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Vesnalui [34]
3 years ago
13

Two distinct coplanar lines that do not intersect are known as

Physics
1 answer:
elena55 [62]3 years ago
7 0

Answer:

Skew Lines

Explanation:

Two distinct lines intersect at most one point; two distinct planes intersect in at most one line. If two coplanar lines do not intersect, they are parallel/ Two lines which are not coplanar cannot intersect and are called skew lines.

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What happens to kinetic energy when you increase the mass?
Rus_ich [418]

Increasing mass increases kinetic energy. This can be seen in the equation KE = 1/2 (m) (v)^2

If you found this helpful, please brainliest me!

5 0
3 years ago
Why isn't direct current used in transformers​
Furkat [3]

Answer:

No, it will not and this has a historical importance. The reason is that transformers work via induction of electrical forces by changes in magnetic fields, so the constat fields produced by dc currents won't work at all

Explanation:

4 0
3 years ago
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Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o
Helen [10]

Answer:

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m.

Explanation:

The Coulomb force between two charges, Q_1 and Q_2, separated by a distance, d, is given

F = k\dfrac{Q_1Q_2}{r^2}

<em>k</em> is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the <em>x</em>-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be <em>Q</em>. It is positive.

Let the distance from charge X be <em>x m.</em> Then the distance from charge Y will be (0.60 - <em>x</em>) m.

Force due to charge X

F_X = k\dfrac{18Q}{x^2}

Force due to charge Y

F_Y = k\dfrac{-27Q}{(0.60-x)^2}

Since both forces are equal and opposite,

F_X = -F_Y

k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}

\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}

2(0.60-x)^2 = 3x^2

2(0.36-1.20x+x^2) = 3x^2

0.72-2.40x+2x^2 = 3x^2

x^2+2.40x-0.72 = 0

Applying the quadratic formula,

x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}

x = -2.7 or x = 0.27

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m

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4 years ago
A mass is set oscillating horizontally, with an amplitude of 11.3 cm and a period of 0.702 sec. What is the velocity of the part
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Answer:

Explanation:

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