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3 years ago

17. Swing arm restraints are intended to prevent a vehicle from falling off a lift.

Engineering

1 answer:

Dimas [21]3 years ago

5 0

This is about usage of Swing arm restraints.

<em><u>B) False</u></em>

There are different safety features that people employ when a vehicle is lifted. However, for this question, we will only talk about swing arm restraints.

Swing arm restraints are lifting restraint devices that are used to prevent a cars arms from shifting or going out of position after that car has been lifted and mounted.

This swing arm restraint does not prevent a vehicle from falling off a lift as it just helps to ensure that the swing arms that are unloaded basically maintain their position.

Read more at; brainly.com/question/17972874

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Engineers in Russia invented a new way to create colorful art with a __________.

Setler [38]

The Engineers in Russia invented a new way to create colorful art with constructivist art.

<h3>What is constructivist art?</h3>

Constructivist art was aimed to reflect modern industrial society and urban space in art. It uses industrial production forms and modest materials for its art production.

The constructivists proposed to replace traditional art's with a focus on construction as Engineers rather than a painter.

Therefore, the Engineers in Russia invented a new way to create colorful art with constructivist art using construction.

Learn more on constructivist here,

brainly.com/question/14054863

3 0

3 years ago

A water pump delivers 3 hp of shaft power when operating. If the pressure differential between the outlet and the inlet of the p

Natali [406]

Answer:

Mechanical Efficiency = 83.51%

Explanation:

Given Data:

Pressure difference = ΔP=1.2 Psi

Flow rate = $V=8ft^3/s\\$

Power of Pump = 3 hp

Required:

Mechanical Efficiency

Solution:

We will first bring the change the units of given data into SI units.

$P=1.2*6.895 = 8.274KPa\\V=8*0.00283=0.226 m^3/s\\P=3*0.746=2.238KW$

Now we will find the change in energy.

Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.

Thus change in energy is

$=(Mass * change in P)/density\\= \frac{M*P}{p}\\\\$

As we know that Mass = Volume x density

substituting the value

Energy = Volume * density x ΔP / density

Change in energy = Volumetric flow x ΔP

Change in energy = 0.226 x 8.274 = 1.869 KW

Now mechanical efficiency = change in energy / work done by shaft

Efficiency = 1.869 / 2.238

Efficiency = 0.8351 = 83.51%

5 0

3 years ago

A pipe is insulated such that the outer radius of the insulation is less than the critical radius. Now the insulation is taken o

irina [24]

Answer:

the heat transfer from the pipe will decrease when the insulation is taken off for r₂< $r_{cr}$

where;

r₂ = outer radius

$r_{cr}$ = critical radius

Explanation:

Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h .

$r_{cr} =\frac{k}{h}$

The rate of heat transfer from the cylinder increases with the addition of insulation for outer radius less than critical radius (r₂< $r_{cr}$ ) 0, and reaches a maximum when r₂ = $r_{cr}$ , and starts to decrease for r₂< $r_{cr}$ . Thus, insulating the pipe may actually increase the rate of heat transfer from the pipe instead of decreasing it when r₂< $r_{cr}$ .

7 0

3 years ago

A coral reef is in danger of being destroyed by a seaside construction project. Which of the following best relates to how an en

Wewaii [24]

Answer:Bio team

Explanation:

4 0

3 years ago

The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe carrying water is 2.00 lb/ft^2. Determine the

soldier1979 [14.2K]

Answer:

a) -8 lb / ft^3

b) -70.4 lb / ft^3

c) 54.4 lb / ft^3

Explanation:

Given:

- Diameter of pipe D = 12 in

- Shear stress t = 2.0 lb/ft^2

- y = 62.4 lb / ft^3

Find pressure gradient dP / dx when:

a) x is in horizontal flow direction

b) Vertical flow up

c) vertical flow down

Solution:

- dP / dx as function of shear stress and radial distance r:

(dP - y*L*sin(Q))/ L = 2*t / r

dP / L - y*sin(Q) = 2*t / r

Where dP / L = - dP/dx,

dP / dx = -2*t / r - y*sin(Q)

Where r = D /2 ,

dP / dx = -4*t / D - y*sin(Q)

a) Horizontal Pipe Q = 0

Hence, dP / dx = -4*2 / 1 - 62.4*sin(0)

dP / dx = -8 + 0

dP/dx = -8 lb / ft^3

b) Vertical pipe flow up Q = pi/2

Hence, dP / dx = -4*2 / 1 - 62.4*sin(pi/2)

dP / dx = 8 - 62.4

dP/dx = -70.4 lb / ft^3

c) Vertical flow down Q = -pi/2

Hence, dP / dx = -4*2 / 1 - 62.4*sin(-pi/2)

dP / dx = -8 + 62.4

dP/dx = 54.4 lb / ft^3

7 0

3 years ago