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3 years ago

How can you safely lift and support a vehicle

Engineering

2 answers:

algol [13]3 years ago

7 0

Answer:

Place the jack under the part of the vehicle that it should contact when raised. If you're using jack stands, place them near the jack. If you place your jack incorrectly, you can injure your car. To find the proper place to position the jack for your particular vehicle, check your owner's manual.

Sveta_85 [38]3 years ago

7 0

Wheel Chocks: It's common sense to have a pair of wheel chocks to hand. These can be bought, or made from sturdy wood. These should act as a failsafe, rather than actually bear any weight and should always be placed under any wheels that aren't elevated.Obviously you can also apply the handbrake, and leave the car in gear for extra protection.

Jack: Before jacking up your car always make sure that you're doing so on solid, level ground. Never gravel, or earth/dirt, and never on a slope! While the jack that comes with your car is fine for changing a wheel at the side of the road in an emergency it's not to be trusted if you want to work under your car.
Bear in mind that a big jack is quite a hefty piece of equipment, if you're going to have to lug it from your shed to where you keep your car you may want to consider a smaller jack – albeit one still capable of lifting your car safely. Never solely rely on the jack to support the car if working under, or near the car.

Axle stands: Once you've got the car to the desired height it's time to place your axle stands. Despite the name they don't have to be placed underneath the axles – but under somewhere strong that can withstand the weight of the car, as with the jacking points.Once in position raise the stands to the correct height (ensuring that the locking pin/lever is in the correct place, then slowly lower the jack so the weight of the car is resting on the stand/s.You can keep the jack under there to be on the safe side, but make sure it's not unbalancing the car in any way, or fighting against the stands.Before you get under the car give it a good shove – if it moves – you need to mount the stands somewhere else.If while working under the car you remove the wheels, it's never a bad idea to put them under the car as an extra safety measure.

Car ramps: Quicker and easier to use than a jack, ramps enable you to get under your car quickly and safely. Always get ramps which can take the weight of your car, always leave the handbrake on/car in gear AND chock the wheels.

Some cars struggle to get up ramps if they have low ground clearance, but low entry ramps are available. If you find that the car simply pushes the ramps out of the way – you can tie a short length of thin rope around the bottom rung of the ramp and lay the rope on the ground so the car drives over it as you approach it.

As the weight of the car pushes down on the rope it will prevent the ramp from sliding backwards. We'd always advise enlisting the help of a friend when you use ramps – just to ensure that the car is going up them straight, and that everything is stable.

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cricket20 [7]

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The power generated by a wind farm is not constant because of irregular wind patterns.

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A piston-cylinder device contains 0.58 kg of steam at 300°C and 0.5 MPa. Steam is cooled at constant pressure until one-half of

Mumz [18]

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a) Tբ = 151.8°C

b) ΔV = - 0.194 m³

c) The T-V diagram is sketched in the image attached.

Explanation:

Using steam tables,

At the given pressure of 0.5 MPa, the saturation temperature is the final temperature.

Right from the steam tables (A-5) with a little interpolation, Tբ = 151.793°C

b) The volume change

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The volume change will be calculated from the mass (0.58 kg), the initial specific volume (αᵢ) and the final specific volume

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ΔV = m(αբ - αᵢ)

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q = 0.5, αₗ = 0.00109 m³/kg, αᵥ = 0.3748 m³/kg

αբ = 0.00109 + 0.5(0.3748 - 0.00109)

αբ = 0.187945 m³/kg

αᵢ = 0.5226 m³/kg

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3 years ago

A reversible power cycle R and an irreversible power cycle I operate between the same hot and cold thermal reservoirs. Cycle I h

anygoal [31]

Answer: Attached below is the missing diagram

answer :

A) 1) Wr > WI, 2) Qc' > Qc

B) 1) QH' > QH, 2) Qc' > Qc

Explanation:

л = w / QH = 1 - Qc / QH and QH = w + Qc

A) each cycle receives same amount of energy by heat transfer

( Given that ; Л1 = 1/3 ЛR )

1) develops greater bet work 

WR develops greater work ( i.e. Wr > WI )

2) discharges greater energy by heat transfer

Qc' > Qc

solution attached below

B) If Each cycle develops the same net work

1) Receives greater net energy by heat transfer from hot reservoir

QH' > QH ( solution is attached below )

2) discharges greater energy by heat transfer to the cold reservoir

Qc' > Qc

solution attached below

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2 years ago

A 10.2 mm diameter steel circular rod is subjected to a tensile load that reduces its cross- sectional area to 52.7 mm^2. Determ

VMariaS [17]

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The percentage ductility is 35.5%.

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Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.

Step1

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Diameter of shaft is 10.2 mm.

Final area of the shaft is 52.7 mm².

Calculation:

Step2

Initial area is calculated as follows:

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Step3

Percentage ductility is calculated as follows:

$D=\frac{A_{i}-A_{f}}{A_{i}}\times100$

$D=\frac{81.713-52.7}{81.713}\times100$

D = 35.5%.

Thus, the percentage ductility is 35.5%.

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3 years ago

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Kaylis [27]

Answer:

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2 years ago

How can you safely lift and support a vehicle￼

How can you safely lift and support a vehicle