A stainless steel ball (=8055 kg/m3, Cp= 480 J/kgK) of diameter D =15 cm is removed from theoven at a uniform temperature of 350oC. The ball is then subjectedto the flow of air at 1 atm pressure and 30oC with a velocityof 6 m/s. The surface temperature of the ball eventuallydrops to 250oC. Determine the average convection heat transfercoefficient during this cooling process and estimate howlong this process has taken.

Answer:

the required expression is [tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]

Explanation:

Given the data in the question;

Q = -0.2[ 1 - [tex]e^(-0.05t)[/tex] ]

ω = 100 rad/s

Torque T = 18 N-m

Electric power input = 2.0 kW

now, form the first law of thermodynamics;

dE/dt = dQ/dt + dw/dt = Q' + w'

dE/dt = Q' + w'  ------ let this be equation 1

w' is the net power on the system

w' = [tex]w_{elect[/tex] - [tex]w_{shaft[/tex]

[tex]w_{shaft[/tex] = T × ω

we substitute

[tex]w_{shaft[/tex]= 18 × 100

[tex]w_{shaft[/tex] = 1800 W

[tex]w_{shaft[/tex] = 1.8 kW  

so

w' = [tex]w_{elect[/tex] - [tex]w_{shaft[/tex]

w' = 2.0 kW - 1.8 kW

w' = 0.2 kW

hence, from equation 1, dE/dt = Q' + w'

we substitute

dE/dt = -0.2[ 1 - [tex]e^{(-0.05t)[/tex] ] + 0.2

dE/dt = -0.2 + 0.2[tex]e^{(-0.05t)[/tex] ] + 0.2

dE/dt = 0.2[tex]e^{(-0.05t)[/tex]

Now, the change in total energy, increment E, as a function of time;

ΔE = [tex]\int\limits^t_0}\frac{dE}{dt} . dt[/tex]

ΔE = [tex]\int\limits^t_0} 0.2e^{(-0.05t)} dt[/tex]

ΔE = [tex]\int\limits^t_0} \frac{0.2}{-0.05} [e^{(-0.05t)}]^t_0[/tex]

[tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]

Therefore, the required expression is [tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]

Answer:

missing question; How long must the heated section be

answer : 5.898 meters

Explanation:

Given parameters:

Diameter of steel pipe = 1.5 inches = 0.0381 m

velocity of air = 2.10 ft/s = 0.64 m/s

Inlet temperature = 25°C

Outlet temperature = 95.5°C

pipe wall temperature = 121.1°C

Average temperature = ( 25 + 95.5 ) / 2 = 60.25°

determine Reynolds number of air at 60.25°

= VD / v = ( 0.64 * 0.0381 ) / ( 1.83 * 10^-5 )

              = 1332.46  this means the flow is laminar

hence mass flow rate ( m ) = eAv = 1.08 * πd^2 * velocity

= 1.08 * 0.0046 * 0.64 = 0.0032 g/s

attached below is the remaining part of the solution

Answer:

The landscape architect uses knowledge of legal requirements to inform design choices.

Explanation:

The landscape architect doesn’t only draw on knowledge of law and government to design grounds for public buildings, but will draw on that knowledge quite often on many projects. The landscape architect’s knowledge of design is not purely intuitive but is a product of study and might involve research. The landscape architect’s knowledge of law and government, however, does come into play in making design choices that are consistent with the law and government regulations.

Answer:

a) the output voltage is 50 V

b)

- the average inductor current is 10 A

- the maximum inductor current is 13 A

- the maximum inductor current is 7 A

c) the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components is 4 A

Explanation:

Given the data in the question;

a) the output voltage

V₀ = V[tex]_s[/tex]/( 1 - D )

given that; V[tex]_s[/tex] = 20 V, D = 0.6

we substitute

V₀ = 20 / ( 1 - 0.6 )

V₀ = 20 / 0.4

V₀ = 50 V

Therefore, the output voltage is 50 V

b)

- the average inductor current

[tex]I_L[/tex] = V[tex]_s[/tex] / ( 1 - D )²R

given that R = 12.5 Ω, V[tex]_s[/tex] = 20 V, D = 0.6

we substitute

[tex]I_L[/tex] = 20 / (( 1 - 0.6 )² × 12.5)

[tex]I_L[/tex] = 20 / (( 0.4)² × 12.5)

[tex]I_L[/tex] = 20 / ( 0.16 × 12.5 )

