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3 years ago

(this is somehow part of my science unit, dont ask why)

Physics

1 answer:

nirvana33 [79]3 years ago

6 0

Answer:

Height of coffee mug = 5 cm

Explanation:

Changing all the data given in to unit cm

50 mm = 5 cm

20 cm = 20 cm

8 m = 800 cm

0.1 km = 100 m = 10000 cm

Since we need to find the height of coffee mug, it's height will not be more than 10 cm.

So 5 cm is the correct choice.

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O How are electricity and magnetism related?

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3 years ago

Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy

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Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

$P_{1}V_{1}=nRT$

$V_{1}=\dfrac{nRT}{P_{1}}$

Put the value into the formula

$V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}$

$V_{1}=2.62\times10^{-3}\ m^3$

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

$T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}$

Put the value into the formula

$T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}$

$T_{2}=233.7\ K$

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

$P_{1}V_{1}=P_{2}V_{2}$

$V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}$

Put the value into the formula

$V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}$

$V_{2}=0.00786\ m^3$

$V_{2}=7.86\times10^{-3}\ m^3$

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

$V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}$

$V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}$

$V_{2}=0.0057\ m^3$

$V_{2}=5.7\times10^{-3}\ m^3$

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

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Answer:

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