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2 years ago

The first attempt to strengthen concrete flexural elements by externally bonded steel plates was attempted in?

Engineering

1 answer:

Vikentia [17]2 years ago

7 0

Answer:

Explanation:

The first reported attempts to strengthen concrete flexural elements by externally bonded steel plates were attempted in France around 1964-65 and practical applications date back to 1966-67 in France and South Africa then followed by Japan and Russia.

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What does this middle button on the middle of my Chevrolet equinox steering wheel mean?

almond37 [142]

Answer:

Depending on how new your vehicle is, to me it looks like some sort of turn on button for forward collision safety feature, but i'm not an expert with this particular vehicle. If you want a better answer, I strongly suggest looking at the owner's manual under which should be located in the front dash compartment on the passengers side. Once you have the manual, look in the appendix until you find controls (or something similar) then go to that page and read about your vehicles control buttons. Your answer should be in the manual.

Something that all automobile owners should do right after purchasing a new vehicle, is reading the owners manual. As boring as it may seem, reading the owner's manual will help you get used to your new car quicker and give you instructions on how to take care of your car so that it lasts.

Have a great day, and I wish you safe traveling for now and forever! :)

7 0

2 years ago

Read 2 more answers

Ben leads a team of a few engineers at a robotics firm. A couple of them would like to improve their skills by taking additional

Anit [1.1K]

Answer:

i dont know

Explanation:

4 0

2 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

Suppose a person (height of 1.8 m) walks in a room installed with a pyroelectric infrared (PIR) motion detector. The distance be

3241004551 [841]

Answer: b is your best option

Explanation:

8 0

3 years ago

generally compound curves are not filtered recommended for A. Road B. water way C. underground road D. rail way

vfiekz [6]

Answer:

C. underground road

Explanation:

Generally compound curves are not filtered and recommended for use in an underground road. However, they are best used in the road, water way, and rail way.

3 0

3 years ago