Describe the laws of liquid pressure and Explain the term fluid. (Please give the correct answer, it's really urgent)

Physics

1 answer:

Alla [95]2 years ago

3 0

Answer:

The laws of liquid pressure are

(i) Pressure inside the liquid increases with the depth from the free surface of the liquid.

(ii) Pressure is same at all points on a horizontal plane, in case of stationary liquid.

(iii) Pressure is same in all directions about a point inside the liquid.

(iv) Pressure at same depth is different in different liquids. It increases with the increase in the density of the liquid.

(v) A liquid will always seek its own level.

A Fluid is any liquid or gas or generally any material that cannot sustain a tangential, or shearing, force when at rest.

Explanation:

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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:

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

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

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

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:

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