A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric conditions, what is the maximum pressure on her hand? What would be the maximum pressure if the "car" were an Indy 500 racer traveling 200 mph?

Answers

Answer 1

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]


Related Questions

The nuclear reactions resulting from thermal neutron absorption in boron and cadmium are 10B5 + 1 n0 ï  7Li3 + 4He2 113Cd48 + 1 n0 ï  114Cd48 + γ[5 MeV] The microscopic thermal absorption cross sections for B-10 and Cd-113 are 3841 b and 20,600 b respectively. Which of these two materials would be the more effective radiation shield? Explain

Answers

Solution :

The nuclear reaction for boron is given as :

[tex]$^{10}\textrm{B}_5 + ^{1}\textrm{n}_0 \rightarrow ^{7}\textrm{Li}_3 + ^{4}\textrm{He}_2$[/tex]

And the reaction for Cadmium is :

[tex]$^{113}\textrm{Cd}_48 + ^{1}\textrm{n}_0 \rightarrow ^{114}\textrm{Cd}_48 + \gamma [5 \ \textrm{MeV}]$[/tex]

We know that it is easier that to shield or stop an alpha particle (i.e. He nucli) as they can be stopped or obstructed by only a few centimetres of the material. However, the gamma rays ( γ ) can penetrate through the material to a greater distance. Therefore, we can choose the first one.

A mixture of octane, C8H18, and air flowing into a combustor has 60% excess air and 1 kmol/s of octane. What is the mole flow rate (kmol/s) of CO2 in the product stream?

Answers

Answer:

8 kmol/s

Explanation:

From the given information:

The combustion reaction equation for Octane in a stoichiometric condition can be expressed as:

[tex]C_{8}H_{18} +12.5(O_2 + \dfrac{79}{21} N_2) \to 9H_2O +8CO_2 + 12.5(\dfrac{79}{21}N_2)[/tex]

[tex]C_{8}H_{18} +12.5(O_2 + 3.76N_2) \to 9H_2O +8CO_2 + 12.5(3.76 \ N_2)[/tex]

In the combustor, it is said that 60% of excess air and 1 mole of Octane is present.

Thus;

the air supplied = 1.6  × 12.5 = 20

The equation can now be re-written as:

[tex]C_{8}H_{18} +20(O_2 + 3.76N_2) \to 9H_2O +8CO_2 + 7.5 \ O_2+ 75.2 \ N_2[/tex] because for 1 mole of Octane, 8 moles of CO2 can be found in the combustion product.

Thus, for 1 kmol/s of Octane also produce 8 kmol/s of CO2.

The mole flow rate in Kmol/s of CO2 in the product stream = 8 kmol/s

Write out simple definitions in words and equations for the following:

a. a1
b. b1
c. S11
d. S12
e. S21
f. S22

Answers

Answer:

a) a1 : This is the incident voltage at port 1

b) b1 : This is the deflected voltage at port 1 ;

      b1 = [tex]S_{21} a_{1} + S_{22} a_{2}[/tex]

c) S11 ; This is the input port voltage reflection coefficient when the input voltage is at port 1

S11 = [tex]\frac{V1^-}{V1^+} |v2^+=0[/tex]

d) S12 : this is the gross voltage gain

S12 = [tex]\frac{V1^-}{V2^+}| v1 ^+[/tex]

e) S21 : This is the forward voltage gain

    S21 = [tex]\frac{V2^-}{V1^+} | v2^+[/tex]

f) S22 : output port voltage reflection coefficient

   S22 = [tex]\frac{v2^-}{v2^+} | v1^+ = 0[/tex][tex]\frac{v2^-}{v2^+} | v1^+ = 0[/tex]

Explanation:

a) a1 : This is the incident voltage at port 1

b) b1 : This is the deflected voltage at port 1 ;

      b1 = [tex]S_{21} a_{1} + S_{22} a_{2}[/tex]

c) S11 ; This is the input port voltage reflection coefficient when the input voltage is at port 1

