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Eva8 [605]
3 years ago
11

Suppose you pour water into a container until it reaches a depth of 21 cm. next, you carefully pour in a 11-cm thickness of oliv

e oil so that it floats on top of the water. what is the pressure at the bottom of the container? (express your answer to four significant digits.)
Physics
1 answer:
ipn [44]3 years ago
8 0
<span>3040. Pascals. The density of olive oil is 0.911 g/cm^3 and the density of water is 1.000 g/cm^3. To calculate the pressure of 1 cm of water, 1000 kg/m^3 * 9.8 m/s^2 * 0.01 m = 98.000 Pa To calculate the pressure of 1 cm of olive oil 911 kg/m^3 * 9.8 m/s^2 * 0.01 m = 89.278 Pa Now to calculate the pressure at the bottom of the container, simply add the products of how many cm of each fluid you have. So 21 * 98.000 Pa + 11 * 89.278 Pa = 2058 Pa + 982.058 Pa = 3040.058 Pa So the pressure at the bottom of the container will be 3040. Pascals.</span>
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Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

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F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

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q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

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The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

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The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

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