution to this equation.
(b) An object moving through the water has an initial velocity of 40 m/s. Two seconds later, the velocity has decreased to 30 m/s. What will the velocity be after ten seconds?
Answers
A find the ganeral solution to this equation
An object moving through the water has an initial velocity of 40 m/s. Two seconds later, the velocity has decreased to 30 m/s. The velocity after ten seconds is 0.
What is velocity?Velocity is defined as the speed at which an object's position changes in relation to time and a frame of reference. Speed is the rate at which an object travels along a path over time, whereas velocity is the speed and direction of an item's motion. In other words, speed is a scalar value, but velocity is a vector.
As given
Initial velocity = 40 m / sec
Velocity after 2 seconds = 30 m / sec
So the velocity after 10 seconds will be = 0 m / sec
That is the object will stop moving.
Thus. an object moving through the water has an initial velocity of 40 m/s. Two seconds later, the velocity has decreased to 30 m/s. The velocity after ten seconds is 0.
To learn more about velocity, refer to the link below:
https://brainly.com/question/28738284
#SPJ2
Your question is incomplete, but probably your complete question was
Related Questions
Request for proposal (RFP) is a type of document that contains the information and proposals mostly through the bidding process. This document is regarding the valuable assets, services, entity, commodity, etc.
Answers
Answer:
Answer to the following is as follows;
Explanation:
A request for proposal is a documentation that invites prospective contractors to submit business opportunities to an agency or corporation interested in procuring a commodities, product, or valuable resource through a bid procedure.
A request for proposal (RFP) is a commercial document that introduces a project, defines it, and invites eligible contractors to compete on its completion.
In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles of the gas are present in the tank? What is the molecular weight of the gas? Assuming that the gas to be a pure element can you identify it?
Answers
Answer:
The molecular weight will be "28.12 g/mol".
Explanation:
The given values are:
Pressure,
P = 10 atm
= [tex]10\times 101325 \ Pa[/tex]
= [tex]1013250 \ Pa[/tex]
Temperature,
T = 298 K
Mass,
m = 11.5 Kg
Volume,
V = 1000 r
= [tex]1 \ m^3[/tex]
R = 8.3145 J/mol K
Now,
By using the ideal gas law, we get
⇒ [tex]PV=nRT[/tex]
o,
⇒ [tex]n=\frac{PV}{RT}[/tex]
By substituting the values, we get
[tex]=\frac{1013250\times 1}{8.3145\times 298}[/tex]
[tex]=408.94 \ moles[/tex]
As we know,
⇒ [tex]Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}[/tex]
or,
⇒ [tex]MW=\frac{m}{n}[/tex]
[tex]=\frac{11.5}{408.94}[/tex]
[tex]=0.02812 \ Kg/mol[/tex]
[tex]=28.12 \ g/mol[/tex]
1. What is the maximum value of the linear density in a crystalline solid (linear density defined as the fraction of the line length occupied by atoms, assumed as spheres and only counted it their center is on the line)?
2. What family of directions has the highest linear density in the FCC system?
3. What family of directions has the highest linear density in the BCC system?
4. What family of planes has the highest planar density in the FCC system?
5. What family of planes has the highest planar density in the BCC system?
6. What family of planes has the highest planar density in the HCP sytem?
Answers
Just because I seen someone else ask but they didn't have enough information.
If a filesystem has a block size of 4096 bytes, this means that a file comprised of only one byte will still use 4096 bytes of storage. A file made up of 4097 bytes will use 4096*2=8192 bytes of storage. Knowing this, can you fill in the gaps in the calculate_storage function below, which calculates the total number of bytes needed to store a file of a given size?