S11 = [tex]\frac{V1^-}{V1^+} |v2^+=0[/tex]

d) S12 : this is the gross voltage gain

S12 = [tex]\frac{V1^-}{V2^+}| v1 ^+[/tex]

e) S21 : This is the forward voltage gain

    S21 = [tex]\frac{V2^-}{V1^+} | v2^+[/tex]

f) S22 : output port voltage reflection coefficient

   S22 = [tex]\frac{v2^-}{v2^+} | v1^+ = 0[/tex][tex]\frac{v2^-}{v2^+} | v1^+ = 0[/tex]

Consider the following ways of handling deadlock: (1) banker’s algorithm, (2) detect
deadlock and kill thread, releasing all resources, (3) reserve all resources in advance,
(4) restart thread and release all resources if thread needs to wait, (5) resource ordering, and (6) detect deadlock and roll back thread’s actions.
a. One criterion to use in evaluating different approaches to deadlock is which
approach permits the greatest concurrency. In other words, which approach allows
the most threads to make progress without waiting when there is no deadlock?
Give a rank order from 1 to 6 for each of the ways of handling deadlock just listed,
where 1 allows the greatest degree of concurrency. Comment on your ordering.
b. Another criterion is efficiency; in other words, which requires the least processor
overhead. Rank order the approaches from 1 to 6, with 1 being the most efficient,
assuming that deadlock is a very rare event. Comment on your ordering. Does
your ordering change if deadlocks occur frequently?


who can answer part B for me?

Answers

Answer:

b

Explanation:

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

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:

C.

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.

Consider the following ways of handling deadlock:
(1) banker’s algorithm,
(2) detect deadlock and kill thread, releasing all resources,
(3) reserve all resources in advance,
(4) restart thread and release all resources if thread needs to wait,
(5) resource ordering, and
(6) detect deadlock and roll back thread’s actions.
b) Another criterion is efficiency; in other words, which requires the least processor overhead. Rank order the approaches from 1 to 6, with 1 being the most efficient, assuming that deadlock is a very rare event. Comment on your ordering

Answers

Answer:

banker’s algorithm,

(2) detect deadlock and kill thread, releasing all resources,

(3) reserve all resources in advance,

(4) restart thread and release all resources if thread needs to wait,

(5) resource ordering, and

(6) detect deadlock and roll back thread’s actions.

b) Another criterion is efficiency; in other words, which requires the least processor overhead. Rank order the approaches from 1 to 6, with 1 being the most efficient, assuming that deadlock is a very rare event. Comment on your ordering

Explanation:

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

An incremental encoder is rotating at 15 rpm. On the wheel there are 40 holes. How many degrees of rotation would 1 pulse be?

Answers

Answer:

1 pulse rotate = 9 degree

Explanation:

given data

incremental encoder rotating = 15 rpm

wheel holes = 40

solution

we get here first 1 revolution time

as 15 revolution take = 60 second

so 1 revolution take = [tex]\frac{60}{15}[/tex]

1 revolution take = 4 seconds

and

40 pulse are there for 1 revolution

40 pulse for 360 degree

so 1 pulse rotate is = [tex]\frac{360}{40}[/tex]

1 pulse rotate = 9 degree

A thin, flat pate that is 0.2m by 0.2m on a side is oriented parallel to an atmospheric airstream having a velocity of 40m/s. The air is at a temperature of T∞ = 20 °C, while the plate is maintained at Ts = 1200 C. The air flows over the top and bottom surfaces of the plate, and measurement of the drag force reveals a value of 0.075 N. What is the rate of heat transfer from both sides of the plate to the air?

Answers

This question is incomplete, the complete question is;

A thin, flat pate that is 0.2m by 0.2m on a side is oriented parallel to an atmospheric airstream having a velocity of 40m/s. The air is at a temperature of T∞ = 20 °C, while the plate is maintained at Ts = 120°C. The air flows over the top and bottom surfaces of the plate, and measurement of the drag force reveals a value of 0.075 N.

