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3 years ago

12

An automobile tire at 30°C has a pressure of 3.00 atm. Temperature decreases to -5°C. Assume that there is no volume change in t

he tire.

1 answer:

Usimov [2.4K]3 years ago

4 0

Answer: decreases, 2.65 atm

Step-by-step explanation: it’s explained on edg.

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<em><u>Solution:</u></em>

<em><u>Given that,</u></em>

$(8a^{-3})^{\frac{-2}{3}$

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<em><u>Use the following law of exponent</u></em>

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Using this, simplify the given expression

$(8a^{-3})^{\frac{-2}{3}} = 8^{\frac{-2}{3}} \times a^{ -3 \times \frac{-2}{3}}\\\\Simplifying\ we\ get\\\\(8a^{-3})^{\frac{-2}{3}} = 8^{\frac{-2}{3}} \times a^2\\\\We\ know\ that\ 8 = 2^3\\\\Therefore\\\\(8a^{-3})^{\frac{-2}{3}} =2^3^{\frac{-2}{3}} \times a^2\\\\(8a^{-3})^{\frac{-2}{3}} =2^{-2} \times a^2\\\\(8a^{-3})^{\frac{-2}{3}} = \frac{a^2}{4}$

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