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

Answers

Answer 1

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

Related Questions

Here are the commonly used Baud rate: 2400,4800,9600,19200,38400, 115200, 460800 There is an inertial measurement unit (IMU) measurement sensor that needs to update 98 bytes data (with extra 2 label bytes) every 10 ms (100Hz), what is the minimum requirement of the baud rate? (1 byte = 8 bits) Which of the above listed Baud rate you can choose to use? (please list all of them) .

Answers

Answer:

115200 and 460800

Explanation:

which of the above listed Baud rate can you choose from

Given Baud rate : 2400,4800,9600,19200,38400, 115200, 460800

The Total bytes = 98 data bytes + 2 extra label bytes for every 10 ms

= 100 bytes for every 10 ms

hence the data rate per second

= [tex]\frac{100 * 8}{10*10^{-3} }[/tex] = 80000

minimum required Baud rate = 80000

Therefore The Baud rate that can be chosen from are : 115200 and 460800

A system samples a sinusoid of frequency 230 Hz at a rate of 175 Hz and writes the sampled signal to its output without further modification. Determine the frequency that the sampling system will generate in its output.

a. 120
b. 55
c. 175
d. 230

Answers

B I got it right so (;

A smooth ceramic sphere (SG 5 2.6) is immersed in a fl ow of water at 208C and 25 cm/s. What is the sphere diameter if it is encountering (a) creeping motion, Red 5 1 or (b) transition to turbulence, Red 5 250,000

Answers

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

Tubular centrifuge is used for recovering cells 60% of the cells or recover data flow rate of 12 l/min with a rotational speed of 4000 RPM what is the RPM to increase the recovery rate of the cells to 95% at the same flow rate

Answers

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.

Air is compressed isothermally from 13 psia and 55°F to 80 psia in a reversible steady-flow device. Calculate the work required, in Btu/lbm, for this compression. The gas constant of air is R.

Answers

Answer:64.10 Btu/lbm

Explanation:

Work done in an isothermally compressed steady flow device is expressed as

Work done = P₁V₁ In { P₁/ P₂}

Work done=RT In { P₁/ P₂}

where P₁=13 psia

P₂= 80 psia

Temperature =°F Temperature is convert to °R

T(°R) = T(°F) + 459.67

T(°R) = 55°F+ 459.67

=514.67T(°R)

According to the properties of molar gas, gas constant and critical properties table, R which s the gas constant of air is given as 0.06855 Btu/lbm

Work = RT In { P₁/ P₂}

0.06855 x 514.67 In { 13/ 80}

=0.06855 x 514.67 In {0.1625}

= 0.06855 x 514.67 x -1.817

=- 64.10Btu/lbm

The required work therefore for this isothermal compression is 64.10 Btu/lbm

Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?

Answers

dutile is the correct answer

Compute the discharge observed at a v-notch weir. The weir has an angle of 90-degrees. The height above the weir is 3 inches.

Answers

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

If you make a mistake in polarity when measuring the value of DC voltage in a circuit with a digital VOM, what will happen? A. The meter will be damaged. B. The meter will read positive voltage only. C. The meter will display a negative sign. D. The meter will display OL which states an overload condition.

Answers

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

Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 58°C, cuando su volumen inicial es de 25 L. Determinar el volumen final

Answers

Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

T = 58 + 273 = 331 K

Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25

Vf = 74,4 litros

Identify how the average friction and heat transfer coefficients are determined in flow over a flat plate.
A) They are determined by differentiating the local friction and heat transfer coefficients at the mid-length of the plate, and then multiplying them by the length of the plate.
B) They are determined by by integrating the local Reynolds number and Nusselt numbers over the entire plate.
C) They are determined by integrating the local friction and heat transfer coefficients over the entire plate, and then dividing them by the length of the plate.
D) They are determined by by differentiating the local Reynolds number and Nusselt numbers at the mid-length of the plate.

