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Svetradugi [14.3K]
3 years ago
15

Drag each label to the correct location on the table. Match to identify permanent and temporary structures.

Engineering
1 answer:
Mama L [17]3 years ago
8 0

Answer:

Permanent Structures:

Motel

Mall

Refinery

Temporary Structures:

Kiosk

Marquee

Exhibition Stand

Explanation: Got a 100 percent on the test

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Swing arm restraints are intended to prevent a vehicle from falling off a lift.
castortr0y [4]

The question is asking whether that statement is true or false. Options are;

A) True

B) False

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

4 0
3 years ago
Bằng lý luận và thực tiễn, bạn hãy trình bày ý kiến của mình về quan điểm sau đây: "Quá trình dạy học là quá trình truyền thụ tr
mrs_skeptik [129]

Answer:

can you please ask in English I can't understand this language

7 0
3 years ago
AVH
BigorU [14]

Answer: preventive maintenance

Explanation:

3 0
3 years ago
The particle travels along the path defined by the parabola y=0.5x2, where x and y are in ft. If the component of velocity along
JulsSmile [24]

Answer:

D=41.48 ft

a=54.43\ ft/s^2

Explanation:

Given that

y=0.5 x²                      

Vx= 2 t

We know that

V_x=\dfrac{dx}{dt}

At t= 0 ,x=0  

x=\int V_x.dt

At t= 3 s

x=\int_{0}^{3} 2t.dt

x=[t^2\left\right ]_0^3

x= 9 ft

When x= 9 ft then

y= 0.5 x 9²  ft

y= 40.5 ft

So distance from origin is

x= 9 ft ,y= 40.5 ft

D=\sqrt{9^2+40.5^2} \ ft

D=41.48 ft

a_x=\dfrac{dV_x}{dt}

Vx= 2 t

a_x= 2\ ft/s^2

At t= 3 s , x= 9 ft

y=0.5 x²    

a_y=\dfrac{d^2y}{dt^2}

y=0.5 x²    

\dfrac{dy}{dt}=x\dfrac{dx}{dt}

\dfrac{d^2y}{dt^2}=\left(\dfrac{dx}{dt}\right)^2+x\dfrac{d^2x}{dt^2}

Given that

\dfrac{dx}{dt}=2t

\dfrac{dx}{dt}=2\times 3

\dfrac{dx}{dt}=6\ ft/s

a_y=\dfrac{d^2y}{dt^2}=6^2+9\times 2\ ft/s^2

a_y=54\ ft/s^2

a=\sqrt{a_x^2+a_y^2}\ ft/s^2

a=\sqrt{2^2+54^2}\ ft/s^2

a=54.43\ ft/s^2

7 0
3 years ago
Were is carbon monoxide located in a car
Semenov [28]

Answer:

all exhaust gases from all gasoline engines

Explanation:

if u look at the back of ur car when its on u can feel the heat from the exhaust and whT ur feeling is the heat coming from the carbon monoxide gases

6 0
4 years ago
Read 2 more answers
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