A layer of viscous fluid of constant thickness (no velocity perpenducilar to plate) flows steadily down an infinite, inclined plane. Determine, with the Navier Stokes equations, the flowrate per unit width as a function of flow height.

Answer:

The RPM to increase the recovery rate of the cells to 95% at the same flow rate is 6,333.3 RPM.

Explanation:

If the tubular centrifuge rotates at about 4,000 revolutions per minute to recover 60% of the cells, in case of wanting to recover 95% of the cells, the following calculation must be carried out to determine the required number of revolutions per minute:

60 = 4,000

95 = X

((95 x 4,000) / 60)) = X

(380,000 / 60) = X

6,333.3 = X

Therefore, as the calculation emerges, the tubular centrifuge will need to rotate at about 6,333.3 revolutions per minute to recover 95% of the cells in the same time.

Answer:

A tax is a monetary payment without the right to individual consideration, which a public law imposes on all taxable persons - including both natural and legal persons - in order to generate income. This means that taxes are public-law levies that everyone must pay to cover general financial needs who meet the criteria of tax liability, whereby the generation of income should at least be an auxiliary purpose. Taxes are usually the main source of income of a modern state. Due to the financial implications for all citizens and the complex tax legislation, taxes and other charges are an ongoing political and social issue.

I dont know I asked this to

Explanation:

Answer:

K = 96 and the design speed of the curve = 50mph

Explanation:

109+00 = 10900

Elevation = 950ft

110+77 = 11077

Elevation = 947.34ft

Station of low point = 110+50 = 11050

To get grade of curve

Gi = 947.34-950/11077-10900

= -2.66/177

= -0.015x100

= -1.5%

Locate of low point (XL)

= 11050-10900

= 150

To get the value of K

XL = |GL| x K

When we substitute values

150 = 1.5 x K

150 = 1.5K

K = 150/1.5

K = 100

The suitable and most nearest value is K = 96

Then we use the standard chart to get the design speed for K = 96

On this chart, the design speed for the curve = 50mph

Therefore K = 96 and speed = 50mph

Answer:

Updated question

The big ben clock tower in London has clocks on all four sides. If each clock has a minute hand that is 11.5 feet in length, how far does the tip of each hand travel in 52 minutes?

The distance traveled by the tip of the minute hand of the clock would be 62.59 ft

Explanation:

Let us assume the shape of the clock is circular.

the minute hand is equal to the radius = 11.5 ft

Diameter = radius x 2

Diameter = 11.5 x 2 = 23 ft

The distance traveled by the tip of the minute hand can be calculated thus;

the fraction of the circumference traveled by the minute hand would be;

52/60 = 0.8667

Circumference of the clock would be;

C = pi x d

where C is the circumference

pi is a constant

d is the diameter

C = 3.14 x 23

C = 72.22 ft

Therefore the fraction of the circumference covered by the minute hand would be;

72.22 ft x 0.8667 = 62.59 ft

Therefore the distance traveled by the tip of the minute hand of the clock would be 62.59 ft

Answer:

P1+1/2pv2/1+pgh1=P2+1/2pv2/2+pgh2

Solution:

Given that :

Volume flow is, [tex]$Q_1 = 1000 \ mm^3/s$[/tex]

So, [tex]$Q_2= \frac{1000}{100}=10 \ mm^3/s$[/tex]

Therefore, the equation of a single straight vessel is given by

[tex]$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$[/tex]    ......................(i)

So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is

[tex]$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4} $[/tex]

or [tex]$d_1=10 \ d_2$[/tex]

Now for parallel pipes

[tex]$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$[/tex]  ...........(ii)

Solving the equations (i) and (ii),

[tex]$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$[/tex]

       [tex]$=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$[/tex]

       [tex]$=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$[/tex]

       [tex]$=\frac{10^6}{10^7}$[/tex]

Therefore,

[tex]$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$[/tex]

or [tex]$H_{f_2}=10 \ H_{f_1}$[/tex]