Answers

Answer:

Explanation:

Let's have

Q = heat transfer surface

∆T = average temperature

F = area of the heat surface

Then the heat transfer coefficient = Q/∆T*F

In a flow over flat plate, the average friction and the heat transfer coefficient are determined by the integration of local friction and also great transfer coefficients over the plate entirely and then dividing by the plates length.

Therefore answer option C is the answer to this question.

I need help please thank for the help on the last one <3

Answers

Answer: D. Monocline (sorry if its wrong but from what i’ve learnt it is D)

What overall material composition would be required to give a material made up of 50wt% mullite and 50wt% alumina at 1400°C?

Answers

Answer: overall composition ⇒ 87 wt% { AL₂O₃] + 13% wt { SiO₂}

Explanation:

Given that;

from the phase diagram SiO₂ - Al₂O₃

alumina at 1400°C

mullite + alumina ranges from 74 - 100% wt

so for 50% mullite and 50wt% alumina

we have;

50/100 = 100 - x / 100 - 74

0.5 = 100 - x / 26

0.5 × 26 = 100 - x

13 = 100 - x

x = 100 - 13

x = 87 wt% { AL₂O₃]

[ 100% - 87% = 13%] 13% wt SiO₂

So overall composition ⇒ 87 wt% { AL₂O₃] + 13% wt { SiO₂}

What is computer programming

Answers

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! :)

A rear wheel drive car has an engine running at 3296 revolutions/minute. It is known that at this engine speed the engine produces 80 hp. The car has an overall gear reduction ratio of 10, a wheel radius of 16 inches, and a 95% drivetrain mechanical efficiency. The weight of the car is 2600 lb, the wheelbase is 95 inches, and the center of gravity is 22 inches above the roadway surface. What is the closest distance the center of gravity can be behind the front axle to have the vehicle achieve its maximum acceleration from rest on good, wet pavement?

Answers

Answer:

the closest distance the center of gravity can be behind the front axle to have the vehicle achieve its maximum acceleration from rest on good, wet pavement is 47.8 in

Explanation:

Given that;

Weight of car W = 2600 lb

power = 80 hp = 44000 lb ft/s

Engine rpm = 3296

gear reduction ratio e = 10

drivetrain efficiency n = 95% = 0.95

wheel radius R = 16 in = 1.3333 ft

Length of wheel base L = 95 in =

coefficient of road adhesion u = 0.60

height of center of gravity above pavement h = 22 in

we know that;

Coefficient of rolling resistance frl = 0.01 for good wet pavement

distance of center of gravity behind the front axle lf = ?

Maximum tractive effort (Fmax) = (uW / L) (lf - frl h) / (1 - uh / L)

First we calculate our Fmax to help us find lf

Power = Torque × 2π × Engine rpm / 60 )

44000 = Torque ( 2π×3296 / 60)

Torque = 127.5 lb ft

Fmax = Torque × e × n / R

so we substitute in our values

Fmax = 127.5 × 10 × 0.95 / 1.333

Fmax = 908.66 lb

Now we input all our values into the initial formula

(Fmax) = (uW / L) (lf - frl h) / (1 - uh / L)

908.66 = [(0.6×2600/95) (lf - 0.01×22)] / [1 - 0.6×22) / 95]

908.66 = (16.42( lf - 0.22)) / 0.86

781.4476 = (16.42( lf - 0.22))

47.59 = lf - 0.22

lf = 47.59 + 0.22

lf = 47.8 in

Therefore the closest distance the center of gravity can be behind the front axle to have the vehicle achieve its maximum acceleration from rest on good, wet pavement is 47.8 in

A roadway is to be designed on a level terrain. The roadway id 500 ft. Five cross-sections have been selected at 0 ft, 125 ft, 250 ft, 375 ft, and 500 ft. the cross sections have areas of 130 ft^2, 140 ft^2, 60 ft^2, 110 ft^2, and 120 ft^2. What is the volume needed along this road assuming a 6% shrinkage?

