Astrology is a sub-discipline of astronomy. ОА. A True ОВ. False

A star that is moving toward an observer has its visible light shifted toward which end of the spectrum? A. blue B. red C. yello

xxTIMURxx [149]

La D mijo es blanco por que al pasar con rapides el color se torna blanco

3 0

2 years ago

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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

You and your friend are pushing a box at the same constant speed. You pushed 20 feet in 10 minutes, your friend was pushing for

anzhelika [568]

Answer:

60 N.C

Explanation:

The box will move in the direction of the push and pull with a force of 60 N. C.

7 0

2 years ago

The charge that passes a cross-sectional area A=10-4 m2 varies with time according

Anarel [89]

Answer:

Current = dQ/dt

or I = dQ/dt

Where I represents current.

Which is the rate of flow of charge.

Q=4 + 2t + t²

dQ/dt = 2 + 2t --- This is the relation that gives the instantaneous current.

At time t=2sec

dQ/dt = I = 2 + 2t

= 2 + 2(2)

=2 + 4

= 6A.

6 0

2 years ago

Michelle wants to Salt from water. She tried pouring the salt water through a piece of filter paper, but the salt did not separa

Katarina [22]

C. Let the water evaporate

3 0

2 years ago