Consider the string length equal to 7. This string is distorted by a function f (x) = 2 sin(2x) - 10sin(10x). What is the wave formed in this string? a. In=12cos (nit ) sin(max) b. 2cos(2t)sin (2x) - 10cos(10t ) sin(10x) c. n 2 sin 2x e' – 10sin 10x e

Answer:

pascals is this = 101352.972 Pa

Explanation:

given data

Atmosphere pressure at sea level = 14.7 lb/in²

we convert 1 Pa = 1 N/m² to lb/ft²

so we convert here  14.7 lb/in² to pascals

we know that 1 lb/ft² = 47.990172 N/m²

so

1 lb/ft² × ft²/(12in)²  = 47.990172  ×  144 N/m²

it will be simplyfy

1 lb/ft²  = 6894.76 N/m²  

so

14.7 lb/in² = 14.7 × 6894.76 N/m²  

14.7 lb/in² =  101352.972 Pa

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.

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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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:

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

This question is incomplete, the complete question is;

Find the magnitude of the steady-state response of the system whose system model is given by

dx(t)/dt + x(t) = f(t)

where f(t) = 2cos8t.  Keep 3 significant figures

Answer: The steady state output x(t) = 0.2481 cos( 8t - 45° )

Explanation:

Given that;

dx(t)/dt + x(t) = f(t)  where f(t) = 2cos8t

dx(t)/dt + x(t) = f(t)

we apply Laplace transformation on both sides

SX(s) + x(s) = f(s)

(S + 1)x(s) = f(s)

f(s) / x(s) = S + 1

x(s) / f(s) = 1 / (S + 1)

Therefore

transfer function = H(s) = x(s)/f(s) = 1/(S+1)

f(t) = 2cos8t →   [ 1 / ( S + 1 ) ]   →  x(t) = Acos(8t - ∅ )

A = Magnitude of steady state output

S = jw

S = j8

so

A = 2 × 1 / √( 8² + 1 ) = 2 / √ (64 + 1 )

A = 2/√65 = 0.2481

∅ = tan⁻¹( 1/1) = 45°

therefore The steady state output x(t) = 0.2481 cos( 8t - 45° )

Answer:

The Answer is C!

Explanation:

dutile is the correct answer

Answer:

commercial freezing

Explanation:

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

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: 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: the discharge observed at a v-notch weir is 66.7 in³/s

Explanation:

Given that;

Notch angle ∅ = 90°

height above the weir is 3 inches { head + head correction factor) h + k = 3 in

Discharge Q = ?

To determine the discharge observed, we us the following expression

Q = 4.28Ctan(∅/2) ( h + k )^5/2

where Q is discharge, C is discharge coefficient, ∅ is notch angle, h is head and k is head correction factor

now we substitute

Q = 4.28 × 1 × tan(90/2) ( 3 )^5/2

Q = 4.28 × 1 × 1 × 15.5884

Q =  66.7 in³/s

Therefore the discharge observed at a v-notch weir is 66.7 in³/s

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 is given below:

Explanation:

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:

-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:

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:

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

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

The image of the load applied to the polystyrene is missing, so i have attached it.

Answer:

a_new = 2.00302 in

b_new = 2.00552

Explanation:

From the image attached, we can see that the load of 500 lb/in is applied in the x-direction while the load of 350 lb/in acts in the y-direction.

Now, formula for stress is;

Stress(σ) = Force/Area

We are not given force and area but the load and plate thickness.

Thus, stress = load/thickness

We are given;

Load in x - direction = 500 lb/in.

Load in y - direction = 350 lb/in.

Thickness; t = 0.25 in

Thus;

σ_x = 500/0.25

σ_x = 2000 ksi

σ_y = 350/0.25

σ_y = 1400 ksi

From Hooke's law for 2 dimensions, strain is given by the formula;

ε_x = (1/E)(σ_x - vσ_y)

ε_y = (1/E)(σ_y - vσ_x)

We are given v_p = 0.25 and Ep = 597 × 10³ psi

Thus;

ε_x = (1/(597 × 10^(3)))(2000 - (0.25 × 1400)

ε_x = 0.00276

ε_y = (1/(597 × 10^(3)))(1400 - (0.25 × 2000)

ε_y = 0.00151

From elongation formula, we know that;

Startin is: ε = ΔL/L

Thus; ΔL = Lε

We are given a = 2 and b = 2

Thus;

ΔL_x = 2 × 0.00276

ΔL_x = 0.00552

ΔL_y = 2 × 0.00151

ΔL_y = 0.00302

New dimensions are;

a_new = 2 + 0.00302

a_new = 2.00302 in

b_new = 2 + 0.00552

b_new = 2.00552

Calculate the normal stress along the x-direction.