Answers

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

In an axial flow compressor air enters the compressor at stagnation conditions of 1 bar and 290 K. Air enters with an absolute velocity of 145 m/s axially into the first stage of the compressor and axial velocity remains constant through the stage. The rotational speed is 5500 rpm and stagnation temperature rise is 22 K. The radius of rotor-blade has a hub to tip ratio of 0.5. The stage work done factor is 0.92, and the isentropic efficiency of the stage is 0.90. Assume for air Cp=1005 kJ/(kg·K) and γ= 1.4

Determine the followings. List your assumptions.

i. The tip radius and corresponding rotor angles at the tip, if the inlet Mach number for the relative velocity at the tip is limited to 0.96.
ii. The mass flow at compressor inlet.
iii. The stagnation pressure ratio of the stage and power required by the first stage.
iv. The rotor angles at the root section.

Answers

Answer:

i) r_t = 0.5101 m

ii) m' = 106.73 kg/s

iii) R_s = 1.26

P = 2359.8 kW

iv) β2 = 55.63°

Explanation:

We are given;

Stagnation pressure; T_01 = 290 K

Inlet velocity; C1 = 145 m/s

Cp for air = 1005 kJ/(kg·K)

Mach number; M = 0.96

Ratio of specific heats; γ = 1.4

Stagnation pressure; P_01 = 1 bar

rotational speed; N = 5500 rpm

Work done factor; τ = 0.92

Isentropic effjciency; η = 0.9

Stagnation temperature rise; ΔT_s = 22 K

i) Formula for Stagnation temperature is given as;

T_01 = T1 + C1/(2Cp)

Thus,making T1 the subject, we havw;

T1 = T_01 - C1/(2Cp)

Plugging in the relevant values, we have;

T1 = 290 - (145/(2 × 1005))

T1 = 289.93 K

Formula for the mach number relative to the tip is given by;

M = V1/√(γRT1)

Where V1 is relative velocity at the tip and R is a gas constant with a value of 287 J/Kg.K

Thus;

V1 = M√(γRT1)

V1 = 0.96√(1.4 × 287 × 289.93)

V1 = 0.96 × 341.312

V1 = 327.66 m/s

Now, tip speed is gotten from the velocity triangle in the image attached by the formula;

U_t = √(V1² - C1²)

U_t = √(327.66² - 145²)

U_t = √86336.0756

U_t = 293.83 m/s

Now relationship between tip speed and tip radius is given by;

U_t = (2πN/60)r_t

Where r_t is tip radius.

Thus;

r_t = (60 × U_t)/(2πN)

r_t = (60 × 293.83)/(2π × 5500)

r_t = 0.5101 m

ii) Now mean radius from derivations is; r_m = 1.5h

While relationship between mean radius and tip radius is;

r_m = r_t - h/2

Thus;

1.5h = 0.5101 - 0.5h

1.5h + 0.5h = 0.5101

2h = 0.5101

h = 0.5101/2

h = 0.2551

So, r_m = 1.5 × 0.2551

r_m = 0.3827 m

Formula for the area is;

A = 2πr_m × h

A = 2π × 0.3827 × 0.2551

A = 0.6134 m²

Isentropic relationship between pressure and temperature gives;

P1 = P_01(T1/T_01)^(γ/(γ - 1))

P1 = 1(289.93/290)^(1.4/(1.4 - 1))

P1 = 0.9992 bar = 0.9992 × 10^(5) N/m²

Formula for density is;

ρ1 = P1/(RT1)

ρ1 = 0.9992 × 10^(5)/(287 × 289.93)

ρ1 = 1.2 kg/m³

Mass flow rate at compressor inlet is;

m' = ρ1 × A × C1

m' = 1.2 × 0.6134 × 145

m' = 106.73 kg/s

iii) stagnation pressure ratio is given as;

