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.
Answer:
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Explanation:
Answer;
Average speed = 47.5 km/hr
Explanation and solution;
Average Speed = Total distance /Total time
Total distance;
-In the first two hours travelling at 40km/hr, total distance traveled is 80km.
(40 × 2) = 80 km
-In next two hours distance covered is 110 km. (55 ×2)
Total distance = 110 + 80 = 190 km
Total time = 2 + 2 = 4 hours
Average Speed = Total distance /Total time
= 190/4
= 47.5 km/hr
Answer:
U₁ = (ϵAV²)/6d
This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.
Explanation:
The energy stored in a capacitor is given by (1/2) (CV²)
Energy in the capacitor initially
U = CV²/2
V = voltage across the plates of the capacitor
C = capacitance of the capacitor
But the capacitance of a capacitor depends on the geometry of the capacitor is given by
C = ϵA/d
ϵ = Absolute permissivity of the dielectric material
A = Cross sectional Area of the capacitor
d = separation between the capacitor
So,
U = CV²/2
Substituting for C
U = ϵAV²/2d
Now, for U₁, the new distance between plates, d₁ = 3d
U₁ = ϵAV²/2d₁
U₁ = ϵAV²/(2(3d))
U₁ = (ϵAV²)/6d
This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.
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