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.
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
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%
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:


a = -6.367 m/s^2
Additional time taken to decelerate: 54.12/6.367 = 8.5 seconds
Total distance traveled:

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
Answer:
C. 1 ft long with a weight of 1 lb