[tex]I_L[/tex] = 20 / 2

[tex]I_L[/tex] = 10 A

Therefore, the average inductor current is 10 A

- the maximum inductor current

[tex]I_{Lmax[/tex] = [V[tex]_s[/tex] / ( 1 - D )²R] + [ V

given that, R = 12.5 Ω, V[tex]_s[/tex] = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

[tex]I_{Lmax[/tex] = [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]

[tex]I_{Lmax[/tex] = [20 / 2 ] + [ 60 / 20 ]    

[tex]I_{Lmax[/tex] = 10 + 3

[tex]I_{Lmax[/tex] = 13 A

Therefore, the maximum inductor current is 13 A

- The minimum inductor current

[tex]I_{Lmax[/tex] = [V[tex]_s[/tex] / ( 1 - D )²R] - [ V

given that, R = 12.5 Ω, V[tex]_s[/tex] = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

[tex]I_{Lmin[/tex] = [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]

[tex]I_{Lmin[/tex] = [20 / 2 ] -[ 60 / 20 ]    

[tex]I_{Lmin[/tex] = 10 - 3

[tex]I_{Lmin[/tex]  = 7 A

Therefore, the maximum inductor current is 7 A

c)  the output voltage ripple

ΔV₀/V₀ = D/RCf

given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz

we substitute

ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )

ΔV₀/V₀ = 0.6 / 100

ΔV₀/V₀ = 0.006 or 0.6%V₀

Therefore, the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components;

under ideal components; diode current = output current

hence the diode current will be;

[tex]I_D[/tex] = V₀/R

as V₀ = 50 V and R = 12.5 Ω

we substitute

[tex]I_D[/tex] = 50 / 12.5

[tex]I_D[/tex] = 4 A

Therefore, the average current in the diode under ideal components is 4 A

Answer:

I think A is correct

The two technicians are right in their given assessment about the use of collapsible steering columns and collapsible brake pedals to reduce the risk of trapping the driver's feet.

This refers to the vehicular movement from one point to another, obeying traffic rules, and respect for other drivers and pedestrians.

Hence, we can see that with regards to safe driving, the opinions of both Technicians A and B are both correct as they both want the safety of the car and the driver.

Read more about safe driving here:

https://brainly.com/question/26084890

#SPj2

Answer:

A = 1.41 m²

Explanation:

The ara can be found out by using Stefan Boltzman Law:

[tex]P = \sigma AT^4[/tex]

where,

A = Area = ?

P = Radiation Power = 1200 W

σ = Stefan Boltzman Constant = 5.6703 x 10⁻⁸ W/m².k⁴

T = surface temperature = 350 k

Therefore:

[tex]1200\ W = (5.6703\ x\ 10^{-8}W/m^2.k^4)(A)(350\ k)^4\\\\A = \frac{1200\ W}{850.9\ W/m^2}[/tex]

A = 1.41 m²

Answer:

Power= Heat energy\ time

heat energy= mc∆T

specific heat of water = 4180

1.5*4180*(60-25)

=219450

time = 10× 60= 600 secs

power = 219450/600

Power= 365.75 Watt

=0.37 KW

Explanation:

the actual has to be bigger because the heater might be required to handle more ...also to accommodate extra energy.

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bit.[tex]^{}[/tex]ly/3a8Nt8n

how am I going to do this, I have a friend that might be able to help I will check

Answer:

Reverse Engineering

Explanation:

Just took the test

Answer:

Product Teardown 28 pieces (1) Plastic packaging: protect and display product for purchase. (4) Exterior screws: hold case halves together. (1) Right case half: acts as part of a handle and contains the rest of the parts. (1) Left case half: acts as part of a handle and contains the rest of the parts.

Explanation:

A product teardown process is an orderly way to know about a particular product and identify its parts, system functionality to recognize modeling improvement and identify cost reduction opportunities. Unlike the traditional costing method, tear down analysis collects information to determine product quality and price desired by the consumers.

Answer:

?

