Multiple Choice

Engineering

1 answer:

katrin [286]2 years ago

8 0

Answer:

Zoning is like a hammer because it is used as a tool for urban planning.

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Some of our modern kitchen cookware is made of ceramic materials. (a) List at least three important characteristics required of

Alisiya [41]

Answer:

It should be Non- toxic

It should possess high Thermal conductivity

It should have the Required Thermal diffusivity

stoneware : This material has good thermal diffusivity and it is quite affordable and it is used in making pizza stones
porcelain: mostly used for mugs and it is non-toxic
Pyrex : posses good thermal conductivity used in oven

C) All the materials are suitable because they serve different purposes when making modern kitchen cookware

Explanation:

A) characteristics required of a ceramic material to be used as a kitchen cookware

It should be Non- toxic
It should possess high Thermal conductivity
It should have the Required Thermal diffusivity

B) comparison of three ceramic materials as to their relative properties

stoneware : This material has good thermal diffusivity and it is quite affordable and it is used in making pizza stones
porcelain: mostly used for mugs and it is non-toxic
Pyrex : posses good thermal conductivity used in ovens

C) material most suitable for the cookware.

All the materials are suitable because they serve different purposes when making modern kitchen cookware

8 0

2 years ago

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:

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