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2 years ago

What are flip flops and what do they look like

Engineering

2 answers:

kondor19780726 [428]2 years ago

8 0

They are what you put in a ground to grow and it turns into a huge tree with flip flops hanging off

Vinil7 [7]2 years ago

4 0

flip flops are shoes and they look like shoes

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I will Brainlist<br> "Burning and Inch". Describe this measurement.

MAVERICK [17]

Answer:

what measurement

Explanation:

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2 years ago

Read 2 more answers

Given a square matrix [A], write a single line MATLAB command that will create a new matrix [Aug] that consists of the original

Liono4ka [1.6K]

Answer:

Consider A is square matrix of order 4 x 4 generated using magic function. Augmented matrix can be generated using:

Aug=[A eye(size(A))]

Above command is tested in MATLAB command window and is attached in figure below

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2 years ago

Consider a drug-eluting balloon catheter deployed into a blood vessel. The balloon is inflated to perfectly adhere to the vessel

GaryK [48]

Answer:

a) Cr = Co - Fx / D

b) ΔC / Δx = ( CR - Cr ) / ( xR - xRo )

Explanation:

A) Derive an expression for the profile c(r) inside the tissue

F = DΔC / X = D ( Co - Cr ) / X ------ 1

where : F = flux , D = drug diffusion coefficient

X = radial distance between Ro and R

Hence : Cr = Co - Fx / D

B) Express the diffusive flux at outer surface of the balloon

Diffusive flux at outer surface = ΔC / Δx = CR - Cr / xR - xRo

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2 years ago

Water is being heated in a closed pan on top of a range whilebeing stirred by a paddle wheel. During the process 30kJ of heat is

My name is Ann [436]

Answer:

38 kJ

Explanation:

The solution is obtained using the energy balance:

ΔE=E_in-E_out

U_2-U_1=Q_in+W_in-Q_out

U_2=U_1+Q_in+W_in-Q_out

=38 kJ

4 0

3 years ago

Engine oil (unused) flows at 1.81 x 10^-3 kg/s inside a 1-cm diameter tube that is heated electrically at a rate of 76 W/m. At a

Ann [662]

Answer:

(a)Tb = 330.12 K (b)Tc =304.73 K (c)19.81 K/m (d) h =60.65 W/m². K

Explanation:

Solution

Given that:

The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s

Diameter of the tube, D = 1cm =0.01 m

Electrical heat rate, q =76 W/m

Wall Temperature, Ts = 370 K

Now,

From the properties table of engine oil we can deduce as follows:

thermal conductivity, k =0.139 W/m .K

Density, ρ = 854 kg/m³

Specific heat, cp = 2120 J/kg.K

(a) Thus

The wall heat flux is given as follows:

qs = q/πD

=76/π *0.01

= 2419.16 W/m²

Now

The oil mean temperature is given as follows:

Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)

Tb =370 - 11/24 * (2419.16 * 0.005/0.139)

Tb = 330.12 K

(b) The center line temperature is given below:

Tc =Ts - 3/4 (qs.R/k)= 370 - 3/4 * ( 2419.16 * 0.005/0.139)

Tc =304.73 K

V = m/ρ (πR²)

Now,

The The axial gradient of the mean temperature is given below:

dTb/dx = 2 *qs/ρ *V*cp * R

=2 *qs/ρ*[m/ρ (πR²) *cp * R

=2 *qs/[m/(πR)*cp

dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120

dTb/dx = 19.81 K/m

(d) The heat transfer coefficient is given below:

h =48/11 (k/D)

=48/11 (0.139/0.01)

h =60.65 W/m². K

8 0

3 years ago