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3 years ago

Can You Please Give Me Some Examples of Density-Independent and Density-Dependent Limiting Factors?

1 answer:

7 0

Density-Dependent:
1. competition.
2. overcrowding.
3. predators.

(These are a few from a test I took, hopefully they help you a bit >.<)

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Question 2 of 10

Mekhanik [1.2K]

the awnser to ur question is D

6 0

2 years ago

Which quantity is a scalar quantity?

Vsevolod [243]

The answer is Area ,area

5 0

2 years ago

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

2 years ago

If a star is 13.7 light years away from earth, how can we tell which elements it is made of?

ICE Princess25 [194]

Light year is a unit of measure of time that makes use of the speed of light and distance between objects to determine the number of years it will take for the light to travel. We can determine what element the object is made up of by the wavelength of the color.

8 0

2 years ago

Wave 1 carries more energy than Wave 2. Wave 3 carries more energy than Wave 2. Keeping speed constant, the greatest amount of e

inn [45]

Answer:

The greatest amount of energy is carried by a wave with the largest frequency.

Explanation:

velocity= frequency X wavelength

If the three waves moves with the same t the speed of light, then those with the largest frequency have more energy.

That is to say, a wave with a higher frequency and a lower wavelength will have the highest energy.

7 0

2 years ago