I need help with this too. (im not good at science or math)

The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

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

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

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

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

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

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

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

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0

3 years ago

As you ride a bicycle on the sidewalk with a speed of

mixer [17]

Answer:

1.2 seconds

Explanation:

distance = ((final speed + initial speed) * time)/2

Here given:

distance: 3.8 meter

initial speed: 6.4 m/s

final speed: 0 m/s

Solving steps:

3.8 = ((0 + 6.4) * time))/2

3.8 = 3.2(time)

time = 3.8/3.2

time = 1.1875 seconds ≈ 1.2 seconds

5 0

1 year ago

What can destroy stuff yet does not have any mass

Artist 52 [7]

If energy could be created or destroyed, all of our ideas of how the world works ... Historically, of course not all the forms of energy were known to begin with. ... too messy or complicated to make sense, we would have had to give up the law. ... can be converted into rest mass and back again (particle physicists do this every )

3 0

2 years ago

A soccer ball is kicked and left

Vedmedyk [2.9K]

Answer:

Explanation:

Considering that this is parabolic motion, we know that the time the ball is in the air begins the instant it leaves the ground, reaches up to its max height, and then begins falling until it reaches the ground. Duh, right? Some important things happen during this trip. There are a few things we need to know in order to even begin the problem. Parabolic motion has x and y coordinates because it is 2-dimmensional; the acceleration in the x dimension is not the same as the acceleration in the y dimension; the velocity of an object at its max height is always 0; the time it takes to reach its max height (where the max height is half the distance the object travels) is half the time it takes to make the whole trip. Yikes. That's a lot to know and much to remember! Don't you just LOVE physics!?

For a. the hang time is the time the ball was in the air. Some of that stuff we talked about above is pertinent to solving this problem. We know that the velocity of the ball is 0 at its max height, and we also know that if we find the time it takes to reach its max height, we can double that number to find how long it was in the air for the whole trip. Use the one-dimensional equation

$v=v_0+at$ to find out how long it took to reach the max height. Even though we don't yet know the max height, we DO know that the velocity at that point is 0. BUT before we do that, since we are working in the y-dimension only, it would behoove us (benefit us) to find the velocity particular to this dimension. We are going to answer c. first, then backtrack.

c. wants the initial vertical velocity. That is found in the magnitude of the "blanket" or generic velocity times the sin of the angle, namely:

$V_y=25sin(45)$ so

$V_y=$ 18 m/s Now we can use that as the initial upwards velocity in part a:

$v=v_0+at$ and filling in:

0 = 18 + (-9.8)t and

-18 = -9.8t so

t = 1.8 seconds. But remember, this is only half the time it was in the air. The whole trip, then, takes 2(1.8) which is

t = 3.6 seconds

That's a and c. Now for b:

b. asks for the x component of the velocity:

$V_x=Vcos\theta$ which works out to be the same as the vertical velocity, since the sin and cos of 45 degrees is the same:

$V_x=25cos45$ and

$V_x=$ 18 m/s

Onto d:

d. wants the max height. Remember, it took 1.8 seconds to get to the max height, so using yet another one-dimensional equation:

Δx = v₀t + $\frac{1}{2}at^2$ where Δx is the displacement, v₀ is the initial upwards velocity, a is the pull of gravity, and t is the time it takes to reach that max height (Δx, our unknown). Filling in:

Δx = $18(1.8)+\frac{1}{2}(-9.8)(1.8)^2$ and if you do the rounding correctly, you'll end up with this:

Δx = 32 - 16 so

the max height, Δx, is 16 meters.

e. wants the range. That translates to the distance the ball traveled. This is found in a glorified version of d = rt, where d is displacement, r is velocity, and t is...well, time (that doesn't change):

Δx = vt so

Δx = 18(3.6) remember that the ball was in the air for a total of 3.6 seconds, so

Δx = 65 meters.

Phew!!!!! That's a lot! I suggest you learn your physics or this will make you insane by the end of the course!

6 0

3 years ago

An automobile of mass 1.46 cross times to the power of blank 10 cubed kg rounds a curve of radius 25.0 m with a velocity of 15.0

Vsevolod [243]

The centripetal force exerted on the automobile while rounding the curve is $1.31\times10^4 N$

Explanation:

given that

$Mass\ of\ the\ automobile\ m =1.46\times 10^3 kg\\radius\ of\ the\ curve\ r =25 m\\velocity\ of\ the\ automobile\ v=15m/s\\$

Objects moving around a circular track will experience centripetal force towards the center of the circular track.

$centripetal\ force=mv^2/r\\=1.46\times10^3\times15^2/25\\=1.46\times 10^3\times 225/15\\=1.31\times 10^4 N$

4 0

2 years ago