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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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As the distance between two objects increases, the gravitational force of attraction between them will
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A pendulum consists of a large balanced mass hanging on the end of a long wire. At the point where a 28-kg pendulum has the grea
Ray Of Light [21]

Answer:

The length of the wire is approximately 67.1 m

Explanation:

The parameters of the pendulum are;

The mass of the pendulum, m = 28 kg

The angle between the pendulum weight and the wire, θ = 89°

The magnitude of the torque exerted by the pendulum's weight, τ = 1.84 × 10⁴ N·m

We have;

Torque, τ = F·L·sinθ = m·g·l·sinθ

Where;

F = The applies force = The weight of the pendulum = m·g

g = The acceleration due to gravity ≈ 9.8 m/s²

l = The length of the wire

Plugging in the values of the variables gives;

1.84 × 10⁴ N·m = 28 kg × 9.8 m/s² × l × sin(89°)

Therefore;

l = 1.84 × 10⁴ N·m/(28 kg × 9.8 m/s² ×  sin(89°)) = 67.0656080029 m ≈ 67.1 m

The length of the wire, l ≈ 67.1 m

6 0
3 years ago
I need help with questions 6-8. Thank you!! Image is attached
Digiron [165]

6) b) 2.7 m/s

7) b) DCA

8) b) B

Explanation:

6)

In a displacement-time plot, the slope of the line is given by

m=\frac{\Delta y}{\Delta x}

where

\Delta y is the change in the y-variable, so it is the displacement

\Delta x is the change in the x-variable, so it is the time elapsed

So, the slope of the line in a displacement-time plot corresponds to the velocity:

v=m=\frac{d}{t}

Therefore, to find the velocity of the object, we have to estimate the slope of its curve.

To estimate the velocity of object B, we have to estimate the slope of the line tangent to curve B at 10 seconds.

By doing an estimate by eye, we see that the displacement of object B changes from -10 m to 0 m when time increases from about 8 s to 12 s, so the velocity is about:

v=\frac{0-(-10)}{12-8}\sim 2.5 m/s

So the closest option is b) 2.7 m/s.

7)

As we said in part A, the velocity of each object is given by the slope of each curve.

Therefore:

- The steeper the curve, the higher the velocity

- The less steep the curve, the lower the velocity

From the graph, we observe that, among A, C and D:

- Curve D has the largest slope (in absolute value), so object D has the largest magnitude of the velocity

- Curve C is less steep than curve C, so object C has the second largest magnitude of velocity

- Curve A is flat, so the slope is zero, so its velocity is zero

So, from greatest magnitude to lowest magnitude of velocity, we have:

b) DCA

8)

In the graph, the overall displacement of each object is given by the change in the y-variable, \Delta y.

This means that the object with largest displacement is the object whose curve has the largest variation in y.

From the graph, we see that:

- Object b has the largest variation in y,  from -15 m to 30 m, so

\Delta y=30-(-15)=45 m

- Then, object D has the second largest displacement (in magnitude), from -15 m to 25 m,

|\Delta y| = 25 -(-15)=40 m

Finally, object C has displacement

\Delta y = 20-(-5)=25 m

While object A has displacement zero. Therefore, the correct option is

b) B

3 0
3 years ago
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