Explanation:

Answer:

4.326 MW

Explanation:

Given :

Discharge through system = [tex]$30 \ m^3/s$[/tex]

Gross head, H = 150 m

Diameter of the pipe = 1.5 m

Length of penstock = 300 m

Head loss due to turbine and draft tube = 7.5 m

Average velocity installed tail race = 0.60 m/s

The power exerted on the system is [tex]$K_S = 17.5 \ mm = 12 \times 10^{-2}$[/tex] m

[tex]$K_S = \frac{QV^2}{V^2}=\frac{QN^2}{(\sqrt{2gH})^2} $[/tex]

[tex]$K_S= \frac{30 \times N^2}{(\sqrt{2 \times 9.81 \times 200})^2}$[/tex]

[tex]$N^2=\frac{12 \times \sqrt{2\times 9.81 \times 200}}{30}$[/tex]

[tex]$N=160$[/tex]

[tex]$u_2=u_1=\frac{2 \pi N}{60} =\frac{2 \pi \times 160}{60}$[/tex]

                         = 16.75 m/s

Head loss (H) = [tex]$\frac{v_2^2}{2g}+\frac{V_{\omega_1}v_1}{g}$[/tex]

       [tex]$7=\frac{(0.75)^2}{2 \times 9.81}+\frac{V_{\omega_1} \times 16.75}{9.81}$[/tex]

       [tex]$V_{\omega_1} = 10.13 \ m/s$[/tex]

The runner power obtained, [tex]$P_R = \rho Q(V_{\omega_1}u_1)$[/tex]

                                                     [tex]$=1000 \times 30 \times (10.13 \times 16.75)$[/tex]

                                                     = 5.090 MW

So the power exerted by the shaft is up to 85% of the runner power due to mechanical losses = 0.85 x 5.090

                               = 4.326 MW

Answer:

Solving

Explanation:

Answer:

Bourdon Tube gauge

Explanation:

The most popular type of pressure gauge in several countries is the Bourdon pressure tube gauge, that is used to determine medium and high loads. Bourdon tube  will measure pressures ranging between 600 mbar - 4,000 bar. While the inner pressure is greater than the exterior pressure, the tube pushes forward, and vise versa.

Answer: True

Explanation:

Answer:

The power required to overcome rolling resistance and aerodynamic drag is 19.623 h.p.

Explanation:

Let suppose that vehicle is moving at constant velocity. By Newton's Law of Motion, the force given by engine must be equal to the sum of the rolling resistance and the aerodynamic drag force of the air. And by definition of power, we have the following formula:

[tex]\dot W = \left(f\cdot W +\frac{\rho\cdot C_{D}\cdot A\cdot v^{2}}{2\cdot g_{c}} \right)\cdot v[/tex] (1)

Where:

[tex]\dot W[/tex]- Power, in pounds-force-feet per second.

[tex]f[/tex] - Rolling resistance coefficient, no unit.

[tex]W[/tex] - Weight of the passanger car, in pounds-force.

[tex]\rho[/tex] - Density of air, in pounds-mass per cubic feet.

[tex]C_{D}[/tex] - Drag coefficient, no unit.

[tex]A[/tex] - Projected frontal area, in square feet.

[tex]v[/tex] - Vehicle speed, in feet per second.

[tex]g_{c}[/tex] - Pound-mass to pound-force ratio, in pounds-mass to pound-force.

If we know that [tex]f = 0.02[/tex], [tex]W = 3,550\,lbf[/tex], [tex]\rho = 0.08\,\frac{lbm}{ft^{3}}[/tex], [tex]C_{D} = 0.34[/tex], [tex]A = 23.3\,ft^{2}[/tex], [tex]v = 80.685\,\frac{ft}{s}[/tex] and [tex]g_{c} = 32.174\,\frac{lbm}{lbf}[/tex], then the power required by the car is:

[tex]\dot W = \left(f\cdot W +\frac{\rho\cdot C_{D}\cdot A\cdot v^{2}}{2\cdot g_{c}} \right)\cdot v[/tex]

[tex]\dot W = 10901.941\,\frac{lbf\cdot ft}{s}[/tex]

[tex]\dot W = 19.623\,h.p.[/tex]

The power required to overcome rolling resistance and aerodynamic drag is 19.623 h.p.

Answer:

- the force of the rope on the tugboat is 14.87 kN

- Force acting on the tugboat is 24.79 kN

Explanation:

 Given the data in the question;

tugboat increases its speed uniformly to 25 km/h

v₁ = 25 km/h = (25 × 1000) / ( 1 × 60min × 60sec )

= 25000m / 3600s

=  6.94 m/s

Now, lets determine the force on the rope using the following relation;

T = [tex]m_b[/tex]v₁ / t₁

[tex]m_b[/tex] is mass of barge( 75 Megagram = 75 × 10³ Kilogram  ), time t₁ is 35 s and v₁ is 6.94 m/s

so we substitute

T = [(75 × 10³) × 6.94 ]  / 35

T = 520500 / 35

T = 14871.43 N

T = 14871.43 / 1000

T = 14.87 kN

Therefore, the force of the rope on the tugboat is 14.87 kN

Now, to determine F acting on the tugboat;

