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Ede4ka [16]
3 years ago
13

A small, positively charged ball is moved close to a large, positively charged ball. which describes how the small ball likely r

esponds when it is released?it will move toward the large ball because like charges repel.it will move toward the large ball because like charges attract.it will move away from the large ball because like charges repel.it will move away from the large ball because like charges attract.
Physics
2 answers:
NNADVOKAT [17]3 years ago
6 0

Answer;

-it will move away from the large ball because like charges repel.

Explanation;

-Electric force is the force that pushes apart two like charges, or that pulls together two unlike charges. The basic law of electrostatics Like charges of electricity repel each other, whereas unlike charges attract each other.

When small, positively charged ball is moved close to a large, positively charged ball it would be pushed away from the large positively charged ball since they are both positively charged. One has to put in energy to try to move the small ball closer to the large ball. The closer one try to move it to the large ball, the more energy one has to put in, so the more electrical potential energy the small ball would have.

MariettaO [177]3 years ago
4 0

Answer:

C. It will move away from the large ball because like charges repel.

Explanation:

I just did the test and got it right

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A particle moves along the x axis. It is intially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration -0.32
Nataliya [291]

Answer:

The position of the particle is -2.34 m.

Explanation:

Hi there!

The equation of position of a particle moving in a straight line with constant acceleration is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the particle at a time t:

x0 = initial position.

v0 = initial velocity.

t = time

a = acceleration

We have the following information:

x0 = 0.270 m

v0 = 0.140 m/s

a = -0.320 m/s²

t = 4.50 s  (In the question, where it says "4.50 m/s^2" it should say "4.50 s". I have looked on the web and have confirmed it).

Then, we have all the needed data to calculate the position of the particle:

x = x0 + v0 · t + 1/2 · a · t²

x = 0.270 m + 0.140 m/s · 4.50 s - 1/2 · 0.320 m/s² · (4.50 s)²

x = -2.34 m

The position of the particle is -2.34 m.

6 0
3 years ago
What is the marble's range if it is fired horizontally from 1.6 m above the ground?
cupoosta [38]

Answer:

The answer is 6.40 meters.

Explanation:

The speed v = √(2gh)

v = √( 2×9.8×6.4) = 11.2 m/s

After, finding the time it takes to hit the ground from a height of 1.6 meters.

time = √(2H÷g)

time = √(2×1.6÷9.8)

time = 0.5714 seconds.

Horizontal distance is speed × time = 11.2 × 0.5714 = 6.40 meters.

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Virty [35]

Answer:

18.24 seconds

Explanation:

First you convert the km/h to m/s, 70km/h=(175/9)m/s,85km/h=(425/18)m/s.

You know it took 10 seconds for the police to reach 85 km/h. Calculate the distance that the car is ahead of the police (175/9)*10=1750/9m. Then by divide 1750/9 with 425/18, you will get the value 8.24. Add the 10 seconds with the 8.24 you will get 18.24 sec which is the total time.

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