What is the rate of heat transfer from both sides of the plate to the air?

Answer:

the rate of heat transfer from both sides of the plate to the air is 236.54 W

Explanation:

Given the data in the question,

first we calculate the  Reynold's number for the flow

Re = pu∞d / Ц

Re = (1.12 × 40 × 0.2) / 1.983 × 10⁻⁵

Re = 451840

Now the Local skin friction coefficient is given as;

Cfx =  T / ( 1/2pu∞²)

Cfx = (Fd/A) / ( 1/2pu∞²)

Cfx = (0.075/(2×0.2×0.2)) / ( 1/2 × 1.12 × 40²)

= 0.9375 / 896

= 0.0010463

Cfx = 1.0463 × 10⁻³

Apply Reynold's- cOLBURN analogy

Cfx/2 = StₓPr^2/3

so

1.0463 × 10⁻³ / 2 = (h/pu∞Cp) × ( 0.711)^2/3

5.2315 × 10⁻⁴ × 1.12 × 40 × 1.005 × 1000 = h(0.711)^2/3

h = 23.554 / 0.7966

h = 29.56 W/m².K

so

The heat transfer rate from both the sides of the plate will be;

Q = 2 × 29.56 × 0.2 × 0.2 × ( 120 - 20 )

Q = 236.54 W

Therefore the rate of heat transfer from both sides of the plate to the air is 236.54 W

1. An asbestos pad is square in cross section, measuring 5 cm on a side at its small end, increasing linearly to 10 cm on a side at the large end. The pad is 15 cm high. If the small end is held at 600 K and the large end at 300 K, what heat‐flow rate will be obtained if the four sides are insulated?2. Solve Problem for the case of the larger cross section exposed to the higher temperature and the smaller end held at 300 K.

Answers

The answer is : 1.73W

If it is desired to lay off a distance of 10,000' with a total error of no more than ± 0.30 ft. If a 100' tape is used and the distance can be measured using full tape measures, what is the maxim error per tape measure allowed?

Answers

Answer:

± 0.003 ft

Explanation:

Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.

Let L' = our distance and L = our tape measure

So, L' = 100L

Now by error determination ΔL' = 100ΔL

Now ΔL' = ± 0.30 ft

ΔL = ΔL'/100

= ± 0.30 ft/100

= ± 0.003 ft

So, the maxim error per tape is ± 0.003 ft

In a p+-n Si junction, the n side has a donor concentration of 1016 cm^-3. If ni = 1010 cm^-3, relative dielectric constant Pr = 12, calculate the depletion width at a reverse bias of 100 V? What is the electric field at the mid-point of the depletion region on the n side?

Answers

Answer:

This graph shows linear

y = f(x) and y = g(x).

Find the solution to the equation f(x) - g(x) = 0

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 an example of an innovative solution to an engineering problem? Explain briefly why you chose this answer.

Answers

Answer:

robotic technology    

Explanation:

Innovation is nothing but the use of various things such as ideas, products, people to build up a solution for the benefit of the human. It can be any product or any solution which is new and can solve people's problems.

Innovation solution makes use of technology to provide and dispatch new solutions or services which is a combination of both technology and ideas.

One such example of an innovative solution we can see is the use of "Robots" in medical science or in any military operations or rescue operation.

Sometimes it is difficult for humans to do everything or go to everywhere. Thus scientist and engineers have developed many advance robots or machines using new ideas and technology to find solutions to these problems.

Using innovations and technologies, one can find solutions to many problems which is difficult for the peoples. Robots can be used in any surveillance operation or in places of radioactive surrounding where there is a danger of humans to get exposed to such threats. They are also used in medical sciences to operate and support the patient.  

is a process that is used to systematically solve problems.

design

engineering

brainstorming

O teamwork

Answers

Answer:

design

Explanation:

Design is a process used to solve problems systematically.