Answers
Answer:
Following are the program to the given question:
def calculate_storage(filesize):#definging a method calculate_storage that takes filesize as a parameter
block_size = 4096#definging block_size that holds value
full_blocks = filesize//block_size#definging full_blocks that divides the value and hold integer part
partial_block_remainder = filesize%block_size#definging partial_block_remainder that holds remainder value
if partial_block_remainder > 0:#definging if that compare the value
return block_size*full_blocks+block_size#return value
return block_size*full_blocks#return value
print(calculate_storage(1)) # calling method by passing value
print(calculate_storage(4096)) # calling method by passing value
print(calculate_storage(4097)) # calling method by passing value
Output:
4096
4096
8192
Explanation:
In this code, a method "calculate_storage" is declared that holds a value "filesize" in its parameters, inside the method "block_size" is declared that holds an integer value, and defines "full_blocks and partial_block_remainder" variable that holds the quotient and remainder value and use it to check its value and return its calculated value. Outside the method, three print method is declared that calls the method and prints its return value.
A stream of ethylene glycol vapor at its normal boiling point and 1atm flowing at a rate of 175 kg/min is to be condensed at constant pressure. The product stream from the condenser is liquid g lycol at the condensation temperature.
a. Calculate the rate at which heat must be transferred from the condenser (kW).
b. If heat were transferred at a lower rate than that calculated in part (A), what would the state of the product stream be? (Dedu ce as much as you can about the phase and the temperature of the stream.)
c. If heat were transferred at a higher rate than that calculated in part (A), what could you deduce about the state of the product stream?
Answers
Answer: hello attached below is the question properly written
a) 2670 Kw
b) product will be made up of vapor and liquid
c) Product will be a super cooled liquid
Explanation:
mass Flow rate ( m ) = 175 kg/min
pressure = 1 atm
molecular weight of ethylene glycol ( mw ) = 62.07 g/mol
enthalpy of vaporization ( ΔHv ) = 56.9 KJ/mol
Using values from the table 8.1 related to the question
a) Determine the rate at which heat must be transferred from condenser
Using values from the table 8.1 related to the question
ΔH = 2670 Kw
b) If heat is transferred at a lower temperature the product will be made up of vapor and liquid
c) If heat was transferred at a higher temperature the product will be a super cooled liquid
A uniform plane electromagnetic wave propagates in a lossless dielectric medium of infinite extent. The electric field in the wave has the instantaneous expression
E(r,t) = (ix √3 - iz) 2 sin(2π.10^8t + 2πx/3 + 2nz/√3 + 30 ), V/m.
Find:
a. iE, the unit vector in the direction of the wave electric field
b. the amplitude Eo of the wave
c. the wavelength of the wave
d. ik, the unit vector in the direction of propagation
Answers
Answer:
Explanation:
From the information given:
The instantaneous expression of the electric field in the wave is:
[tex]E(r,t)= (i_x \sqrt{3} -i_z) 2 \ sin (2 \pi*10^8t + 2 \pi x/3+2 \pi z /\sqrt{3} + 30 ^0) , \ V/m[/tex]
To determine the unit vector in line with the wave electric field, we take the first term in E(r,t) for [tex]I_E^\to[/tex] as:
[tex]I_E^\to = i_x \sqrt{3}-i_z \\ \\ I_E^\to = \dfrac{i_x \sqrt{3}-i_z}{\sqrt{3 +1}} \\ \\ \mathbf{ I_E = \dfrac{i_x\sqrt{3} -i_z}{2}}[/tex]
The amplitude is denoted by the numerical value after the first term, which is:
[tex]\mathbf{E_o = 2}[/tex]
The wavelength can be determined by using the expression:
[tex]\beta =\dfrac{2 \pi}{\lambda }[/tex]
from the given instantaneous expression:
