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Likurg_2 [28]
3 years ago
13

The internal pressure on a bottle of soda is 5550 torr. the bottle cap is designed to withstand a pressure of 7.2 atm. (a) calcu

late the pressure in atm
Physics
2 answers:
Mademuasel [1]3 years ago
6 0

This is a conversion problem. 760 Torr is equal to 1 atmosphere. so we can can solve now. 5,550torr*(1atmosphere/760torr)=7.3atmospheres. since the two torrs cancel out with cross multiplication we are left in atmosphere units.

Kryger [21]3 years ago
4 0

Answer:

The internal pressure on the bottle of soda is 7.303atm

Explanation:

We know that ''torr.'' is actually Torricelli a pressure unit. In order to find this pressure in ''atm'' (which is another pressure unit) we need to know the conversion factor.

The conversion factor is the following :

1atm=760torr.

If we use the conversion factor to find the pressure in atm :

(5550torr.).(\frac{1atm}{760torr.})=7.303atm

We found out that 5550torr. are equal to 7.303atm

You can find the conversion factor on any units table.

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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

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