Thus the answer is option A). 10

Answer:

hello your question has a missing part below is the missing part

Consider the string length equal to [tex]\pi[/tex]

answer : 2cos(2t) sin(2x) - 10cos(10t)sin(10x)

Explanation:

Given string length = [tex]\pi[/tex]

distorted function f(x) = 2sin(2x) - 10sin(10x)

Determine the wave formed in the string

attached below is a detailed solution of the problem

Answer:

For 31 P NMR spectra

low temperature

there is two types of 19f seen in low temperature and they are

therefore in low temperature the 31p couples with the two types of 19F seen ( [tex]b_{f} and c_{f}[/tex]to form a triplet and this couples more with [tex]a_{f}[/tex] to form a doublet. i.e. one (1) peak

High temperature

At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak

For 19 P NMR spectra

low temperature

In low temperature [tex]a_{f}, b_{f} , c_{f}[/tex] is fixed  and the environment where [tex]b_{f} and c _{f}[/tex] is the same hence a peak is formed and another peak is formed by [tex]a_{f}[/tex] that makes the number of peaks = 2 peaks

High temperature

In high temperature [tex]a_{f}, b_{f} , c_{f}[/tex]  exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all

Explanation:

For 31 P NMR spectra

low temperature

there is two types of 19f seen in low temperature and they are

therefore in low temperature the 31p couples with the two types of 19F seen ( [tex]b_{f} and c_{f}[/tex]to form a triplet and this couples more with [tex]a_{f}[/tex] to form a doublet. i.e. one (1) peak

High temperature

At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak

For 19 P NMR spectra

low temperature

In low temperature [tex]a_{f}, b_{f} , c_{f}[/tex] is fixed  and the environment where [tex]b_{f} and c _{f}[/tex] is the same hence a peak is formed and another peak is formed by [tex]a_{f}[/tex] that makes the number of peaks = 2 peaks

High temperature

In high temperature [tex]a_{f}, b_{f} , c_{f}[/tex]  exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all

Answer:

-9.5° C

Explanation:

See attachment for calculations.

On the concluding parts, from the attachment, we have that

√[(297000 * 4)/(1056)] = 297 - T(l), and solving further, we get

297 - T(l) = √(1188000/1056)

297 - T(l) = √1125

297 - T(l) = 33.5

T(l) = 297 - 33.5

T(l) = 263.5

When you convert back to °C, we have

263.5 - 273 = -9.5° C

Answer:

Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3/s and at a velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Fig. PI3-39. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction.

Step-by-step solution:

Step 1 of 5

Given data:-

The velocity of water is .

The water flow rate is.

Answer:

commercial freezing

Explanation:

smaller ice crystals are formed this causes less damage to cell membranes so the quality is less effected

Answer:

a) volume flow rate of air at the inlet is 0.0471 m³/s

b) the velocity of the air at the exit is  8.517 m/s

Explanation:

Given that;

The electrical power Input W_elec = -1400 W = -1.4 kW

Inlet temperature of air T_in = 22°C

Inlet pressure of air p_in = 100 kPa

Exit temperature T_out = 47°C

Exit area of the dyer is A_out = 60 cm²= 0.006 m²

cp = 1.007 kJ/kg·K

R = 0.287 kPa·m3/kg·K

Using mass balance

m_in = m_out = m_air

W _elec = m_air ( h_in - h_out)

we know that h = CpT

so

W _elec = m_air.Cp ( T_in - T_out)

we substitute

-1.4 = m_air.1.007 ( 22 - 47 )

-1.4 =  - m_air.25.175

m_air = -1.4 / - 25.175

m_ air = 0.0556 kg/s

a) volume flow rate of air at the inlet

we know that

m_air = P_in × V_in

now from the ideal gas equation

P_in = p_in / RT_in

we substitute our values

= (100×10³) / ((0.287×10³)(22+273))

= 100000 / 84665

P_in = 1.18 kg/m³

therefore inlet volume flowrate will be;