Human beings have specific needs and desires, which require a design process to interpret those needs and make them real from a product or service.

Design uses specific methods and techniques integrating ideals, creativity, technology and innovation to satisfy users' needs and solve problems.

A duck is cooked in the kitchen oven for 4 hours. Knowing that the oven, powered by 220 V, absorbs a current of 20 A and uses energy costing 0.048 € / kWh, how much does it cost to cook the duck?​

Answers

Explanation:

cooking of duck will cost 48000

by the help of the method of rate × A + €

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  

so

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

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

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:

A.

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.

B.

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 .

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."

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.

The temperature controller for a clothes dryer consists of a bimetallic switch mounted on an electrical heater attached to a wall-mounted insulation pad. The switch is set to open at 70 °C, the maximum dryer air temperature. In order to operate the dryer at a lower air temperature, sufficient power is supplied to the heater such that the switch reaches 70 °C (Tset) when the air temperature T is less than Tset. If the convection heat transfer coefficient between the air and the exposed switch surface of 30 mm2 is 25 W/m2 ꞏ K.


Required:

How much heater power Pe is required when the desired dryer air temperature is T[infinity]=50°C?

Answers

Answer:

[tex]0.015\ \text{W}[/tex]

Explanation:

[tex]T_{set}[/tex] = Set temperature = [tex]70^{\circ}\text{C}[/tex]

[tex]T_\infty[/tex] = Air temperature = [tex]50^{\circ}\text{C}[/tex]

A = Surface area = [tex]30\ \text{mm}^2=30\times 10^{-6}\ \text{m}^2[/tex]

h = Convection heat transfer coefficient = [tex]25\ \text{W/m}^2\text{K}[/tex]

Heater power is given by

[tex]P_e=hA(T_{set}-T_\infty)\\\Rightarrow P_e=25\times 30\times 10^{-6}(70-50)\\\Rightarrow P_e=0.015\ \text{W}[/tex]

The required heater power is [tex]0.015\ \text{W}[/tex]

How will the delay and active power per device change as you increase the doping density of both the N- and the P-MOSFET?

Answers

Answer:

hello your question is incomplete attached below is the missing part of the  question

Consider an inverter operating a power supply voltage VDD. Assume that matched condition for this inverter. Make the necessary assumptions to get to an answer for the following questions.

answer : Nd ∝ rt

Explanation:

Determine how the delay and active power per device will change as the doping density of N- and P-MOSFET increases

Pactive ( active power ) = Efs * F

Pactive = [tex]\frac{q^2Nd^2*Xn^2}{6Eo} * f[/tex]

also note that ; Pactive ∝ Nd2 (

tD = K . [tex]\frac{Vdd}{(Vdd - Vt )^2}[/tex]  since K = constant

Hence : Nd ∝ rt

For two different air velocities, the Nusselt number for two different diameter cylinders in cross flow is the same. The average heat transfer coefficient for the smaller-diameter cylinder is:_______

a. The same as that of the larger-diameter cylinder
b. Larger than that of the larger-diameter cylinder
c. Smaller than that of the larger-diameter cylinder

Answers

Answer:

B. Larger than that of the larger-diameter cylinder

Explanation:

By looking at the equation for the Nusselt number for a cylinder in cross flow, Nu = hd/k, and assuming both cylinders are made of the same material, you can see that the heat transfer coefficient h will have to be much larger for the smaller cylinder. You can verify this by using random numbers for each variable (shown below).

Verification:

ds = 1

db = 10

k = 12

Nu = 10

h small = ?

h big = ?

Nu = hd/k

10 = h small * ds / 12

120 = 1 * h small

h small = 120

Nu = hd/k

10 = h big*db / 12

120 = 10 * h big

h big = 12

This shows that the heat transfer coefficient for the smaller diameter, ds, must be bigger than the heat transfer coefficient for the larger diameter, db.

