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Marta_Voda [28]
3 years ago
7

PLEASE HELP 50 POINTS SPACE QUESTION

Physics
1 answer:
rodikova [14]3 years ago
5 0

Earth's distance from the sun doesn't change enough to cause seasonal differences. Instead, our seasons change because Earth tilts on its axis.

hope it helps.

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What are the two parts of a force pair?
iragen [17]

These two forces are called action and reaction forces and are the subject of Newton's third law of motion.

<em>Have a luvely day!</em>

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Which of the following best defines mass?
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"<span>The amount of matter in a substance" is the one among the following choices given in the question that best defines mass. The correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer that has actually come to your desired help.</span>
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What is a stable and unstable equilibrium
Citrus2011 [14]

Answer:

here

Explanation:

Equilibrium is a state of a system which does not change. ... An equilibrium is considered stable (for simplicity we will consider asymptotic stability only) if the system always returns to it after small disturbances. If the system moves away from the equilibrium after small disturbances, then the equilibrium is unstable.

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3 years ago
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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Can an object have two types of energy at one time?
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Yes, of course, a good example would be us.

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