V_in = m_air / P_in

= 0.0556 / 1.18

= 0.0471 m³/s

the volume flow rate of air at the inlet is 0.0471 m³/s

b) velocity of the air at the exit

the mass flow rate remains unchanged across the duct

m_ air = P_in.A_in.V_in = P_out.A_out.V_out

still from the ideal gas equation

P_out = p_out/ RT_out   ( assume p_in = p_out)

P_out = (100×10³) / ((0.287×10³)(47+273))

P_out  = 1.088 kg/m³

so the exit velocity will be;

V_out = m_air / P_out.A_out

we substitute our values

V_out = 0.0556 / ( 1.088 × 0.006)

= 0.0556 / 0.006528

= 8.517 m/s

 Therefore the velocity of the air at the exit is  8.517 m/s

Answer:

51112.5 ft^3

Explanation:

Determine the volume needed along the road when we assume a 6% shrinkage

shrinkage factor = 1 - shrinkage  = 1 - 0.06 =  0.94

first we have to calculate the volume between the cross sectional areas (i.e. A1 ---- A5 ) using average end area method

Volume between A1 - A2

= (125 ft - 0 ft) * [(130 ft^2 + 140 ft^2) / 2]

 = 125 ft * 135 ft^2

= 16875 ft^3

Volume between A2 - A3

= (250 ft - 125 ft) * [(140 ft^2 + 60 ft^2) / 2]

= 125 ft * (200 ft^2 / 2)

= 12500 ft^3

Volume between A3 - A4

= (375 ft - 250 ft) * [(60 ft^2 + 110 ft^2) / 2]

= 125 ft * (170 ft^2 / 2)

= 10625 ft^3

Volume between A4 - A5

(500 ft - 375 ft) * [(110 ft^2 + 120 ft^2) / 2]

 = 125 ft * 115 ft^2

= 14375 ft3

Hence the total volume along the 500 ft road

= ∑ volumes between cross sectional areas

=  16875 ft^3 + 12500 ft^3 + 10625 ft^3 + 14375 ft^3 = 54375 ft^3

Finally the volume needed along this road is calculated as

Total volume * shrinkage factor

= 54375 * 0.94  = 51112.5 ft^3

Answer:

[tex]10.8\ \text{lb/ft^2}[/tex]

[tex]101.96\ \text{lb/ft}^2[/tex]

Explanation:

[tex]v_1[/tex] = Velocity of car = 65 mph = [tex]65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}[/tex]

[tex]\rho[/tex] = Density of air = [tex]0.00237\ \text{slug/ft}^3[/tex]

[tex]v_2=0[/tex]

[tex]P_1=0[/tex]

[tex]h_1=h_2[/tex]

From Bernoulli's law we have

[tex]P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}[/tex]

The maximum pressure on the girl's hand is [tex]10.8\ \text{lb/ft^2}[/tex]

Now [tex]v_1[/tex] = 200 mph = [tex]200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}[/tex]

[tex]P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2[/tex]

The maximum pressure on the girl's hand is [tex]101.96\ \text{lb/ft}^2[/tex]

Solution :

Sieve Size (in)                   Weight retain(g)

3                                         1.62

2                                         2.17

[tex]$1\frac{1}{2}$[/tex]                                       3.62

[tex]$\frac{3}{4}$[/tex]                                        2.27

[tex]$\frac{3}{8}$[/tex]                                        1.38

PAN                                    0.21

Given :

Sieve       weight       % wt. retain    % cumulative       % finer

size        retained                               wt. retain

No. 4        59.5            10.225%          10.225%            89.775%

No. 8        86.5            14.865%          25.090%           74.91%

No. 16       138              23.7154%        48.8054%         51.2%

No. 30      127.8           21.91%              70.7154%          29.2850%

No. 50      97               16.6695%         87.3849%         12.62%

No. 100     66.8            11.4796%         98.92%              1.08%

Pan            6.3               1.08%              100%                   0%

                581.9 gram

Effective size = percentage finer 10% ([tex]$$D_{20}[/tex])

0.149 mm, N 100, % finer 1.08

0.297, N 50 , % finer 12.62%

x  ,   10%

[tex]$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$[/tex]