[tex]\beta = \dfrac{2 \pi}{3}x+\dfrac{2 \pi}{\sqrt{3}}z[/tex]
[tex]\beta = \sqrt{\dfrac{2 \pi}{(3)^2}+\dfrac{(2 \pi}{(\sqrt{3})^2}}[/tex]
[tex]\beta = \sqrt{\dfrac{2 \pi}{9}+\dfrac{2 \pi}{{3}}}[/tex]
Factorizing 2π
[tex]\beta =2 \pi \sqrt{\dfrac{1}{9}+\dfrac{1}{{3}}}[/tex]
[tex]\beta =2 \pi \sqrt{\dfrac{9+3}{9*3}}}[/tex]
[tex]\beta =2 \pi \sqrt{\dfrac{12}{27}}}[/tex]
[tex]\beta =2 \pi \sqrt{\dfrac{4*3}{9*3}}}[/tex]
[tex]\beta =2 \pi \sqrt{\dfrac{4}{9}}}[/tex]
[tex]\beta =2 \pi\times {\dfrac{2}{3}}}[/tex]
recall from the expression using in calculating wavelength:
[tex]\beta =\dfrac{2 \pi}{\lambda }[/tex]
∴
equating both together, we have:
[tex]\dfrac{2 \pi}{\lambda }= 2 \pi\times {\dfrac{2}{3}}}[/tex]
[tex]\lambda = \dfrac{3}{2}[/tex]
λ = 1.5 m
In line with the wave direction; unit vector [tex]i_k[/tex] can be computed as follows:
[tex]i_k = - [ \beta_1x +\beta_2z]/\beta[/tex]
where;
[tex]\beta_1 = \dfrac{2 \pi }{3} \ ; \ \beta_2 = \dfrac{2 \pi }{\sqrt{3}} \ ; \ \beta = \dfrac{2 \pi \times 2}{3} ;[/tex]
∴
[tex]i_k = - \Big[\dfrac{2 \pi}{3}x + \dfrac{2 \pi}{\sqrt{3}} z\Big]\times \dfrac{1}{\dfrac{2 \pi *2}{3}}[/tex]
[tex]i_k = - \Big[\dfrac{x}{2} + \sqrt\dfrac{{3}}{4}} z\Big][/tex]
[tex]i_k = - \Big[\dfrac{1}{2}x + \sqrt{\dfrac{3}{4} }z\Big][/tex]
[tex]\mathbf{i_k = - \Big[0.5x +0.86 z\Big]}[/tex]
The steps for proper studing for the exam. -1 (use: First, Then, Next, After that, Finally)
Answers
Answer:
here is your answer
Explanation:
1. First observe the syllabus for all subjects.
2. Then gather all your books
3. For a particular exam read the book related to it ( if possible read early in the morning.
4. After that write the things you learnt by reading.
5. Finally give your exams by giving it all
HOPE IT HELPS......
OM NAMO SHIVAYE
The AGC control voltage: ___________
a. varies as the signal strength of the received signal varies.
b. a negative feedback voltage.
c. is actually the dc voltage component produced by the mixing action in the AM demodulator stage.
d. is produced by an RC circuit having a much larger time constant than that of the detector.
e. all of the above
Answers
Answer:
The AGC circuit operates with an input voltage range of 60 dB (5 mV p-p to 5 V p-p), with a fixed output voltage of 250 mV p-p.
Explanation:
Ma puteti ajuta cu un argument de 2 pagini despre inlocuirea garniturii de etansare de pe pistonul etrierului de franare la un autoturism ?
Answers
Answer:
can you translate
Explanation:
what Is that?
An alternating copolymer is known to have a number-average molecular weight of 100,000 g/mol and a degree of polymerization of 2210. If one of the repeat units is ethylene, which of styrene, propylene, tetrafluoroethylene, and vinyl chloride is the other repeat unit?
a. vinyl chloride
b. propylene
c. tetrafluoroethylene
d. styrene
Answers
Answer:
d. Styrene
Explanation:
An alternating copolymer repeat unit types if the other is Styrene. The equation to calculate m is :
m = Mn / Dp
m = 100,000 / 2210 = 45.25g/ mol
alternating copolymer is chain fraction than of each repeat unit type then chain fraction is styrene and unknown repeat units.
In the figure below, this “double” nozzle discharges water (at 10°C, density= 1000 kg/m3) into the atmosphere at a rate of 0.50 m3/s. The pressure at the inlet is to be 315612 Pa. If the nozzle is lying in a horizontal plane. Jet A is 10 cm in diameter, jet B is 12 cm in diameter, and the pipe (1) is 30 cm in diameter. The x-component of force (Rx) acting through the flange bolts is required to hold the nozzle in place is:
Answers
Solution :
Given data :
p = 315612 Pa
[tex]$V_1=7.07 \ m/sec$[/tex]
At exit of B,
p = [tex]$P_{atm}$[/tex]
[tex]$V_B = 26.1 \ m/sec$[/tex]
At exit of A,
[tex]p=P_{atm}[/tex]
[tex]$V_{A} = 26.1 \ m/s$[/tex]
We need to determine X component of force ([tex]$R_x$[/tex]) to hold in its place.
