Answer:
Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.
Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.
Explanation:
Part a
When Road is Level
The stopping sight distance is given as

Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is this case is 0 as the road is level
Substituting values

So the minimum stopping sight distance is 563.36 ft.
Part b
When Road has a maximum grade of 4%
The stopping sight distance is given as

Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade
For upgrade of 4%, Substituting values

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>
For downgrade of 4%, Substituting values

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>
As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.
Hello. The answer to your question is ''<span>hypothesis''. I hope this helps! </span>
<span>Near the equator, the patterns of convection currents are called
Hadley Cells.</span>
Hadley Cells refers to the low-latitude overturning movements that
have air increasing at the equator and air dropping at roughly latitude of 30
degree and these cells are also responsible for the trade winds in the Tropics
and control low-latitude patterns of weather.
Answer:
0.176m from the flagpole, westward.
Explanation:
Let the Eastward be the positive direction. So initially runner A is at position -6km, running with velocity of 9km/h while runner B is at position 5km running at a velocity of -8km/h. We can conduct the following equation for their distances over the same time t


When A an B meets, they are at the same position and at the same time. So





So where they meet is 0.176m from the flagpole, westward.
Answer:
Explanation:
conversion of 2.5 days to seconds
2.5 days =2.5 days*24 hours*3600 seconds
2.5 days=216000seconds
b) conversion of 25 km to mm
1 km=1000000 mm
therefore
25 km=25*1000000=25000000 mm
c) conversion 22 g to kg
1000 g=1 kg
therefore
22 g=22/1000=0.022 kg