[tex]$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$[/tex]

x = 0.2634 mm

Effective size, [tex]$D_{10} = 0.2643 \ mm$[/tex]

Now, N 16 (1.19 mm)  ,  51.2%

N 8 (2.38 mm)  ,  74.91%

x,  60%

[tex]$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$[/tex]

x = 1.6317 mm

[tex]$\therefore D_{60} = 1.6317 \ mm$[/tex]

Uniformity co-efficient = [tex]$\frac{D_{60}}{D_{10}}$[/tex]

   [tex]$Cu= \frac{1.6317}{0.2643}$[/tex]

Cu = 6.17

Now, fineness modulus = [tex]$\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$[/tex]

[tex]$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$[/tex]

= 4.41

which lies between No. 4  and No. 5 sieve [4.76 to 4.00]

So, fineness modulus = 4.38 mm

Answer:

a) Real Power (kW) = 1.5 kW

b) Reactive Power (kvar) is 1.1663 KVAR

c) Apparent Power (kVA) is 1.9 KVA

d) the Power Factor cos∅ is 0.7894

e) the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω

Explanation:

Given that;

V = 380v

i = 5A

P = 1500 W

determine;

a) Real Power (kW)

P = 1500W = 1.5 kW

therefore Real Power (kW) = 1.5 kW

b) Reactive Power (kvar)

p = V×i×cos∅

cos∅ = p / Vi

cos∅ = 1500 / ( 380 × 5 ) = 0.7894

∅ = cos⁻¹ (0.7894)

∅ = 37.87°

Q = VIsin∅

Q = 380 × 5 × sin( 37.87° )

Q = 1.1663 KVAR

Therefore Reactive Power (kvar) is 1.1663 KVAR

c) Apparent Power (kVA)

S = P + jQ

= ( 1500 + J 1166.3 ) VA

S = 1900 ∠ 37.87° VA

S = 1.9 KVA

Therefore Apparent Power (kVA) is 1.9 KVA

d) Power Factor

p = V×i×cos∅

cos∅ = p / Vi

cos∅ = 1500 / ( 380 × 5 ) = 0.7894

Therefore the Power Factor cos∅ is 0.7894

e) The impedance Z in polar and rectangular form

Z = 380 / ( S∠-37.87) = V/I

Z = ( 60 + j 46.647) Ω

Z = 76 ∠ 37.87° Ω

Therefore the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω

Answer:

C. The meter will display a negative sign.

Explanation:

If you use an analog voltmeter and you measure voltage with reverse polarity you will damage it. But in this case we are using a digital multimeter. This kind of multimeter is designed to be able to deal with positive and negative voltages

This question is incomplete, the complete question is;

All of the questions for this quiz involve a drone with a mass of 1.2 kg.

The drone hovers for 29 seconds at a height of 53 meters above the ground while taking a picture of a crowd. What is the vertical impulse by the lift force (combined force of the four rotors) during this time (in Newton-seconds)?

Answer: the vertical impulse is 341.388 Ns

Explanation:

Given that;

mass m = 1.2 kg

t = 29 sec

h = 53 m

vertical impulse = ?

Vertical impulse by lift force is given as;

Vertical Impulse = F × t

= mg × t

so we substitute

Vertical impulse = 1.2 ×9.81 × 29

= 341.388 Ns

therefore the vertical impulse is 341.388 Ns

Answer:

Yes, they can all be present in the same malware because each of them perform slightly differing functions.

Explanation:

Backdoor is a software which when placed into your computer will permit hackers to easily gain reentry into your computer. This can happen even after you have already patched the flaw that they have used to hack your system before.

A bot is a program that does the same task in a continuous manner akin to when you use a blender by pressing the button to blend what you have put into it.

A keylogger is a part of a hidden software that monitors and records everything you type on your computer keyboard after which it writes it onto a file, with the hopes of capturing relevant information such as your bank account number and even passwords and other sensitive means of identification.

A Spyware is somehow similar to a keylogger just that it steals information from your computer and sends it to someone else.

