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3 years ago

_____ can alter wind flow and create a rain shadow effect. Beaches Lakes Mountains Oceans

Physics

2 answers:

Ket [755]3 years ago

5 0

The correct answer is Mountains. Mountains can alter wind flow and create a rain shadow effect. A mountain is a large landform and is high surrounding a limited area. It is generally steeper and higher than a hill.

olga55 [171]3 years ago

4 0

Answer: Mountains can alter wind flow and create a rain shadow effect.

Explanation:

Rain shadow effects :An area having less precipitation of less rain fall due to the effect of the topographic (physical features of that area ) barriers( to prevent the movement) like mountain or mountains range.

Sides of mountain:

Windward side: Area or side where high rainfall occurs. Wind prevailing loose their moisture on this side which leads to high rainfall on this side.

Leeward side : Area or side where low rainfall occurs.

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3 years ago

Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

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$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$

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