[tex]Ft_1[/tex] = ( [tex]m_t[/tex] + [tex]m_b[/tex] )[tex]v_1[/tex]

we solve for F

F = ( [tex]m_t[/tex] + [tex]m_b[/tex] )[tex]v_1[/tex] / [tex]t_1[/tex]

where [tex]m_t[/tex] is mass of tugboat (50 Megagram = 50 × 10³ Kilogram  )

so we substitute

F = [( (50 × 10³) + (75 × 10³) )6.94] / 35

F = [ 125000 × 6.94 ) / 35

F = 867500 / 35

F = 24785.7 N

F = 24785.7 / 1000

F = 24.79 kN

Therefore, Force acting on the tugboat is 24.79 kN

Answer:

  12.44Ω

Explanation:

For a question such as this, the assumption must be that the resistance is inversely proportional to the square of the diameter. The resistance will be ...

  20Ω × (1.262/1.6)² ≈ 12.44Ω

Answer:

[tex]ohms[/tex]

Explanation:

hope it helps

Answer:

please give me brainlist and follow

Explanation:

Mild steel can be converted into high carbons steel by which of the following heat treatment process? Explanation: Case hardening, also referred as carburizing increases carbon content of steel, thus, imparting hardness to steel.

Answer:

Ans: To see if the results are similar

Answer:

  entering drivers

Explanation:

Drivers entering a traffic circle must yield the right-of-way to those already in the traffic circle. (Among other things, this rule helps avoid gridlock.)

Answer:

Answer:

Explanation:

One approach to this would be to create the function with two parameters. One would be the element to be added or deleted while the other would be a char variable that would indicate the process to be completed (adding or deleting). This char can have a value of '+' or '-' accordingly. Then within the function, we would use an IF/Else statement to detect whether the char variable had a value of '+' or '-'. If the value was '+' we would execute the insertion process of the code but if the value was '-' we would only execute the deletion process of the code. Therefore, allowing both processes to be available within a single function.

Answer:

They do not cause phase-distortion of an input signal. ( option 2 )

Explanation:

Linear-phase filters are desirable in many applications simply because  they do not cause phase-distortion of an input signal.  

The signals inputted into a Linear-phase filter ( FIR type ) are delayed equally as frequency is delayed in the pass band but they are are distorted.

The main special function of the Linear-phase filter is to reduce signal distortion.

Answer:

attached below

Explanation:

Applying the state transition Formula

state - next = ( state *2 + in ) % 5

how this works : remainder of previous cycle is doubled to enable the calculation of the new remainder.  

Input of current cycle is represented as either 1 or 0

since the dividing number = 5 . possible remainders = 1,2,3,4,0

each remainder is represented as :

S0 = zero remainder , S1 = 1 remainder , S2 = 2 remainder, S3 = 3 remainder,

S4 = 4 remainder

Answer:

The plotted diagram of line distance versus the transmission speed is uploaded below.

Explanation:

Given the data in the question;

Bit length B = 1000 bits

propagation velocity V = 2 × 10⁸ m/s

now, we know that the bit length of a link is expressed as;

B = R × d/V

where V is propagation velocity

d is the distance

R is the transmission speed

B is bit length

so we substitute

1000 = R × d/(2 × 10⁸)

1000 = Rd/(2 × 10⁸)

2 × 10¹¹ = Rd

R = 2 × 10¹¹ / d

R = 2E+11

Hence, we plot the transmission speed versus line distance; as shown in the image BELOW.

From the plot, if the transmission speed increases, the distances between stations decreases and vise versa.

Hence, both are inversely proportional.

The graph plot of line distance versus the transmission speed for a bit length of 1000 bits is; plotted below with Lν = 2 × 10¹¹ m/s

We are given;

Bit length; B = 1000 bits

Propagation velocity; V = 2 × 10⁸ m/s

Formula for bit length of a link is expressed as;

B = ν × L/V

where;

V is propagation velocity

L is the distance

ν is the transmission speed

B is bit length

Thus, plugging in the relevant values gives;

1000 = ν × L/(2 × 10⁸)

Thus;

Lν = 1000 × 2 × 10⁸

Lν = 2 × 10¹¹ m/s

Thus, find the attached image of a graph showing the line distance versus the transmission speed for a bit length of 1000 bits.

Read more about Bit length at; https://brainly.com/question/16612919