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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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Two point charge
timama [110]

Answer:

Showing results for Two point charge q, separated by 1.5cm have change value of +2.0 and -4.0AND/C respectively what is the magnitude of the Electric force midway between them?

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4 0
1 year ago
Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros
kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

v^2-u^2=2gh

where

v = 0 is the velocity at the maximum height

u is the initial velocity

g=-9.8 m/s^2 is the acceleration of gravity

Solving for u,

u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s

The time needed to reach the maximum height can now be found by using the equation

v=u+gt

Solving for t,

t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

v'^2-u^2=2gh'

we find

v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s

And the corresponding time is

t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s

So the time to go from h' to h is

\Delta t = t-t'=0.52-0.32=0.20 s

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

\Delta t=2\cdot 0.20 = 0.40 s

(b) 0.08 s

This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:

v'^2-u^2=2gh'

v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(0.20)}=4.74 m/s

And the corresponding time is

t'=\frac{v'-u}{g}=\frac{4.74-5.14}{-9.8}=0.04s

So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:

\Delta t = 2\cdot 0.04 =0.08 s

8 0
3 years ago
Which of the below statements are true?
andrezito [222]

Answer:

B

Explanation:

Displacement is the distance from the start point to the endpoint, displacement disregard the path taken or the amount traveled.

if you start at point A, then go to point B, and back to point A, the displacement is zero because you started and ended at the same point.

for this question, pretend you started at point A, went east 20 km to point B, and then west 8 km to point C, your displacement is 12 km. 12 km is the distance between point A and point C.

4 0
2 years ago
The lift does 3000 J of work in 5 seconds. What is the power of the lift?
diamong [38]

Answer:

600

Explanation:

p=Work/time

3000/5=600 is power

8 0
2 years ago
A 5.0 kg object moving at 5.0 m/s. KE = mv2 times 1/2
steposvetlana [31]

Answer: KE = 62.5J

Explanation:

Given that

Mass of object = 5kg

kinetic energy KE = ?

velocity of object = 5m/s

Since kinetic energy is the energy possessed by a moving object, and it depends on the mass (m) of the object and the velocity (v) by which it moves. Therefore, the object has kinetic energy.

i.e K.E = 1/2mv^2

KE = 1/2 x 5kg x (5m/s)^2

KE = 0.5 x 5 x 25

KE = 62.5J

Thus, the object has 62.5 joules of kinetic energy.

5 0
2 years ago
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