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3 years ago

Which word from the passage best explains what the web in the passage symbolizes

Engineering

1 answer:

frez [133]3 years ago

6 0

Answer:

I cant see the passage

Explanation:

I cant see the passage

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Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi

qaws [65]

Answer:

$\Delta V = 209.151\,L$ , $\Delta C = 217.517\,USD$

Explanation:

The drag force is equal to:

$F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A$

Where $C_{D}$ is the drag coefficient and $A$ is the frontal area, respectively. The work loss due to drag forces is:

$W = F_{D}\cdot \Delta s$

The reduction on amount of fuel is associated with the reduction in work loss:

$\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s$

Where $F_{D,1}$ and $F_{D,2}$ are the original and the reduced frontal areas, respectively.

$\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s$

The change is work loss in a year is:

$\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)$

$\Delta W = 2.043\times 10^{9}\,J$

$\Delta W = 2.043\times 10^{6}\,kJ$

The change in chemical energy from gasoline is:

$\Delta E = \frac{\Delta W}{\eta}$

$\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}$

$\Delta E = 6.81\times 10^{6}\,kJ$

The changes in gasoline consumption is:

$\Delta m = \frac{\Delta E}{L_{c}}$

$\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }$

$\Delta m = 154.772\,kg$

$\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }$

$\Delta V = 209.151\,L$

Lastly, the money saved is:

$\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )$

$\Delta C = 217.517\,USD$

4 0

3 years ago

All of these are uses of microwaves except...

Talja [164]

Answer:D

Explanation:

5 0

3 years ago

Water at 200C flows through a pipe of 10 mm diameter pipe at 1 m/s. Is the flow Turbulent ? a. Yes b. No

Degger [83]

Answer:

Yes, the flow is turbulent.

Explanation:

Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.

Given:

Diameter of pipe is 10mm.

Velocity of the pipe is 1m/s.

Temperature of water is 200°C.

The kinematic viscosity at temperature 200°C is $1.557\times10^{-7}$ m2/s.

Calculation:

Step1

Expression for Reynolds number is given as follows:

$Re=\frac{vd}{\nu}$

Here, v is velocity, $\nu$ is kinematic viscosity, d is diameter and Re is Reynolds number.

Substitute the values in the above equation as follows:

$Re=\frac{vd}{\nu}$

$Re=\frac{1\times(10mm)(\frac{1m}{1000mm})}{1.557\times10^{-7}}$

Re=64226.07579

Thus, the Reynolds number is 64226.07579. This is greater than 2000.

Hence, the given flow is turbulent flow.

5 0

3 years ago

A one-dimensional plane wall of thickness 2L=80 mm experiences uniform thermal generation of q= 1000 W/m^3 and is convectively c

Eduardwww [97]

Answer:

$h=1.99998\ W/m^2.C$

$k=33.333\ W/m.C$

Explanation:

Considering the one dimensional and steady state:

From Heat Conduction equation considering the above assumption:

$\frac{\partial^2T}{\partial x^2}+\frac{\dot e_{gen}}{k}=0$ Eq (1)

Where:

k is thermal Conductivity

$\dot e_{gen}$ is uniform thermal generation

$T(x) = a(L^2-x^2)+b$

$\frac{\partial\ T(x)}{\partial x}=\frac{\partial\ a(L^2-x^2)+b}{\partial x}=-2ax\\\frac{\partial^2\ T(x)}{\partial x^2}=\frac{\partial^2\ -2ax}{\partial x^2}=-2a$