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nata0808 [166]
2 years ago
15

compute the electric field experienced by a test a charge q = +0.08µC from a source charge q = +15µC in a vacuum when the test c

harge is placed 0.02 m away from the other charge.
Physics
1 answer:
Fudgin [204]2 years ago
3 0

<u>here</u><u> </u><u>is</u><u> </u><u>your</u><u> </u><u>answer</u><u> </u><u>hope</u><u> </u><u>it's</u><u> </u><u>helpful</u><u> </u><u>for</u><u> </u><u>you</u><u> </u><u>thankyou</u><u> </u><u>I</u><u> </u>

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At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107
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Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

K = - V\cdot \frac{dP}{dV}

Where:

K - Bulk module, measured in pascals.

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-\frac{K \,dV}{V} = dP

This resultant expression is solved by definite integration and algebraic handling:

-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP

-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}

\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}

\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }

The final volume is predicted by:

V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }

If V_{o} = 1\,m^{3}, P_{o} - P_{f} = -10132500\,Pa and K = 2.3\times 10^{9}\,Pa, then:

V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }

V_{f} \approx 0.996\,m^{3}

Change in volume due to increasure on pressure is:

\Delta V = V_{o} - V_{f}

\Delta V = 1\,m^{3} - 0.996\,m^{3}

\Delta V = 0.004\,m^{3}

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

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