R_s = (1 + ηΔT_s/T_01)^(γ/(γ - 1))

R_s = (1 + (0.9 × 22/290))^(1.4/(1.4 - 1))

R_s = 1.26

Work is;

W = C_p × ΔT_s

W = 1005 × 22

W = 22110 J/Kg

Power is;

P = W × m'

P = 22110 × 106.73

P = 2359800.3 W

P = 2359.8 kW

iv) We want to find the rotor angle.

now;

Tan β1 = U_t/C1

tan β1 = 293.83/145

tan β1 = 2.0264

β1 = tan^(-1) 2.0264

β1 = 63.73°

Formula for Stagnation pressure rise is given by;

ΔT_s = (τ•U_t•C1/C_p) × tan(β1 - β2)

Plugging in the relevant values;

22 = (0.92 × 293.83 × 145/1005) × (tan 63.73 - tan β2)

(tan 63.73 - tan β2) = 0.5641

2.0264 - 0.5641 = tan β2

tan β2 = 1.4623

β2 = tan^(-1) 1.4623

β2 = 55.63°

What is the Bernoulli formula?

Answers

Answer:

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

A cylindrical bar of metal having a diameter of 19.2 mm and a length of 207 mm is deformed elastically in tension with a force of 52900 N. Given that the elastic modulus and Poisson's ratio of the metal are 61.4 GPa and 0.34, respectively, determine the following:

a. The amount by which this specimen will elongate in the direction of the applied stress.
b. The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.

Answers

Answer:

1)ΔL = 0.616 mm

2)Δd = 0.00194 mm

Explanation:

We are given;

Force; F = 52900 N

Initial length; L_o = 207 mm = 0.207 m

Diameter; d_o = 19.2 mm = 0.0192 m

Elastic modulus; E = 61.4 GPa = 61.4 × 10^(9) N/m²

Now, from Hooke's law;

E = σ/ε

Where; σ is stress = force/area = F/A

A = πd²/4 = π × 0.0192²/4

A = 0.00009216π

σ = 52900/0.00009216π

ε = ΔL/L_o

ε = ΔL/0.207

Thus,from E = σ/ε, we have;

61.4 × 10^(9) = (52900/0.00009216π) ÷ (ΔL/0.207)

Making ΔL the subject, we have;

ΔL = (52900 × 0.207)/(61.4 × 10^(9) × 0.00009216π)

ΔL = 0.616 × 10^(-3) m

ΔL = 0.616 mm

B) Poisson's ratio is given as;

υ = ε_x/ε_z

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

υ = (Δd/d_o) ÷ (ΔL/L_o)

Making Δd the subject gives;

Δd = (υ × d_o × ΔL)/L_o

We are given Poisson's ratio to be 0.34.

Thus;

Δd = (0.34 × 19.2 × 0.616)/207

Δd = 0.00194 mm

A gas stream contains 18.0 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is reduced and some of the hexane is liquefied. The hexane mole fraction in the gas stream leaving the condenser is 0.0500. Liquid hexane condensate is recovered at a rate of 1.50 L/min.

(a) What is the flow rate of the gas stream leaving the condenser in mol/min? (Hint : First calculate the molar flow rate of the condensate and note that the rates at which C6H14 and N2 enter the unit must equal the total rates at which they leave in the two exit streams.)

(b) What percentage of the hexane entering the condenser is recovered as a liquid?

