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jeyben [28]
3 years ago
5

Why do you think the collapse of ecosystem was unable to recover?

Physics
1 answer:
inessss [21]3 years ago
8 0

Answer: When ecosystems collapse, they rapidly lose their structure and function, with dramatic changes to their size or extent, or the species that comprise them. These losses tend to homogenize and simplify the ecosystem – fewer species, fewer habitats, and fewer connections between the two. This is why when an ecosystem failure is fatal, it's unable to recover

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3. A fighter jet covers a distance of 895m while accelerating from 22m/s to 35m/s. How long (in
larisa86 [58]

Answer:

31.404 seconds

Explanation:

To answer this equation, SUVAT is your best option utilizing and rearranging the known values to solve for the unknown.

here we have the values for

s=895

u=22

v=35

t= the unknown value

in this instant the equation s=0.5 x (u+v)t is the best equation to use

so we sub in the known values

895=0.5 x (22+35)t

rearrange to solve for t

895=28.5t

895/28.5=t

t=31.404 seconds (rounded to 3 decimal places)

6 0
3 years ago
Why do aeroplanes take longer to fly west than east?
PolarNik [594]

Every planet/moon has global wind that are mostly determined by the way the planet/moon rotates and how evenly the Sun illuminates it. On the Earth the equator gets much more Sun than the poles. resulting in warmer air at the equator than the poles and creating circulation cells (or "Hadley Cells") which consist of warm air rising over the equator and then moving North and South from it and back round.

The Earth is also rotating. When any solid body rotates, bits of it that are nearer its axis move slower than those which are further away. As you move north (or south) from the equator, you are moving closer to the axis of the Earth and so the air which started at the equator and moved north (or south) will be moving faster than the ground it is over (it has the rotation speed of the ground at the equator, not the ground which is is now over). This results in winds which always move from the west to the east in the mid latitudes.

8 0
3 years ago
You are doing an experiment to determine your reaction time. Your friend holds a ruler. You place your fingers near the sides of
olga2289 [7]
I'm guessing that you mean like this:
-- The ruler is held with zero at the bottom, and the centimeter markings
    increase as you go up the ruler.
-- You place your fingers with the ruler and the zero mark between them.
-- The number where you catch the ruler is the distance it has fallen.

Then, all we have to find is the time it takes for the ruler to fall 11.3 cm .

Here's the formula for the distance an object falls from rest
in a certain time:

                Distance = (1/2) (gravity) (time)²

On Earth, the acceleration of gravity is 9.8 m/s².
So we can write ...

                              11.2 cm  =  (1/2) (9.8 m/s²) (time)²
or
                         0.112 meter  =  (4.9 m/s²) (time)²                      

Divide each side
by  4.9 m/s² :        (0.112 m) / (4.9 m/s²)  =  time²

                            (0.112 / 4.9)  sec²  =  time²

Square root
each side:            time = √(0.112/4.9  sec²)

                                  =  √ 0.5488 sec²

                                  =      0.74 second     (rounded)  

4 0
3 years ago
Read 2 more answers
Suppose that on earth you can throw a ball vertically upward a distance of 1.20 m. Given that the acceleration of gravity on the
tatuchka [14]

Answer:

7.04 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement on Earth = 1.2 m

a = Acceleration due to gravity on Moon = 1.67 m/s²

a = Acceleration due to gravity Earth= 9.81 m/s²

Accelration going up is considered as negetive

Initial Velocity of the ball

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 1.2-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 1.2}\\\Rightarrow u=4.85\ m/s

Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-4.85^2}{2\times -1.67}\\\Rightarrow s=7.04\ m

The displacement of the ball on the moon is 7.04 m

6 0
3 years ago
If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g
EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 × 10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M_1, M_2 are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}

\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}

\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}

\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2

Since M_2 = 2m_2 and r = R/3

\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5

So gravity would have been decreased by a factor of 4.5  

8 0
3 years ago
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