From figure,
[tex]$\sum F_x = m_0'V_{0x} - m_iV_{ix} $[/tex]
[tex]$=F_x+P_1A_1\sin 30=-mVA-mV_1 \sin 30$[/tex]
[tex]$=F_x=-pA_1\sin 30-m_AV_AA-m_B \sin30$[/tex]
Substitute all the values,
[tex]$=F_x=[-315612 \times \frac{\pi}{4}(0.3)^2 \sin 30]-[26.1 \times 1000 \times 26.1 \frac{\pi}{4}(0.1)^2]-[7.07 \times 1000\times 0.5 \sin 30]$[/tex][tex]$=F_x = -11154.64-5350.21-1767.28$[/tex]
[tex]$F_x = -18.2733 \ kN$[/tex]
Therefore, the force required to hold the nozzle in its place along horizontal direction.
[tex]$F_x = -18.2733 \ kN$[/tex]
Draw the logic circuit for each of the following. For each gate, determine if it generates either EVEN or ODD parity bit and finds the output for the given input data: (Remember: A XOR generates EVEN parity bit. A XNOR generates Odd Parity bit, whatever how many inputs they have.)
Data Inputs Which kind parity bit can it generate?
4-input XOR, input data-1001 Even Parity Bit ODD Parity Bit
5-input XOR, input data-10010
6-input XOR, input data-101001
7-input XOR, input data 1011011
Answers
Answer:
a) 4-input XOR, input data-1001 = 0 Even parity Bit
b) 5-input XOR, input data-10010 = 0 Even parity Bit
c) 6-input XOR, input data-101001 = 1 Even parity Bit
d) 7-input XOR, input data 1011011 = 1 Even parity Bit
Explanation:
a) 4-input XOR, input data-1001 ; generates 0 Even parity Bit
b) 5-input XOR, input data-10010 ; generates 0 Even parity Bit
c) 6-input XOR, input data-101001 ; generates 1 Even parity Bit
d) 7-input XOR, input data 1011011 ; generates 1 Even parity Bit
Attached below is the Logic circuits of the data inputs
The pressure gage on a 2.5-m3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank (mass in kg) if the temperature is 28°C and the atmospheric pressure is 97 kPa.
Answers
Answer:
[tex]n=5.36kg[/tex]
Explanation:
From the question we are told that:
Volume [tex]V=2.5m^3[/tex]
Pressure[tex]\rho=500Kpa[/tex]
Temperature [tex]T=28^o[/tex]
Atmospheric pressure [tex]\rho_{atm} =97 kPa.[/tex]
Generally the equation for an Ideal gas is mathematically given by
[tex]PV=nRT[/tex]
Therefore
[tex]n=\frac{500*2.5}{8.314*28}[/tex]
[tex]n=5.36kg[/tex]
It is essential to wait until the end of the project to check if the sponsor/customer requirements and expectations have been met regarding the quality of the project deliverables.
T/F
Answers
Answer:
False.
Explanation:
Project management can be defined as the process of designing, planning, developing, leading and execution of a project plan or activities using a set of skills, tools, knowledge, techniques and experience to achieve the set goals and objectives of creating a unique product or service.
Generally, projects are considered to be temporary because they usually have a start-time and an end-time to complete, execute or implement the project plan.
The fundamentals of Project Management includes;
1. Project initiation.
2. Project planning.
3. Project execution.
4. Monitoring and controlling of the project.
5. Adapting and closure of project.
Basically, the specifications of a project or manufacturing process outlines the minimum requirements and quality that are acceptable. Thus, it must be adhered to strictly in order to achieve a successful and desired outcome.
As a rule, it is essential to check at every stage of a project if the sponsor or customer requirements and expectations have been met regarding the quality of the project deliverables.
