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3 years ago

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Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0*109 V and the quantity of charg

e of the charge transferred I s30 C? (a )What is the charge in energy of that transferred charge? (b) If all the energy released could be used to accelerate a 1000kg car from rest. What would be its final speed?

1 answer:

nata0808 [166]3 years ago

3 0

Answer:

a) $U_{e} = 3 \times 10^{10}\,J$ , b) $v \approx 7745.967\,\frac{m}{s}$

Explanation:

a) The potential energy is:

$U_{e} = Q \cdot \Delta V$

$U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)$

$U_{e} = 3 \times 10^{10}\,J$

b) Maximum final speed:

$U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }$

The final speed is:

$v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }$

$v \approx 7745.967\,\frac{m}{s}$

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A certain carbon monoxide molecule consists of a carbon atom mc = 12 u and an oxygen atom mo = 17 u that are separated by a dist

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Answer:

a) x_{cm} = m₂/ (m₁ + m₂) d , b) x_{cm} = 52.97 pm

Explanation:

The expression for the center of mass is

$x_{cm}$ = 1 / M ∑ $x_{i}$ $m_{i}$

Where M is the total masses, mI and xi are the mass and position of each element of the system.

Let's fix our reference system on the oxygen atom and the molecule aligned on the x-axis, let's use index 1 for oxygen and index 2 for carbon

x_{cm} = 1 / (m₁ + m₂) (0+ m₂ x₂)

Let's reduce the magnitudes to the SI system

m₁ = 17 u = 17 1,661 10⁻²⁷ kg = 28,237 10⁻²⁷ kg

m₂ = 12 u = 12 1,661 10⁻²⁷ kg = 19,932 10⁻²⁷ kg

d = 128 pm = 128 10⁻¹² m

The equation for the center of mass is

x_{cm} = m₂/ (m₁ + m₂) d

b) let's calculate the value

x_{cm} = 19.932 10⁻²⁷ /(19.932+ 28.237) 10⁻²⁷ 128 10-12

x_{cm} = 52.97 10⁻¹² m

x_{cm} = 52.97 pm

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3 years ago

A cylinder with a piston contains 0.200 mol of oxygen at 2.00×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The

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Answer:

A. $W=600\ J$

B. $Q=2112\ J$

C. $\Delta U=1512\ J$

D. $W=0\ J$

Explanation:

Given:

no. of moles of oxygen in the cylinder, $n=0.2$

initial pressure in the cylinder, $P_i=2\times 10^5\ Pa$

initial temperature of the gas in the cylinder, $T_i=360\ K$

<em>According to the question the final volume becomes twice of the initial volume.</em>

<u>Using ideal gas law:</u>

$P.V=n.R.T$

$2\times 10^5\times V_i=0.2\times 8.314\times 360$

$V_i=0.003\ m^3$

A.

<u>Work done by the gas during the initial isobaric expansion:</u>

$W=P.dV$

$W=P_i\times (V_f-V_i)$

$W=2\times 10^5\times (0.006-0.003)$

$W=600\ J$

C.

<u>we have the specific heat capacity of oxygen at constant pressure as:</u>

$c_v=21\ J.mol^{-1}.K^{-1}$

Now we apply Charles Law:

$\frac{V_i}{T_i} =\frac{V_f}{T_f}$

$\frac{0.003}{360} =\frac{0.006}{T_f}$

$T_f=720\ K$

<u>Now change in internal energy:</u>

$\Delta U=n.c_p.(T_f-T_i)$

$\Delta U=0.2\times 21\times (720-360)$

$\Delta U=1512\ J$

B.

<u>Now heat added to the system:</u>

$Q=W+\Delta U$

$Q=600+1512$

$Q=2112\ J$

D.

Since during final cooling the process is isochoric (i.e. the volume does not changes). So,

$W=0\ J$

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3 years ago