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cestrela7 [59]
3 years ago
14

A submerged submarine alters its buoyancy so that it initially accelerates upward at 0.325 m/s^2. What is the submarine's averag

e density at this time? (Hint: the density of sea water is 1.025x10^3 kg/m^3.)
Physics
1 answer:
scoundrel [369]3 years ago
7 0
<span>You are given a submerged submarine accelerating upward at 0.325 m/s</span>² and the density of sea water is 1.025x10³ kg/m³. The submarine's average density at this time is 22 kg/m³.
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The resultant in the x-direction:

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Two point charges, 3.4 μC and -2.0 μC , are placed 5.0 cm apart on the x axis. Assume that the negative charge is at the origin,
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The electric field is zero at x = -16.45cm

Data;

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<h3>The Electric Field at point 0</h3>

As the 3μC is larger than -2.0μC  and the charges are opposite sign. The electric field will be zero at the negative axis.

Let the point be at x.

For an electric field to be equal to zero;

k(\frac{q_1}{d_1})^2 + k(\frac{q_2}{d_2})^2 = 0\\\frac{3.4}{(5-x)^2} - \frac{2}{x^2} = 0\\

Let's solve for x using mathematical methods.

\frac{3.4x^2 - 2(5-x)^2}{(5-x)^2x^2}= 0\\ 3.4x^2 - 2(5-x^2) = 0\\3.4x^2 - 50 - 2x^2 + 20x = 0\\1.4x^2 +20x - 50 = 0

Solving the above quadratic equation;

x = -16.45cm

The electric field is zero at x = -16.45cm

Learn more on electric field at a point here;

brainly.com/question/1592046

brainly.com/question/14372859

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When all parts of a circuit are composed of conducting materials,
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