A root kit is a bad software that is capable of modifying the operating system or other privileged access devices in order to gain continuous access into your system for the purpose of gathering of information and/or reducing the system’s functionality.

Yes, they can all be present in the same malware because each of them perform slightly differing functions.

Answer: the mass flux of pollutant through the tube due to advection is 0.0283 mg/cm².s

Explanation:

Given that;

Diameter of tube = 3 cm, radius r = 1.5 cm

water flow is 100 cm³/s

pollutant concentration = 2 mg/L

first we find the rate of flow of pollutant

we know that

1 L = 1000 cm³

xL = 100 cm³

100Lcm³ = xL1000cm³

xL = 100/1000

xL = 1/10 L

so 100cm³ = 1/10 L

now pollutant concentration in 100 cm³ = 1/10L × 2mg/L = 0.2 mg

Rate of flow of pollutant = 0.2 mg/s

Mass flux density is the pollutant mass per unit time per unit area

so Area of tube = πr² = 3.14 × 1.5² = 7.065 cm²

So

Mass flux = 0.2 / 7.065

Mass flux = 0.0283 mg/cm².s

Therefore, the mass flux of pollutant through the tube due to advection is 0.0283 mg/cm².s

Answer:

a. 4[tex]\mu m[/tex]

b. 1 m

Explanation:

According to the question, the data is as follows

The Density of water at 20 degrees celcius is 1000 kg/m^3

Viscosity is 0.001kg/m/.s

Velocity V = 25 cm/s

V = 0.25 m/s

Now

a. The creeping motion is

As we know that

Reynold Number = (Density of water × V × d) ÷ (Viscosity)

1 = (1,000 × 0.25 × d) ÷ 0.0001

d = (1 × 0.001) ÷ (1,000 × 0.25)

= 4E - 06^m

= 4[tex]\mu m[/tex]

b. Now the sphere diameter is

Reynold Number = (Density of water × V × d) ÷ (Viscosity)

250,000 = (1,000 × 0.25 × d) ÷ 0.0001

d = (250,000 × 0.001) ÷ (1,000 × 0.25)

= 1 m

Answer:

Computer programming is where you learn and see how computers work. People do this for a living as a job, if you get really good at it you will soon be able to program/ create a computer.

Explanation:

Hope dis helps! :)

Answer:

a) the  rate of cooling provided is 18604.8 Btu/h

b) the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm

c) the second-law efficiency of the evaporator is 37.39%

Explanation:

Given that;

Temperature of sink TL = 50°F = 510 R

Temperature at evaporator inlet TI = 10°F = 470 R

mass flow rate m" = 0.08 lbm/s

quality of refrigerant at evaporator inlet x1 = 0.3

quality of refrigerant at evaporator exit x2 = 1.0

T₀ = 77°F = 537 R

h1 = 107.5 Btu/lbm

s1 = 0.2851 Btu/lbm·R,

h2 = 172.1 Btu/ lbm,

s2 = 0.4225 Btu/lbm·R.

a) rate of cooling provided, in Btu/h.

QL = m"( h2 - h1)

we substitute

QL = 0.08( 172.1 - 107.5

= 0.08 × 64.6

= 5.168 Btu/s

we convert to Btu/h

5.168 × 60 × 60

QL = 18604.8 Btu/h

Therefore the  rate of cooling provided is 18604.8 Btu/h

b) The rate of exergy destruction in the evaporator

Entropy generation can be expressed as;

S_gen = m"(s2 - s1) - QL/TL

so we substitute

S_gen = 0.08( 0.4225 -  0.2851  ) - 5.168 / 510

= 0.010992 - 0.01013

S_gen = 0.00086 Btu/ibm.R

now the energy destroyed expressed as;

X_dest = T₀ × S_gen

so

X_dest =  537 × 0.00086

X_dest = 0.46 Btu/Ibm

Therefore the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm

c)  The second-law efficiency of the evaporator.

