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3 years ago

Which is an example of an abiotic factor in a pond environment?

Physics

2 answers:

Allushta [10]3 years ago

7 0

The water <span>is an example of an abiotic factor in a pond environment</span>

vesna_86 [32]3 years ago

3 0

"Non-Living" things are known as "Abiotic factor" in the ecosystem.

In short, Your Answer would be: Option A) The Water

Hope this helps!

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A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observ

siniylev [52]

Answer:

$m=1.53kg$

Explanation:

To solve this problem we use the Hooke's Law:

$F=k*\Delta x$ (1)

F is the Force needed to expand or compress the spring by a distance Δx.

The spring stretches 0.2cm per Newton, in other words:

1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm

The force applied is due to the weight

$F=mg$

We replace in (1):

$mg=k*\Delta x$

We solve the equation for m:

$m=k*\Delta x/g=5*3/9.81=1.53kg$

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2 years ago

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Mashcka [7]

Answer: C. At the bottom

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2 years ago

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2 years ago

How much work must be done to bring three electrons from a great distance apart to 5.0×10^−10 m from one another (at the corners

Inessa05 [86]

Answer:

1.38 x 10^-18 J

Explanation:

q = - 1.6 x 10^-19 C

d = 5 x 10^-10 m

the potential energy of the system gives the value of work done

The formula for the potential energy is given by

$U =\frac{Kq_{1}q_{2}}{d}$

So, the total potential energy of teh system is

$U =\frac{Kq_{1}q_{2}}{d}+\frac{Kq_{2}q_{3}}{d}+\frac{Kq_{1}q_{3}}{d}$

As all the charges are same and the distance between the two charges is same so the total potential energy becomes

$U =3\times \frac{Kq^{2}}{d}$

K = 9 x 10^9 Nm^2/C^2

By substituting the values

$U =3\times \frac{9\times 10^{9}\times \ 1.6 \times 1.6 \times 10^{-38}}{5\times 10^{-10}}$

U = 1.38 x 10^-18 J

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3 years ago

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SpyIntel [72]

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2 years ago

Read 2 more answers