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Hitman42 [59]
2 years ago
12

Provide one example of a bad collision, and suggest an engineering solution to avoid the collision.

Engineering
1 answer:
JulsSmile [24]2 years ago
7 0

Answer:

1). Keep your distance. Drive far enough behind the car in front of you so you can stop safely. ...

Drive strategically. Avoid situations that could force you to suddenly use your brakes. ...

Don't get distracted. ...

Don't drive when drowsy or under the influence.

2). By far the deadliest accident type is the head-on collision. Head-on collisions consider both vehicle's speed at the time of the crash, which means even an accident at lower speeds can be catastrophic

Explanation:

first is how to avoid the collision and second is bad collision

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If a student doesn't major in Engineering as an undergraduate, They could still find a successful
34kurt

Answer:

True

Explanation:

It could either be true or false because you dont really have to be great at something just to do it, you could try new things too.

5 0
2 years ago
Read 2 more answers
Compare the use of a low-strength, ductile material (1018 CD) in which the stress-concentration factor can be ignored to a high-
kicyunya [14]

Answer:

Step 1 of 3

Case A:

AISI 1018 CD steel,

Fillet radius at wall=0.1 in,

Diameter of bar

From table deterministic ASTM minimum tensile and yield strengths for some hot rolled and cold drawn steels for 1018 CD steel

Tensile strength

Yield strength

The cross section at A experiences maximum bending moment at wall and constant torsion throughout the length. Due to reasonably high length to diameter ratio transverse shear will be very small compared to bending and torsion.

At the critical stress elements on the top and bottom surfaces transverse shear is zero

Explanation:

See the next steps in the attached image

5 0
3 years ago
An asphalt concrete mixture includes 94% aggregate by weight. The specific gravities of aggregate and asphalt are 2.65 and 1.0,
garik1379 [7]

Answer:

2.0%

Explanation:

Percentage of aggregate = 94%

Specific gravity = 2.65

Specific gravity of asphalt = 1.9

Density of mix = 147pcf = 147lb/ft³

Total weight of mix: (volume = 1ft³)

= (147lb/ft³)(1ft³)

= 147lb

Percentage weight of asphalt in<u> mix:</u>

100% - 94%

= 6%

Weight of asphalt binders

= 6% x 147lb

= 8.82lb

Weight of aggregate in mix:

= 94% x 147

= 138.18lb

Specific weight of asphalt binder:

(Gab)(Yw)

Yw = specific Weight of water

= 62.4lb

Gab = specific gravity of asphalt binder

= 1.0

(62.4lb)(1.0)

= 62.4 lb/ft³

Volume of asphalt in binder:

8.82/62.4

= 0.14ft³

Specific weight of binder in mix:

2.65 x 62.4lb/ft³

= 165.36 lb/ft³

Volume of aggregate:

= 138.18/165.36

= 0.84ft³

Volume of void in the mix:

1ft³ - 0.84ft³ - 0.14ft³

= 0.02ft³

<u>The percentage of void in total mix:</u>

VTM = (0.02ft³/1ft³)100

= 2.0%

8 0
3 years ago
Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a
Virty [35]

Answer:

a) The additional time required for the truck to stop is <u>8.5 seconds</u>

b) The additional distance traveled by the truck is <u>230.05 ft</u>

Explanation:

Since the acceleration is constant, the average speed is:

(final speed - initial speed) / 2 = 0.75 v0

Since travelling at this speed for 8.5 seconds causes the vehicle to travel 690 ft, we can solve for v0:

0.75v0 * 8.5 = 690

v0 = 108.24 ft/s

The speed after 8.5 seconds is: 108.24 / 2 = 54.12 ft/s

We can now use the following equation to solve for acceleration:

v^2 - u^2 = 2*a*s

54.12^2 - 108.24^2 = 2*a*690

a = -6.367 m/s^2

Additional time taken to decelerate: 54.12/6.367 = 8.5 seconds

Total distance traveled:

v^2 - u^2 = 2*a*s

0 - 108.24^2 = 2 * (-6.367) * s

solving for s we get total distance traveled = 920.05 ft

Additional Distance Traveled: 920.05 - 690 = 230.05 ft

5 0
3 years ago
A foot-pound is described as the amount of twisting force applied to a shaft by a perpendicular
Orlov [11]

Answer:

C. 1 ft long with a weight of 1 lb

7 0
2 years ago
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