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alexira [117]
3 years ago
13

Consider a spherical Gaussian surface and three charges: q1 = 1.60 μC , q2 = -2.61 μC , and q3 = 3.67 μC . Find the electric flu

x through the Gaussian surface if it completely encloses (a) only charges q1 and q2, (b) only charges q2 and q3, and (c) all three charges.

Physics
1 answer:
Olenka [21]3 years ago
4 0

Answer:

Explanation:

<h3>Guass Law: Also known as  "Gauss's flux theorem" is the total of the electric flux "φ" out of a closed surface is equal to the charge "Q" enclosed divided by the permittivity εο. Solution is attached.</h3>

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3 years ago
When its 75 kW (100 hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rat
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Answer:

0.2289

Explanation:

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P= 700*9.81*2.5=17167.5 W= 17.1675 kW

To express it as a fraction of 75 kw then 17.1675/75=0.2289 or 22.89%

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3 years ago
during conversion water to ice,forces of attraction between molecules. OPTIONS:1.)Decreases 2.)Increases 3.)Does not change 4.)C
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because water is loosely packed but when it is cold it becomes closely packed in order to form ice and thus the force attraction between them also increase.

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3 years ago
A bowling ball of mass m = 1.7 kg drops from a height h = 14.2 m. A semi-circular tube of radius r = 6.2 m rest centered on a sc
marshall27 [118]

Answer:

W_net = mg + 2mgh/r

Explanation:

The forces applied in this motion of the bowling ball are both gravitational and centripetal forces.

Now, gravitational force is; F_g = mg

While centripetal force is; F_c = mv²/r

Since we want to express the net force in terms of the variables in the statement and we are not given "v", let's find an expression of v with the variables given.

Now, from Newton's equation of motion, at initial velocity of 0, v² = 2gh.

Thus;

F_c = 2mgh/r

Where;

m is ball mass

r is tube radius

h is fall height

Thus, the net force will be;

F_net = F_g + F_c

Now, Net force would be equal to the net weight that will be read on the scale.

Thus;

W_net = F_net = F_g + F_c

W_net = mg + 2mgh/r

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