When a voltage (v=353 sin (251t+30) is applied to two elements impedance a current (i =7.07 cos 251 t) is passing. Find the nature and the value of the elements and the circuit power
Answers
Answer:
A.C. voltage, V= V0 sin ωt As,t = πω = 12.2πω = 12T, therefore, first half cycle (T/2). Hence, average value of AC voltage, Eav = 2V0π.A copper block receives heat from two different sources: 5 kW from a source at 1500 K and 3 kW from a source at 1000 K. It loses heat to atmosphere at 300 K. Assuming the block to be at steady state, determine (a) the net rate of heat transfer in kW; (b) the rate of entropy generation in the system's universe
Answers
Answer:
a) Zero
b) the rate of entropy generation in the system's universe = ds/dt = 0.2603 KW/K
Explanation:
a) In steady state
Net rate of Heat transfer = net rate of heat gain - net rate of heat lost
Hence, the rate of heat transfer = 0
b) In steady state, entropy generated
ds/dt = - [ Qgain/Th1 + Qgain/Th2 - Qlost/300 K]
Substituting the given values, we get –
ds/dt = -[5/1500 + 3/1000 – (5+3)/300]
ds/dt = - [0.0033 + 0.003 -0.2666]
ds/dt = 0.2603 KW/K
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
Answers
78950W the answer
Explanation:
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
An L2 steel strap having a thickness of 0.125 in. and a width of 2 in. is bent into a circular arc of radius 600 in. Determine the maximum bending stress in the strap.
Answers
Answer:
the maximum bending stress in the strap is 3.02 ksi
Explanation:
Given the data in the question;
steel strap thickness = 0.125 in
width = 2 in
circular arc radius = 600 in
we know that, standard value of modulus of elasticity of L2 steel is; E = 29 × 10³ ksi;
Now, using simple theory of bending
1/p = M/EI
solve for M
Mp = EI
M = EI / p ----- let this be equation 1
The maximum bending stress in the strap is;
σ = Mc / I -------let this be equation 2
substitute equation 1 into 2
σ = ( EI / p)c / I
σ = ( c/p )E
so we substitute in our values
σ = ( (0.125/2) / 600 )29 × 10³
σ = 0.00010416666 × 29 × 10³
σ = 3.02 ksi
Therefore, the maximum bending stress in the strap is 3.02 ksi
A circuit diagram for a lighting circuit is shown in Figure 6.
Figure 6
230 V AC
A
RL1
+
B T 12 V
04
4
Explain the function of the relay RL1 in the lighting circuit shown in Figure 6.
[2 marks)
Answers
Answer:
is there a picture of the figure?
Problem
In the clevis shown in Fig. find the minimum bolt diameter and the minimum thickness of each yoke that will support a load P= 14 kips without exceeding a shearing stress of 12 ksi and a bearing stress of 20 ksi
Answers
Answer:
In the clevis shown in Fig. 1-11b, find the minimum bolt diameter and the minimum thickness of each yoke that will support a load P = 14 kips without exceeding a shearing stress of 12 ksi and a bearing stress of 20 ksi.
127-clevis-double-shear-bolt.gif
Solution 127
Hide Click here to show or hide the solution
127-fbd-clevis-double-shear-bolt.gifFor shearing of rivets (double shear)
P=τA
14=12[2(14πd2)]
d=0.8618in → diameter of bolt answer
For bearing of yoke:
P=σbAb
14=20[2(0.8618t)]
t=0.4061in → thickness of yoke answer
The calculated value of the thermal conductivity of the carbon nano tube was found as: KCN = 3113 W/m-K, however, the theoretical value of the thermal conductivity of the wire is actually: K = 4500 W/m-K and the island separation is 5 μm (this is the actual spacing between the two islands). The difference between the measured and theoretical values is due to the contact resistance between the nano tube and the islands in the experiment.
Required:
a. Calculate the thermal contact resistance (Rtd) that exists between the carbon nano tube and the top surfaces of the heated and sensing islands.
b. Using the value of thermal contact resistance calculated in part A, calculate the fraction of the total resistance between the heated and sensing islands that is due to the thermal contact resistances for island separation distance of 5, 10, 15, and 20 μm.
