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poizon [28]
2 years ago
15

2. When performing an alignment, what action should be taken immediately after putting a vehicle on the rack?

Engineering
1 answer:
mash [69]2 years ago
3 0
D...................
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because it will keep the tires in much better shape and it can improve the handling and keep your  car from pulling to one side

Explanation:

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What are flip flops and what do they look like
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flip flops are shoes and they look like shoes
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The rolling process is governed by the frictional force between the rollers and the workpiece. The frictional force at the entra
adell [148]

Answer:

b)false

Explanation:

Rolling is a process in which work piece passes through rolls to produce desired out put of the work piece.Rolling  is a metal forming process.

We know that friction force is responsible for motion of work piece between rolls.If friction force is so small at the entrance side then work piece will not enter in the forming zone and forming process will not occurs.So the friction force should be high at the entrance side and low at the exit side.

So given statement is wrong.

3 0
2 years ago
At the instant shown, slider block B is moving with a constant acceleration, and its speed is 150 mm/s. Knowing that after slide
Xelga [282]

Answer:

a) aA = - 13.33 mm/s²

aB = - 20 mm/s²

b) aD = - 13.33 mm/s²

c) vB = 70 mm/s

d) xB = 440 mm

Explanation:

Given

The initial speed of B is: v₀B = 150 mm/s

Distance moved by A is: xA = 240 mm

Velocity of A is: vA = 60 mm/s

Assuming:

Displacement of blocks are denoted by:

A = xA

B = xB

C = xC

D = xD

From the pic shown, the total length of the cable is:

xB + (xB - xA) + 2*(d - xA) = L

⇒ 2*xB - 3*xA = L - 2*d

where L - 2*d is constant. Differentiating the above equation with respect to time:

d(2*xB)/dt - d(3*xA)/dt = 0

⇒ 2*vB - 3*vA = 0    (i)

Substituting in equation (i)

2*(150 mm/s) - 3*vA = 0

⇒ v₀A = 100 mm/s  (initial speed of A)

Then, we use the equation

vA² = v₀A² + 2*aA*xA

Substituting the values in above equation:

(60 mm/s)² = (100 mm/s)² + 2*aA*(240 mm)

⇒ aA = - 13.33 mm/s²

If  2*vB - 3*vA = 0

Differentiating the above equation with respect to time:

d(2*vB)/dt - d(3*vA)/dt = 0

⇒ 2*aB - 3*aA = 0    (ii)

Substituting in equation (ii)

2*aB - 3*(- 13.33 mm/s²) = 0

⇒ aB = - 20 mm/s²

b) From the pic shown,

xD - xA = constant

If we apply

d(xD)/dt - d(xA)/dt = 0

⇒ vD - vA = 0

then

d(vD)/dt - d(vA)/dt = 0

⇒ aD - aA = 0

⇒ aD = aA = - 13.33 mm/s²

c) We use the formula

vB = v₀B + aB*t

Substituting the values in above equation:

vB = 150 mm/s + (- 20 mm/s²)*(4 s)

⇒ vB = 70 mm/s

d) We apply the equation

xB = v₀B*t + 0.5*aB*t²

Substituting the values in above equation:

xB = (150 mm/s)*(4 s) + 0.5*(- 20 mm/s²)*(4 s)²

⇒ xB = 440 mm

4 0
3 years ago
A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate
kherson [118]

Explanation:

thermal expansion ∝L = (δL/δT)÷L ----(1)

δL = L∝L + δT ----(2)

we have δL = 12.5x10⁻⁶

length l = 200mm

δT = 115°c - 15°c = 100°c

putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

= 0.25 MM

L₂ = L + δ L

= 200 + 0.25

L₂ = 200.25mm

12.5X10⁻⁶ *115-15 * 20

= 0.025

20 +0.025

D₂ = 20.025

as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0

3 0
3 years ago
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