For two different air velocities, the Nusselt number for two different diameter cylinders in cross flow is the same. The average heat transfer coefficient for the smaller-diameter cylinder is " Smaller than that of the larger-diameter cylinder." (Option C)

What does the information confirm about this?

The given information states that for two different air velocities, the Nusselt number for two different diameter cylinders in cross flow is the same.

The Nusselt number is a dimensionless number that relates the convective heat transfer rate to the conductive heat transfer rate.

Since the Nusselt number is the same for both cylinders, and the heat transfer coefficient is directly related to the Nusselt number, the smaller-diameter cylinder will have a smaller average heat transfer coefficient compared to the larger-diameter cylinder.

Thus option C is the right answer.

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An inductor has an inductance of 0.025 H and a wire resistance of . How long will it take the current to reach its full Ohm’s law value?

Answers

Answer:

Time constant t = 0.0083 (Approx)

Explanation:

Given:

L = 0.025

Missing resistance r = 3Ω

Find:

Time constant t

Computation:

Time constant t = L / r

Time constant t = 0.025 / 3

Time constant t = 0.0083 (Approx)

Using the inductance - resistance relation to calculate the time constant , the time constant would be 0.08333

Given the Parameters :

Inductance in Henry = 0.025 H

Resistance of wire in ohms, R = 3 ohms

The time taken for current to reach the wire can be calculated thus :

Time constant = (Inductance, L) ÷ resistance, R

Time constant = 0.025 ÷ 3 = 0.083333

Therefore, the time constant for current to reach is 0.08333

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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°

We put capacitors on our voltage supplies in order to filter out high frequency noise. Which is better. a 10uF capacitor or a 0.1uF capacitor? Why?

Answers

Answer:

10uF

Explanation:

A higher value of capacitance is the best option when we are trying to filter power supply outputs in other to reduce hum.

The greater the capacitance or the voltage of a circuit is, the more energy it can the particular circuit can store. When capacitors are being connected in series, the total value of the capacitance reduces but contrarily, the voltage of the same system increases anyway. Connecting circuits in parallel helps to keep the voltage rating the same but on the other hand, it increases the total capacitance.

A 10 μF capacitor is better.

This is because, to filter out high frequency noise, our capacitor is connected in parallel with the voltage supply. This parallel connection causes the capacitance of the circuit to increase but the voltage stays constant.

Since there is an increase in capacitance, this causes the circuit to filter out high frequency noise.

So, a high value capacitance connected in parallel with the voltage source is a better filter for high frequency noise.

So, the 10 μF capacitor is better.

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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

Is it possible to have an iron-carbon alloy for which the mass fractions of total ferrite and proeutectoid cementite are 0.846 and 0.049, respectively? Why or why not?

Answers

Answer:

Yes it is possible.

Explanation:

This problem is about to possibility to have alloy of iron-carbon for which mass fraction of ferrite, [tex]$W_{\alpha} = 0.846$[/tex]  and proeutectoid cementite, [tex]$W_{Fe_3C}=0.049$[/tex]

An alloy formation is possible when the composition values of the two alloy are equal.

Now writing the expression for the mass fraction of total ferrite, we have

[tex]$W_{\alpha}=\frac{C_{Fe_3C}-C_0}{C_{Fe_3C}-C_{\alpha}}$[/tex]

[tex]$0.846}=\frac{6.70-C_0}{6.70-0.022}$[/tex]

[tex]$5.649588 = 6.70 - C_0$[/tex]

[tex]$\therefore C_0 = 1.05 $[/tex] wt. % of C

Now write the expression for the mass fraction of the proeutectoid cementite :

[tex]$W_{Fe_3C}=\frac{C_1-0.76}{5.94}$[/tex]

[tex]$0.049=\frac{C_1-0.76}{5.94}$[/tex]

[tex]$C_1 = 1.05$[/tex] % wt. C

Since, [tex]$C_0 =C_1$[/tex], it is possible to have an alloy of iron - carbon.

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