Energy expended is expressed as;

X_exp = m"(h1 - h2) - m"T₀(s1 - s2)

we substitute

= 0.08( 107.5 - 172.1 ) - [0.08 × 537 ( 0.2851 - 0.4225 )

= -5.168 - [ - 5.9027)

= -5.168 + 5.9027

= 0.7347 Btu/s

Now second law efficiency is expressed as;

nH = 1 - (X_dest / X_esp)

= 1 - ( 0.46 / 0.7347 )

= 1 - 0.6261

= 0.3739

nH = 37.39%

Therefore the second-law efficiency of the evaporator is 37.39%

Answer:

common fate

Explanation:

The gestalt effect may be defined as the ability of our brain to generate the whole forms from the groupings of lines, points, curves and shapes. Gestalt theory lays emphasis on the fact that whole of anything is much greater than the parts.

Some of the principles of Gestalt theory are proximity, similarity, closure, symmetry & order, figure or ground and common fate.

Common fate : According to this principle, people will tend to group things together which are pointed towards or moving in a same direction. It is the perception of the people that objects moving together belongs together.

dutile is the correct answer

Answer:

The friction factor is 0.303.

Explanation:

The flow velocity ([tex]v[/tex]), measured in meters per second, is determined by the following expression:

[tex]v = \frac{4\cdot \dot V}{\pi \cdot D^{2}}[/tex] (1)

Where:

[tex]\dot V[/tex] - Flow rate, measured in cubic meters per second.

[tex]D[/tex] - Diameter, measured in meters.

If we know that [tex]\dot V = 0.01\,\frac{m^{3}}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the flow velocity is:

[tex]v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}[/tex]

[tex]v \approx 5.093\,\frac{m}{s}[/tex]

The density and dinamic viscosity of the glycerin at 20 ºC are [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex] and [tex]\mu = 1.5\,\frac{kg}{m\cdot s}[/tex], then the Reynolds number ([tex]Re[/tex]), dimensionless, which is used to define the flow regime of the fluid, is used:

[tex]Re = \frac{\rho\cdot v \cdot D}{\mu}[/tex] (2)

If we know that [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex], [tex]\mu = 1.519\,\frac{kg}{m\cdot s}[/tex], [tex]v \approx 5.093\,\frac{m}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the Reynolds number is:

[tex]Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }[/tex]

[tex]Re = 211.230[/tex]

A pipeline is in turbulent flow when [tex]Re > 4000[/tex], otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor ([tex]f[/tex]), dimensionless, is determined by the following expression:

[tex]f = \frac{64}{Re}[/tex]

If we get that  [tex]Re = 211.230[/tex], then the friction factor is:

[tex]f = \frac{64}{211.230}[/tex]

[tex]f = 0.303[/tex]

The friction factor is 0.303.

Answer:

0.75 kg

Explanation:

c = Damping coefficient = 3 kg/s

x = Displacement of spring = 0.5 m

F = Force = 1.5 N

From Hooke's law we get

[tex]F=kx\\\Rightarrow k=\dfrac{F}{x}\\\Rightarrow k=\dfrac{1.5}{0.5}\\\Rightarrow k=3\ \text{N/m}[/tex]

In the case of critical damping we have the relation

[tex]c^2-4mk=0\\\Rightarrow m=\dfrac{c^2}{4k}\\\Rightarrow m=\dfrac{3^2}{4\times 3}\\\Rightarrow m=0.75\ \text{kg}[/tex]

Mass that would produce critical damping is 0.75 kg.

0.75 kg is the mass that would produce critical damping. As spring with an m-kg mass and a damping constant 3 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons.

A change in time and position is referred to as an object's velocity. When there is no movement of the object, the velocity of the object is said to be 0.

For any body in planar motion, the velocity is always instantaneously 0 at some point in the plane of motion (if it were rigidly connected to the body). This place is known as the instantaneous center of zero velocity, or IC.

Example: The gravitational pull of the earth pushes the ball away from the thrower when a ball is thrown upwards on Earth at a constant speed. The speed of the ball increases until it reaches its maximum, at which point it starts to plummet.

Thus, it is 0.75 kg.

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