Answers
Answer:
a) 1,607,973.9 K/W
b)
i) 0.3082 = 30.82%
ii) 0.1821 = 18.21%
iii) 0.1293 = 12.93%
iv) 0.1002 = 10.02%
Explanation:
Value of thermal conductivity ( calculated value ) KCN = 3113 W/m-k
Thermal conductivity ( theoretical value ) K = 4500 W/m-k
Island separation = 5 μm
a) Determine the thermal contact resistance
Resistance due to contact between carbon nano tube and top surfaces can be determined using the relation below
( I / A*K ) + 2Rc = ( l / A*KCN ) ------- ( 1 )
where ; I = 5 * 10^-6 m
A = π * ( 14 * 10^-9 )^2 m^2 = 153.93 * 10^-18 , K = 4500 , KCN = 3113
input values into equation 1 above
hence Rc = 1,607,973.9 K/W
b) Determine fraction of total resistance between heated and sensing
fraction of total resistance ; f1 = [tex]\frac{2 Rc}{I/KA + 2Rc}[/tex]
where : Rc = 1607973.9, K = 4500, A = 153.93 * 10^-18 ,
i) for I = 5 * 10^-6 m
fraction = 0.3082 = 30.82%
ii) for I = 10 * 10^-6 m
fraction = 0.1821 = 18.21%
iii) for I = 15 * 10^-6 m
fraction = 0.1293 = 12.93%
iv) for I = 20*10^-6
fraction = 0.1002 = 10.02%
The sample calculation for iron oxide in the IDEAS section of this experiment used known atomic weights to calculate an empirical formula. However, early chemists did not have any references in which they could look up atomic weights. Instead, they guessed at the formulas of compounds and measured the percent compositions of elements in compounds in order to calculate atomic weights. Calculate an atomic weight for iron using the hypothetical formula Fe101 and the composition data given in the example in the IDEAS section. You may assume the atomic weight of oxygen is known from other sources to be 16 amu.
Answers
Answer:
37.33 grams
Explanation:
The missing information embedded in the idea section is attached in the image below:
The aim of this question is to determine the atomic wt. of Iron (Fe) from the hypothetic formula:
Fe₁O₁
Here, we know that the mole ratio can be written as:
[tex]\dfrac{O}{Fe}=\dfrac{1}{1}[/tex]
Suppose we assume that the atomic wt. of Fe = β(unknown)???
Then the grams of O and Fe that is contained in Fe₁O₁ can be expressed as:
For O:
1 × 16 grams of Oxygen = 16 grams of O
For Fe:
1 × β grams of Fe = β grams of Fe
Now, let's take a look at the idea experiment, the mole solution can be computed as:
[tex]\dfrac{O}{Fe} = \dfrac{3}{2} \\ \\ \text{It implies that} \implies \dfrac{(3\times 16) \text{grams of O}}{(2 \times 56 ) \ \text{grams of Fe}}[/tex]
Equating both expressions above, we have:
[tex]\implies \dfrac{16}{ \beta} = \dfrac{3\times 16}{2\times 56}[/tex]
[tex]{ \beta} = \dfrac{(2\times 56)\times 16}{ 3\times 16}[/tex]
[tex]\mathbf{{ \beta} = 37.33 \ grams}[/tex]
An assembly line has 3 fail safe sensors and one emergency shutdown switch.The line should keep moving unless any of the following conditions arise:
(1) If the emergency switch is pressed
(2) If the senor1 and sensor2 are activated at the same time.
(3) If sensor 2 and sensor3 are activated at the same time.
(4) If all the sensors are activated at the same time
Suppose a combinational circuit for above case is to be implemented only with NAND Gates. How many minimum number of 2 input NAND gates are required.
Answers
Answer:
1 NAND gate
Explanation:
The minimum number of 2 input NAND gates that can be used to implement the combinational circuit = 1
The only true combinations conditions that can produce a false result ( i.e. condition/result different from the expected result as stated in the question )
Sensor 2 activated + Emergency switch pressed = False ( Line will keep moving )
A confined aquifer with a transmissivity of 300 m2/day and a storativity of 0.0005 and a well radius of 0.3 m. Find the drawdown in the well at 100 days if the following pumping schedule is followed after a long period of time of no pumping.
