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2 years ago

6

Kaya just bought a house and realizes that the chimney needs to be totally torn down and rebuilt. She needs to call an expert si

nce she cannot do that work herself. What is the name of the expert that Kaya needs to call?
apprentice

bricklayer

contractor

engineer

1 answer:

lisabon 2012 [21]2 years ago

3 0

a bricklayer, a bricklayer builds houses repairs walls and chimneys etc.

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Why do engineers play a variety of roles in the engineering process?

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Answer: The engineering design process emphasizes open-ended problem solving and encourages students to learn from failure. This process nurtures students abilities to create innovative solutions to challenges in any subject! In addition to their involvement in design and development, many engineers work in testing, production, or maintenance. These engineers supervise production in factories, determine the causes of a component's failure, and test manufactured products to maintain quality.

Explanation:

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2 years ago

____ grinders are used to grind diameters, shoulders, and faces much like the lathe is used for turning, facing, and boring oper

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Answer:

Cylindrical

Explanation:

<em>A cylindrical grinder </em><em>is a tool for shaping the exterior of an item. Although cylindrical grinders may produce a wide range of forms, the item must have a central axis of rotation. Shapes such as cylinders, ellipses, cams, and crankshafts are examples of this.</em><em> Cylindrical grinding</em><em> machines are specialized grinding machines that are used to process cylinders, rods, and similar workpieces. The cylinders revolve in one direction between two centers, while the grinding wheel or wheels are close together and rotate in the other direction.</em>

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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

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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:

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

$\Delta C = 217.517\,USD$

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3 years ago

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3 years ago

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Answer:

Explanation:

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