[tex]\to \sigma_x =\frac{500}{t}=\frac{500}{0.25}= 2000 \ \frac{lb}{in^2}\\\\[/tex]

Calculate the normal stress along the y-direction.

[tex]\to \sigma_y =\frac{350}{t}=\frac{350}{0.25}= 1400 \ \frac{lb}{in^2}[/tex]

Calculate the strain along the x-direction.

[tex]\to \varepsilon_y=\frac{1}{E}[\sigma_x-v(\sigma_y-\sigma_z)]\\\\[/tex]

        [tex]=\frac{1}{597\times 10^3} [2000-0.25(1400+0)] \\\\= 2.764\times 10^{-3}\\\\[/tex]

Calculate the strain along the y-direction.

[tex]\to \varepsilon_y=\frac{1}{E}[\sigma_x-v(\sigma_x+\sigma_z)][/tex]

        [tex]=\frac{1}{597\times 10^3}[1400 -0.25 (2000+0)]\\\\=1508\times 10^{-3}[/tex]

Calculate the strain along the z-direction.

[tex]\to \varepsilon_x= \frac{1}{E}[\sigma_z-v(\sigma_x +\sigma_y)][/tex]

        [tex]=\frac{1}{597\times 10^3} [0-0.25(2000+1400)] \\\\=-1.424\times 10^{-3}[/tex]

Calculate the new dimensions.

[tex]\to b' =b+ \varepsilon_xb\\\\[/tex]

       [tex]= 2+2.764\times 10^{-3} \times 2\\\\= 2.005528 \ in\\\\[/tex]

[tex]\to a' = a + \varepsilon_y a\\\\[/tex]

        [tex]= 2+1.508\times 10^{-3} \times 2\\\\= 2.003016\ in\\\\[/tex]

[tex]\to c'= c + \varepsilon_x c\\\\[/tex]

        [tex]= 0.25 +(-1.424\times 10^{-3}) \times 0.25\\\\= 0.249644\ in\\\\[/tex]

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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:

q = (ρg/μ)(sin θ)(h³/3)

Explanation:

I've attached an image of a figure showing the coordinate system.

In this system: the velocity components v and w are equal to zero.

From continuity equation, we know that δu/δx = 0

Now,from the x-component of the navier stokes equation, we have;

-δp/δx + ρg(sin θ) + μ(δ²u/δy²) = 0 - - - - - (eq1)

Due to the fact that we have a free surface, it means we will not have a pressure gradient in the x-component and so δp/δx = 0

Then our eq 1 is now;

ρg(sin θ) + μ(δ²u/δy²) = 0

μ(δ²u/δy²) = -ρg(sin θ)

Divide both sides by μ to get;

(δ²u/δy²) = -(ρg/μ)(sin θ)

Integrating both sides gives;

δu/δy = -(ρg/μ)(sin θ)y + b1 - - - - (eq2)

Now, the shear stress is given by the formula;

τ_yx = μ[δu/δy + δv/δx]

From the diagram, at the free surface,τ_yx = 0 and y = h

This means that δu/δy = 0

Thus, putting 0 for δu/δy in eq 2, we have;

0 = -(ρg/μ)(sin θ)h + b1

b1 = h(ρg/μ)(sin θ)

So, eq 2 is now;

δu/δy = -(ρg/μ)(sin θ)y + h(ρg/μ)(sin θ)

Integrating both sides gives;

u = -[(y²/2) × (ρg/μ)(sin θ)] + h(ρg/μ)(sin θ)y + b2 - - - eq3

Because u = 0 when y = 0, it means that b2 = 0 also because when we plug 0 for u and y into eq3, we will get b2 = 0.

Thus, we now have:

u = -[(y²/2) × (ρg/μ)(sin θ)] + h(ρg/μ)(sin θ)y

Factorizing like terms, we have;

u = (ρg/μ)(sin θ)[hy - y²/2] - - - (eq 4)

The flow rate per unit width is gotten by Integrating eq 4 between the boundaries of h and 0 to give;

∫u = (h,0)∫(ρg/μ)(sin θ)[hy - y²/2]

q = (ρg/μ)(sin θ)[hy²/2 - y³/6] between h and 0

q = (ρg/μ)(sin θ)[h³/2 - h³/6]

q = (ρg/μ)(sin θ)(h³/3)

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:

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:

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:

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Explanation:

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