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Firlakuza [10]
3 years ago
6

To understand the electric force between charged and uncharged conductors and insulators. When a test charge is brought near a c

harged object, we know from Coulomb's law that it will experience a net force (either attractive or repulsive, depending on the nature of the object's charge). A test charge may also experience an electric force when brought near a neutral object. Any attraction of a neutral insulator or neutral conductor to a test charge must occur through induced polarization. In an insulator, the electrons are bound to their molecules. Though they cannot move freely throughout the insulator, they can shift slightly, creating a rather weak net attraction to a test charge that is brought close to the insulator's surface. In a conductor, free electrons will accumulate on the surface of the conductor nearest the positive test charge. This will create a strong attractive force if the test charge is placed very close to the conductor's surface.Consider three plastic balls (A, B, and C), each carrying a uniformly distributed charge equal to either +Q, -Q or zero, and an uncharged copper ball (D). A positive test charge (T) experiences the forces. The test charge T is strongly attracted to A, strongly repelled from B, weakly attracted to C, and strongly attracted to D. Assume throughout this problem that the balls are brought very close together. What is the nature of the force between balls A and B? a. Strongly attractive b. Strongly repulsive c. Weakly attractive d. Neither attractive nor repulsive
Physics
1 answer:
Alchen [17]3 years ago
6 0

Answer:

the correct  is a  Strongly ATTRACTIVE

Explanation:

For this exercise we must use that charges of the same sign repel and charges of the opposite sign attract, the attraction is strong if they are charged or weak if the charges are induced.

Let's apply this to our case.

The test load T is attracted by the sphere A, this implies that the charges are of different sign

the test charge T is repelled by the sphere B, therefore the charges are of equal sign

As the test charge cannot change the sign, this implies that the spheres A and B are of different sign, therefore attractive forces.

Now let's analyze the intensity, as in the exercise they indicate that spheres A and B are charged and are insulators, these charges cannot move, so the attraction must be Strong.

When reviewing the statements, the correct one is a  Strongly ATTRACTIVE

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The sun’s gravitational attraction and the planet’s inertia keeps planets moving is circular orbits.

Explanation:

The planets in the Solar System move around the Sun in a circular orbit. This motion can be explained as a combination of two effects:

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F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M is the mass of the Sun

m is the mass of the planet

r is the distance between the Sun and the planet

This force acts as centripetal force, continuously "pulling" the planet towards the centre of its circular orbit.

2) The inertia of the planet. In fact, according to Newton's first law, an object in motion at constant velocity will continue moving at its velocity, unless acted upon an external unbalanced force. Therefore, the planet tends to continue its motion in a straight line (tangential to the circular orbit), however it turns in a circle due to the presence of the gravitational attraction of the Sun.

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8 0
3 years ago
Which of the following is not an example of work?
barxatty [35]
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8 0
3 years ago
A car travels a distance of 400 m in 5 seconds. Calculate its average velocity.
ikadub [295]

Answer:

80 m/s

Explanation:

x = 400 m

t = 5 s

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5 0
2 years ago
Read 2 more answers
Infer the direction of the net force acting on a car as it slows down and turns right.
Kobotan [32]
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6 0
3 years ago
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A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it
natka813 [3]

Answer:

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Explanation:

The rotated angle is given by:

\theta=\omega0*t+1/2*\alpha*t^2

Since this is a quadratic equation it can be solved using:

x=\frac{-b \± \sqrt{b^2-4*a*c}  }{2*a}

Rewriting our equation:

1/2*\alpha*t^2+\omega0*t-\theta=0

t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Since \sqrt{\omega0^2+2*\theta*\alpha} >\omega0 we discard the negative solution.

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

8 0
3 years ago
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