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3 years ago

8

What is NOT an example of Refraction. Remember to not give me an example.

1 answer:

MAXImum [283]3 years ago

7 0

There is reflection if that might work

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How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

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

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

4 years ago

Which of the following was NOT one of Ghana's chief trading exports?

valina [46]

The answer is b. ivory

3 0

3 years ago

Bob, of mass m, drops from a tree limb at the same time that Esther, also of mass m, begins her descent down a frictionless slid

galben [10]

Answer:

Explanation:

They have the same kinetic energy

5 0

3 years ago

A circular loop of wire with a radius of 20 cm lies in the xy plane and carries a current of 2 A, counterclockwise when viewed f

Zina [86]

To solve this problem we will apply the concept related to the magnetic dipole moment that is defined as the product between the current and the object area. In our case we have the radius so we will get the area, which would be

$A=\pi r^2$

$A =\pi (0.2)^2$

$A =0.1256 m^2$

Once the area is obtained, it is possible to calculate the magnetic dipole moment considering the previously given definition:

$\mu=IA$

$\mu=2(0.1256)$

$\mu=0.25 A\cdot m^2$

Therefore the magnetic dipole moment is $0.25A\cdot m^2$

6 0

3 years ago

A current of 12 A flows for 20 minutes into an electric cooker. How much charge has the cooker used?

cluponka [151]

Time= 20 minutes= 20*60=1200 s
Charge=current*time=12*1200=14400 C

7 0

3 years ago