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?​

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

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

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]

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%

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.

Answer:

can you translate

Explanation:

what Is that?

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.

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

Answer:

The volume for this is 29.7

Explanation:

Trust me on this I'm an expert

Answer:

is there a picture of the figure?

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.

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

Answer:

ay man ima be real, i just need the points yo

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.

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

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]

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

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:

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]

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:

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

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

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]

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

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 )

Answer:

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]