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tekilochka [14]
3 years ago
10

Select the correct answer. The most frequent maintenance task for a car is: A. Oil changes B. Tire replacements C. Coolant chang

es D. Brake replacements
Engineering
2 answers:
earnstyle [38]3 years ago
7 0

Answer:

<h2> A. Oil changes</h2>

  • <em>the</em><em> </em><em>most</em><em> </em><em>frequent</em><em> </em><em>maintenance</em><em> </em><em>task</em><em> </em><em>for</em><em> </em><em>a</em><em> </em><em>car</em>
abruzzese [7]3 years ago
5 0

Answer:

  A. Oil changes

Explanation:

It depends on the car and its usage and environment. Usually oil is supposed to be changed every few months, more often if the car is driven a lot. Coolant changes may be indicated as seasons change, so will generally occur less frequently than oil changes.

Tire and brake replacement depend on usage and driving habits. Some owners may never have to replace either one, if they trade their car every year or two. Folks who drive with their foot on the brake pedal may have to replace brakes relatively often.

The most frequent task is generally oil changes.

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Calculate the number of 12 V batteries (capacity 120 Ah) needed to run a 3 kW DC motor that operates in 240 V. How many hours th
Mrac [35]

Answer:

  • 20 batteries
  • 9.6 hours

Explanation:

To obtain 240 V from 12 V batteries they must be connected in series. The number needed is ...

  240/12 = 20 . . . batteries needed

__

The current draw will be ...

  (3000 W)/(240 V) = 12.5 A

Then the time available from the battery stack is ...

  (120 Ah)/(12.5 A) = 9.6 h

The motor can run 9.6 hours from the series connection.

3 0
3 years ago
A hospitality organization wants to clear an area of forestland to start a new hotel. The environmental activists have filed a c
o-na [289]

Answer:

B. Public

Explanation:

A ____________ is any group that has an actual or potential interest in or impact on an organization's ability to achieve its objectives.

a. People

b. Public

c. Profile

d. Pro-association

4 0
3 years ago
Why is it important for engineers to consider both short and long term implications of their work?
polet [3.4K]
The right answer should be A).
7 0
3 years ago
The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation.
zysi [14]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity  G = 11 × 10³ Ksi =  11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft J\tau

shaft J\tau = \dfrac{\pi}{2}r^4

where ;

r = 1 in /2

r = 0.5 in

shaft J \tau = \dfrac{\pi}{2} \times 0.5^4

shaft J\tau = 0.098218

Now; the angle of twist at  B with respect to D  is calculated by using the expression

\phi_{B/D} = \sum \dfrac{TL}{JG}

\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}

where;

T_{CD} \ \  and \ \  L_{CD} are the torques at segments CD and length at segments CD

{T_{BC} \  \ and  \ \ L_{BC}} are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

L_{BC}} = 2.5  in

J\tau = 0.098218

G =  11 × 10⁶ lb/in²

T_{BC = -60 lb.ft

T_{CD = 0 lb.ft

L_{CD = 5.5 in

\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times  12 )}{ (0.9818)(11 \times 10^6)}

\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}

\phi_{B/D} = \dfrac{-21600}{1079980}

\phi_{B/D} = − 0.02 rad

To degree; we have

\phi_{B/D}  = -0.02 \times \dfrac{180}{\pi}

\mathbf{\phi_{B/D}  = -1.15^0}

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For  the angle of twist of C with respect to D; we have:

\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}

\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}

\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times  12 )}{ (0.9818)(11 \times 10^6)}

\phi_{C/D} = \dfrac{21600}{1079980}

\phi_{C/D} = 0.02 rad

To degree; we have

\phi_{C/D}  = 0.02 \times \dfrac{180}{\pi}

\mathbf{\phi_{C/D}  = 1.15^0}

3 0
3 years ago
(a) A 10.0-mm-diameter Brinell hardness indenter produced an indentation 2.3 mm in diameter in a steel alloy when a load of 1000
vampirchik [111]

Answer:

(a)  We are asked to compute the Brinell hardness for the given indentation. for HB, where P= 1000 kg, d= 2.3 mm, and D= 10 mm.  

Thus, the Brinell hardness is computed as

HB=2P/\pi D{D-\sqrt{D^2-d^2}

=2*1000hg/\pi (10mm)[10mm-\sqrt{(1000^2-(2.3mm)^2} ]

(b)    This  part  of  the  problem  calls  for  us  to  determine  the  indentation diameter d which  will  yield  a  270  HB  when P=  500  kg.  

d=\sqrt{D^2-[D-\frac{2P}{(HB)\pi D} } ]^2\\=\sqrt{(10mm)^2-[10mm-\frac{2*500}{450( \pi10mm)} } ]^2

6 0
4 years ago
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