Answers

Answer:

A. 72.34mol/min

B. 76.0%

Explanation:

We start by converting to molar flow rate. Using density and molecular weight of hexane

= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17

= 988.5/86.17

= 11.47mol/min

n1 = n2+n3

n1 = n2 + 11.47mol/min

We have a balance on hexane

n1y1C6H14 = n2y2C6H14 + n3y3C6H14

n1(0.18) = n2(0.05) + 11.47(1.00)

To get n2

(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)

0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min

0.18n2-0.05n2 = 11.47-2.0646

= 0.13n2 = 9.4054

n2 = 9.4054/0.13

n2 = 72.34 mol/min

This value is the flow rate of gas that is leaving the system.

n1 = n2 + 11.47mol/min

72.34mol/min + 11.47mol/min

= 83.81 mol/min

Amount of hexane entering condenser

0.18(83.81)

= 15.1 mol/min

Then the percentage condensed =

11.47/15.1

= 7.59

~7.6

7.6x100

= 76.0%

Therefore the answers are a.) 72.34mol/min b.) 76.0%

Please refer to the attachment .

A cylinder 10 mm in diameter is pulled with a stress of 150 MPa. The diameter elastically decreased by 0.007 mm. Determine Poisson's ratio if the material has a elastic modulus of 100 GPa.

Answers

Answer:Poisson's Ratio,μ = 0.46

Explanation:

Poisson's Ratio is calculate as

μ = transverse/ longitudinal strain

μ = - εt / εl

where

μ = Poisson's ratio

εt = transverse strain

εl = longitudinal strain

Transverse strain can be expressed as

εt = change in diameter / initial diameter

where

εt =transverse strain

change in diameter=0.007mm

initial diameter = 10mm

εt =0.007mm/ 10mm= 0.0007

Longitudinal strain can be expressed as

εl=Stress/ elastic modulus = σ/ E

= Stress = 150 MPa , converting to GPa becomes 150/1000 = 0.15 GPa

εl= 0.15 GPa / 100 GPa= 0.0015

Poisson's Ratio,μ = transverse/ longitudinal strain

( 0.0007 /0.0015) = 0.46 =0.46

A tube of diameter 3 cm and length 3 m has a water flow of 100 cm3/s. If the pollutant concentration in the water is constant at 2 mg/L, find the mass flux (mg/cm2-s) of pollutant through the tube due to advection.

Answers

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

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

Saturated water vapor at 150°C is compressed in a reversible steady-flow device to 1000 kPa while its specific volume remains constant. Determine the work required in kJ/kg.

Answers

Answer:

work required = 205.59 kJ/kg

Explanation:

Given data:

Temperature of water vapor = 150°c

final pressure ( P2 ) = 1000 kPa

specific volume = constant

Determine work required in kJ/kg

we apply the equation below to resolve the problem

[tex]w_{rev} = v ( P1 - P2 )[/tex] ---- ( 1 )

next we have to find the value of the specific volume and saturation pressure of water vapor at 150°c using the saturated water-temperature table

v = specific volume = 0.39248 m^3/kg

P1 = saturation pressure = 476.16 kPa

substitute values into equation 1

[tex]w_{rev} =[/tex] 0.39248 ( 476.16 - 1000 )

= -205.59 kj/kg

hence work required = 205.59 kJ/kg

The seers were of the opinion that_____ . *

a healthy mind guides a healthy body.

the healthy body needs no exercise.

a healthy mind resides in a healthy body.

the healthy mind resides in every body.

Answers

Answer:

✔️a healthy mind resides in a healthy body.

Explanation:

The seers were of the opinion that "a healthy mind resides in a healthy body."

Just like the English translation of a famous quotation from Thales, pre-Socratic Greek philosopher puts it "a sound mind in a sound body"; which tries to demonstrate the close connections that exists in bodily well-being and one's ability to enjoy life.

The seers were actually of the opinion that a healthy mind resides in a healthy body. It implies that there is connection between the body and the mind. When the body catches an illness, the mind and other parts of the body are affected. When our minds are not healthy, it affects the effective functioning of the body.

So, a healthy mind will definitely be found in a healthy body.

✔️a healthy mind resides in a healthy body.

Explanation:

The seers were of the opinion that "a healthy mind resides in a healthy body."

So, a healthy mind will definitely be found in a healthy body.

Which of the following is an example of a tax

Answers

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:

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