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Ghella [55]
3 years ago
12

How might the construction of a new recreational path between the river and the railroad track affect the stability of the railr

oad?
Physics
1 answer:
Alex_Xolod [135]3 years ago
3 0
Had to look for the missing details and here is my answer.

This question is based on Hazard City: Landslide Hazard Assessment. And based on the given illustration, how the construction of a new recreational path between the <span>railroad track and the river would influence the railroad stability is that it would make the railroad more vulnerable to landslide due to increasing slope. Hope this helps.</span>
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Regeneration means that an organism regrows a lost part, so that the original function is restored.

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NNADVOKAT [17]

Here's what you need to know about waves:

Wavelength = (speed) / (frequency)

Now ... The question gives you the speed and the frequency,
but they're stated in unusual ways, with complicated numbers.

Frequency:  How many each second ?
The thing that's making the waves is vibrating 47 times in 26.9 seconds.
Frequency = (47) / (46.9 s) =  1.747... per second.  (1.747... Hz)

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4 0
3 years ago
Can someone please help me with this physics question? I'm desperate!
Lelu [443]

Answer:

a) 2·√10 seconds

b) Linda should be approximately 30.6 meters

c) Jenny's speed at the 100-m mark is approximately 6.325 m/s

Explanation:

The speed with which Linda is running = 8.6 m/s

The point Jenny starts = The 80-m mark

The acceleration of Jenny = 1.0 m/s²

a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, <em>t</em>, is given as follows

Δs = u·t + (1/2)·a·t²

Δs = Distance = 100-m - 80-m = 20-m

u = The initial velocity of Jenny = 0

a = Jenny's acceleration = 1.0 m/s²

∴ 20 = 0×t + (1/2) × 1 × t² = t²/2

20 = t²/2

t = √(20 × 2) = 2·√10

The time it takes Jenny to run from the 80-m mark to the 100-m mark = 2·√10 seconds

b) The distance Linda runs in t = 2·√10 seconds, d = v × t

Given that Linda's velocity, v = 8.6 m/s, we have;

d = 8.0 × 2·√10 = 16·√10

The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters

Therefore, Linda should be approximately (50.6 - 20) meters = 30.6 meters behind Jenny when Jenny starts running

c) Jenny's speed at the 100 m mark is given as follows;

v = u + a·t

t = 2·√10 seconds, a = 1.0 m/s², u = 0

∴ v = 0×t + 1.0×2·√10 = 2·√10 ≈ 6.325

Jenny's speed at the 100-m mark ≈ 6.325 m/s

3 0
2 years ago
When applying a horizontal force of 30N, an object of mass 6 kg accelerates at 4m/s2. The force of friction on the surface must
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Using Newton's Second Law, F = ma, where F is the net force

So the net force is:

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mr Goodwill [35]
An organism scientific name consist of : C. its genus name and its species name
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hope this helps

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