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3 years ago

What voltage must be applied to an 6 nF capacitor to store 0.14 mC of charge? Give answer in terms of kV.

Physics

1 answer:

levacccp [35]3 years ago

8 0

Answer:

$V=23.3kV$

Explanation:

Definition of the capacitance C, where a voltage V is applied and a charge Q is stored:

$Q=C*V$

We solve to find V:

$V=Q/C=0.14*10^{-3}C/6*10^{-9}F)=2.33*10^{4}V=23.3kV$

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Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum

cestrela7 [59]

Answer:

$T_{2}=278.80 K$

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

$(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$ .

Now, let's use the ideal gas equation to the initial and the final state:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$

Let's recall that the term nR is a constant. That is why we can match these equations.

We can find a relation between the volumes of the initial and the final state.

$\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}$

Combining this equation with the first equation we have:

$(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$

$(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}$

Now, we just need to solve this equation for T₂.

$T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}$

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

$p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40$

Finally, T2 will be:

$T_{2}=278.80 K$

6 0

3 years ago

Provides most of the energy used in the world today.

lara [203]

fossil fuels is used the most often in the world.

6 0

3 years ago

A flat sheet of ice has a thickness of 2.20 cm. It is on top of a flat sheet of crystalline quartz that has a thickness of 1.50

kolezko [41]

Answer:

$Distance_{vaccum}=5.19cm$

Explanation:

The speed of light in these mediums shall be lower than that in vacuum thus the total time light needs to cross both the media are calculated as under

Total time = Time taken through ice + Time taken through quartz

Time taken through ice = Thickness of ice / (speed of light in ice)

$T_{ice}=\frac{2.20\times 10^{-2} \times \mu _{ice}}{V_{vaccum}}$

$T_{quartz}=\frac{1.50\times 10^{-2} \times \mu _{quartz}}{V_{vaccum}}$

Thus in the same time the it would had covered a distance of

$Distance_{vaccum}=Totaltime\times V_{vaccum}\\\\Distance_{vaccum}=10^{-2}[2.20\mu _{ice+1.50\mu _{quartz}}]$

we have

$\mu _{ice}=1.309\\\\\mu _{quartz}=1.542$

Applying values we have

$Distance_{vaccum}=10^{-2}[2.20\times 1.309+1.50\times 1.542]$

$Distance_{vaccum}=5.19cm$

6 0

3 years ago

Describe rotational motion

ratelena [41]

Rotational motion may be described analytically for bodies undergoing pure rotation.

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3 years ago

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra

Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

$m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}$