Period
1 2 3 4
Time (days) 0-20 20-50 50-90 90-100
Q (m3/day) 500 300 800 0
Answers
Answer:
8.4627 m
Explanation:
Transmissivity( T ) = 300 m^2/day
Storativity( S ) = 0.0005
well radius ( r ) = 0.3m
Determine the drawdown in well at 100 days
Drawdown at 100 days = ∑ Drawdown at various period
We will use the equation : S = Q / U*π*T [ -0.5772 - In U ] ----- ( 1 )
where : Q = discharge , T = transmissivity
S = drawdown ,
U = r^2*s / 4*T*t --- ( 2 )
r = well radius , S = Storativity, t = time period
i) During 0-20
U1 = r^2*s / u*π*t = 1.875 * 10^-9
Input values into equation 1
S1 = 2.5885
ii) During 20-50
U2 = r^2*s / 4*π*t = 0.3^2 * 30 / u * 300 * 30 = 1.25 * 10^-9
input values into equation 1
S2 = 1.5854 m
iii) During 50 -90
U3 = r^2*s / 4*π*t = 9.375 * 10^-10
input values into equation 1
S3 = 4.2888 m
iv) During 90-100
U4 = 0
s4 = 0
Drawdown at 100 days = ∑ Drawdowns at various period
= s1 + s2 + s3 + s4 = 2.5885 + 1.5854 + 4.2888 + 0
= 8.4627 m
find the volume of the pond with the following dimension length 40m breadth 10m height 1.2m depth 0.9m express in both meters and feet
Answers
Answer:
The volume for this is 29.7
Explanation:
Trust me on this I'm an expert
làm giúp tôi hệ thống truyền lực trên xe toyota
Answers
Answer:
ay man ima be real, i just need the points yo
To convert from the U.S. Customary (FPS) system of units to the SI system of units. A first-year engineering student records three separate measurements as 653 lb, 69.0 mi/h, and 297(106)ft2. Suppose this engineering student has to turn in the results, but the professor only accepts results given in SI units.
Required:
What is the area measurement, 293 (106) ft^2, in SI units?
Answers
This question is incomplete, the complete question is;
To convert from the U.S. Customary (FPS) system of units to the SI system of units. A first-year engineering student records three separate measurements as 653 lb, 69.0 mi/h, and 293 × 10⁶ ft². Suppose this engineering student has to turn in the results, but the professor only accepts results given in SI units.
Required:
What is the area measurement, 293 × 10⁶ ft², in SI units?
293 × 10⁶ ft² = ?km²
Answer:
the area measurement is 27.221 km²
Explanation:
Given the data in the question;
What is the area measurement, 293 × 10⁶ ft², in SI units
we are to the result of the measured area from ft² to km²
we know that;
1 meter = 3.2808 ft
1 km = 1000 m
1 ft = (1 / 3.2808)m
1 m = ( 1/1000 ) km
since our measured are is 293 × 10⁶ ft²
hence
A = 293 × 10⁶ × [ (1 / 3.2808)m ]²
A = 27221252.74 m²
A = 27221252.74 × [ ( 1/1000 ) km ]²
A = 27.221 km²
Therefore, the area measurement is 27.221 km²
Which of the following conditions would completely shut down a circuit
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quy trình sản xuất bao bì plastic dạng túi
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a) Volume Fraction of glass fiber and epoxy matrix
b) The density of the composite
c) The ratio of the load carried by the fiber to that carried by
the matrix.
Note
The density of epoxy =1.3 gm/cm3
The density of glass =2.2 gm/cm3 E of epoxy = 2.75 GPa Eofglass =380GPa
From what year did 3.5G enter Vietnam?
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Answer:
From what year did 3.5G enter Vietnam?
Explanation:
Vietnam does not have 3.5G network. MobiFone's first trial with 3.5G technology on this band was able to cover the whole waters of Vung Tau and Con Dao in August 2014.
Answer: 2011
For more info
Search “Making 3G Work in Vietnam Presented by: Marc Daniel Einstein Senior Industry Analyst October 5th, 2008 Hanoi, Vietnam.”